A Method of Finding the Latitude of a Place, by means of Two Altitudes of the Sun and the Time Elapsed Betwixt the Observations. By the Rev. W. Lax, A. M. Lowndes's Professor of Astronomy in the University of Cambridge

VII. A Method of finding the Latitude of a Place, by Means of two Altitudes of the Sun and the Time elapsed betwixt the Observations. By the Rev. W. Lax, A. M. Lowndes's Professor of Astronomy in the University of Cambridge. Read January 10, 1799. I hope the following method of determining the latitude, by means of two altitudes of the sun and the time elapsed betwixt the observations, will be found not less convenient for nautical purposes than the rules which are commonly employed. But I would rather recommend it in those cases where rigid accuracy is required, and the astronomer is provided with no better instrument for taking the sun's altitude than a Hadley's sextant of the most improved construction. The process will be neither difficult nor tedious; and, if the observations are made with proper exactness, I conceive the latitude will generally be obtained within a few seconds of the truth. With these expectations, I have ventured to reduce the method into its present form; and submit it, with the utmost deference, to this learned Society. In the spherical triangle, whose sides are the complements of the latitude, declination, and altitude, let \( z \) represent the angle at the pole, and \( t \) its tangent; \( Z \) the azimuth, and \( T \) its tangent; \( L \) the latitude, and \( \lambda \) its cosine, radius being unity; then, if the altitude and declination remain constant, we shall have \( \dot{L} = \lambda T \dot{z} \), and, consequently, \( \dot{L} \) will vary as \( T \dot{z} \), when the increment of \( \lambda \), com- pared with \( \lambda \) itself, is inconsiderable. Hence, if the abscisse of the curve \( ABCD \) (fig. 1.) be always proportional to \( z \), and its ordinate to \( T \), the area \( GB \) intercepted betwixt any two of these ordinates may represent the increment of the latitude corresponding to the increment of the time \( EG \). Let \( abcd \) (fig. 2.) be another curve, whose abscisse \( ae \) is always equal to \( AE \) in the preceding figure, but whose ordinate \( eb \) is proportional to \( t \), the tangent of the hour-angle; then will the area \( gb \) vary as \( GB \), at small distances from the meridian, and, of course, may represent the increment of the latitude. Now, to prove this, we have only to shew that \( T \) and \( t \), when both are small, bear to each other a given ratio. Let \( S \) and \( \Sigma \) be the sine and cosine of the azimuth; \( s \) and \( \sigma \) the sine and cosine of the angle at the pole; then will \( \frac{\dot{T}}{T} = \dot{Z} \cdot \frac{1 + T^2}{T} \), and \( \frac{i}{t} = \dot{z} \cdot \frac{1 + t^2}{t} \); \( \dot{Z} = \frac{\dot{S}}{\Sigma} \), and \( \dot{z} = \frac{\dot{s}}{\sigma} \). But, since the complements of the declination and altitude remain constant, whilst the latitude is made to vary, \( \dot{S} \) will be to \( \dot{s} \) as \( S \) to \( s \); and, therefore, we shall have \( \frac{\dot{T}}{T} : \frac{i}{t} :: \frac{S}{\Sigma} \times \frac{1 + T^2}{T} : \frac{s}{\sigma} \times \frac{1 + t^2}{t} :: 1 + T^2 : 1 + t^2 :: \) the square of the secant of the azimuth: the square of the secant of the hour-angle, which may be considered as a ratio of equality, when the angles are very small. The fluxions, therefore, of the tangents are as the tangents themselves; and, consequently, they must always preserve the same ratio towards each other. Let us now suppose that an altitude of the sun is taken at the distance $ae$ from the meridian, but that, in consequence of an error in the assumed latitude, the calculated time is $ag$; and that, with a lat. differing from the former by one minute, we compute again, and the time is found equal to $af$; then will the area $gc$ be to $gb$ as one minute to the whole error in latitude. Let another altitude be taken at the distance $ae$ from noon, and let the times computed with the two different latitudes that were employed before be $ag$ and $af$; then, in this case likewise, the area $gc$ will be to the area $gb$, as one minute to the error in latitude. Now the latter curve is the "figura tangentium," whose quadrature is given by Cotes, in his Harmonia Mensurarum, and the expression for which is extremely simple. For the fluxion of the area is $\frac{ti}{1+t^2}$, and the area itself $= \log_1 + t^2 = \log_1$ secant of the angle at the pole. The difference of the log. secants, or log. cosines, will of course be equal to the area intercepted betwixt the tangents which correspond to them. Hence a table might easily be constructed with a double argument,—the distance from noon, and the variation in time arising from the different suppositions of latitude,—which might immediately exhibit the logarithm of the area corresponding to any particular base $eg$ supposed to be given. A second table might have for its argument the difference of the logarithms of the area $gb$ and the area $gc$, (which is likewise conceived to be known) and discover at once, in degrees, minutes, and seconds, the correction to be made in the assumed latitude. This correction, as it appears from a comparison of the signs of $\dot{L}$ and $\dot{\lambda}$ in the equation $\dot{L} = \lambda T \dot{\lambda}$, must be added or subtracted, according as the distance from noon obtained by computation is too great or too. little, when the azimuth is less than 90 degrees; but the contrary, when the azimuth exceeds a right angle. Tables of the above description shall be constructed, if this method be received with approbation; and, in the mean time, it is proposed to subjoin a short specimen which is already completed. I have presumed that we are able to determine $eg$, the error in time arising from an error in the assumed latitude, at either of the observations; and hence it becomes necessary, before we can avail ourselves of the principles which have been laid down, to point out the manner in which this may be accomplished. The clock gives us the whole interval betwixt the observations (supposed to be made on different sides of the meridian) equal to $ae + ae$, and by computation we obtain $ag + ag$, and thence we deduce $eg + eg$ the whole error in time. Now the area $gb$ is equal to $gb$; and therefore, if we make a rough division of the whole error, without any regard to accuracy, in the inverse ratio of the hour-angles at the two observations, and, entering the first table with these times, mark the area corresponding to each at their respective distances from noon, and increase the one and diminish the other equally, till we get the areas of the same magnitude, this, we may conclude, is the proper value of each. If the table were constructed to every second of time, we might ascertain these logarithmic areas merely from inspection; but, as it will be advisable to confine it within narrower limits, we shall sometimes find it necessary, as in other tables, to deduce their ultimate value by the rule of three. When we have increased one portion of time and diminished the other, till the difference of their corresponding areas becomes a minimum, we must divide this difference betwixt them, in the proportion of their respective increments in the next interval of time, and subtract or add the part assigned to each, according as it is greater or less than the other. The table, however, might easily be carried to such an extent, that exactness in this division could never be required; but, on the contrary, it would be quite sufficient, when the hour-angles were nearly equal, to add the areas together, and take half the sum for the value of each. From these principles may be deduced the following practical rule for determining the latitude of a place. When the sun comes within fifteen degrees of the meridian, in the morning, let his altitude be taken, and the time of the observation be accurately marked; and let another altitude be taken after he has passed the meridian, whilst his distance from it is less than fifteen degrees; and let the time of this observation likewise be noted. Then, with the supposed latitude of the place, compute the times corresponding to each of the altitudes in terms of the log. cosi of the hour-angle, and take the difference of the intervals, shewn by the clock, and determined by calculation, and divid it betwixt the observations in the manner explained above. Compute the log. cosine of the hour-angle a second time, with the greatest altitude and the latitude increased or diminished by a minute, according as it appears, from a comparison of the intervals, to have been too little or too great; and take the difference betwixt this log. cosine and that which resulted from the first operation, when the same altitude was employed. Having thus obtained the two areas $gb$ and $gc$, we must subtract their logarithms from each other, and with their difference entering the second table we shall find the degrees, minutes, and seconds, by which the assumed latitude is to be increased or diminished. It will be needless, perhaps, to suggest that, in the higher latitudes, we may extend the limits above specified a few degrees farther from the meridian, without offering any material violence to the theory, as it has hitherto been explained; and that, on the other hand, when the declination and latitude are nearly equal and of the same denomination, it will be expedient to confine our observations within a much shorter distance from noon. But it will afterwards be demonstrated that, whatever be the magnitude of the hour-angles, or however nearly the latitude and declination may approach towards each other, we can always secure, with very little additional trouble, an exact conclusion. We may remark that the latitude, determined in this manner, will be nearly equivalent, in point of accuracy, to the mean result of two meridian altitudes. For we know that the increment of latitude : increment of altitude :: radius : cosine of azimuth; and, since the cosine of a small angle differs so little from the radius, this may be considered, within the limits which I have prescribed, as a ratio of equality. If, therefore, one altitude of the sun were taken, and we could ascertain the error in time arising from an error in the assumed latitude without the aid of a second observation, the latitude would be discovered with nearly the same precision as if it had been deduced from the meridian altitude. But, by means of a second observation made on a different side of noon, we obtain a second error in time of the same kind; and this being added to the former, and their sum divided in a just proportion betwixt the two observations, the same effect will be produced, with respect to the accuracy of the result, as if two latitudes had been deduced from meridian altitudes, and a mean betwixt them had been taken. Mr. Lax's Method of finding the Latitude of a Place, I might perhaps be allowed to say more; for I am satisfied, from experience, that I can take an altitude of the sun with greater exactness, when he is in any other situation, than when he is upon the meridian. If we could ascertain, within a few seconds, or even within a minute, the time when he attains his greatest altitude, there would then be no reason why an observation should not be made with the same degree of certainty in this, as in other cases; but we are generally obliged to keep our eye steadfastly fixed, for several minutes, upon the two images, and it is well known that, in such circumstances, the best eyes are apt to be deceived. Besides, it is impossible to preserve the contact of the limbs by perpetually moving the index, whilst the sun continues to ascend so very slowly. We are compelled to wait till they are evidently separated, and then, by one turn of the screw, to bring them into contact again, which must necessarily be a source of some inaccuracy. It is for the first of these reasons that, in taking an altitude of the sun, when he is near the meridian, I have found it advisable, not, in the usual manner, to bring the images almost to touch each other, and then to wait till they actually do so, but to bring them at once into contact, with such a degree of velocity as would make them sensibly overlap, or separate, whilst the clock beats a second. But I consider it as one of the principal advantages of this method, that we can avail ourselves of any number of altitudes, and, of course, approximate as near as we please to a true conclusion with so little additional labour. If there be an equal number of observations made on each side of the meridian, we must combine them together by pairs, according to the preceding instructions, and thus determine the different logarithmic values of $gb$. Having then added them all together, and taken a mean betwixt them, we have only to compute a single incremental area $gc$ with any of the altitudes and the lat. varied one minute, and, subtracting its log. from the mean log. value of $gb$, we shall obtain a very accurate correction of the assumed latitude. But, if there be more observations on one side of the meridian than on the other, when all the pairs have been united, and the areas resulting from them found, we may combine the supernumerary observations on either side with any of those which are made on the opposite side. The properest, however, for this purpose, is the observation which is made at the least distance from the meridian. I should hope, moreover, the practical astronomer will think it a circumstance of some moment, that the principal part of the work consists in finding the time, an operation which he is obliged so frequently to perform. Any of the three methods which are usually adopted upon this occasion might easily be applied to the tables which have been described; but I will venture to recommend a different rule, which I conceive to be better adapted to our purpose than any of the others, and to which the directions before given had a particular reference. Let $a$ be the sine of the altitude; $\gamma$ the cosine of the hour-angle; $d$ the sine, $\delta$ the cosine, and $r$ the tangent of declination; $l$ the sine, $\lambda$ the cosine, and $s$ the tangent of the latitude. Then $\gamma = \frac{a - dl}{\delta \lambda} = \frac{a}{\delta \lambda} - \frac{dl}{\delta \lambda} = \frac{ars}{\delta \lambda rs} - rs = \frac{a}{dl} - 1 \cdot rs,$ when radius is unity, but $\frac{am^3}{dl} - m^2 \cdot \frac{rs}{m^3}$, when radius is $m$ $= \frac{rs}{m^3} \times 9.$ into the square of the tangent of the arc whose secant is $\frac{am^3}{dl}$. Hence we deduce the following rule for deter- MDCCXCIX. mining the log. cosine of the angle at the pole. From the log. sine of the altitude increased by three times the log. radius, subtract the sum of the log. sines of the latit. and declination; take half of the remainder, and, considering it as the log. secant of an arc, find the log. tangent corresponding; multiply this by 2, and add the log. tangents of the latit. and declin. and reject thrice the log. radius; the sum will be the log. cosine of the angle required. But, when the declin. and latit. are of a different denomination, it is evident that our expression becomes \( m^2 + \frac{a m^3}{d l} \cdot \frac{r s}{m^3} \), which is equal to \( \frac{r s}{m^3} \times \frac{1}{\sqrt{a m^3}} \). In this case, therefore, having found the log. value of \( \frac{a m^3}{d l} \), and divided it by 2, we must consider the quotient as the log. tangent of an arc, whose log. secant being taken, we are to proceed as in the former case. The advantages of this rule are obvious. We obtain the angle in terms of the log. cosine; and, consequently, when we have calculated the second time with the new latit. we have only to subtract one result from the other, and we immediately determine the area corresponding to the difference of the times. Besides, in the second computation, fewer of the elements are changed by this rule, than by any of those which are usually employed; and this is a consideration of much importance. But, if we are disposed to adopt the following method of ascertaining the incremental area \( g c \), this advantage will be found still greater. Let us resume the expression \( y = \frac{a - dl}{\delta \lambda} \), and we shall have \( y = \frac{1}{\delta} \times \frac{-d \lambda' - a \lambda + dl \lambda'}{\lambda \lambda'} (\lambda' be- ing the succeeding value of $\lambda$) = $\frac{1}{\delta} \times \frac{d \lambda^2 \dot{\lambda}}{l} - a \dot{\lambda} + d l \dot{\lambda}$ $\frac{d \lambda^2 \dot{\lambda}}{\delta l \lambda^2} - a \dot{\lambda} + d l \dot{\lambda} = \frac{d \lambda \dot{\lambda}}{\delta l \lambda^2} - a \dot{\lambda} \times \frac{\dot{\lambda}}{\lambda^2} = \frac{r}{s} - \gamma \cdot \frac{\dot{\lambda}}{\lambda^2}$ (taking $\dot{\lambda}$ positive instead of negative, as it ought to be when $l$ is positive) $\gamma \rightarrow \frac{r}{s} \cdot \frac{\dot{\lambda}}{\lambda^2}$; and, consequently, $\frac{\dot{\gamma}}{\gamma} = 1 - \frac{r}{s \gamma} \cdot \frac{\dot{\lambda}}{\lambda^2}$, when radius is unity, but $= m^2 - \frac{r m^3}{s \gamma} \cdot \frac{\dot{\lambda}}{\lambda^2 m^2}$, when radius is $m$. Now $\frac{\dot{\gamma}}{\gamma}$ may be considered as the increment of the hyperbolic log. of $\gamma$, and therefore, with its proper modulus, may represent the area which is the object of our investigation. We may suppose the other side of the equation to be the square of the cosine of the arc whose sine is $\frac{r m^3}{s \gamma}$ into the increment of the hyperbolic log. of $\lambda^2$, divided by the square of the radius; and if, instead of taking this log. with the hyperbolic, we take it with Briggs's modulus, we must then consider $\frac{\dot{\gamma}}{\gamma}$ as the increment of the log. of $\gamma$, according to the same system. But $\frac{\dot{\lambda}}{\lambda^2}$ being equal to $\frac{s L}{m^2}$ (when $L$ is only one minute), it will vary as $s$; and, therefore, if its value be determined according to Briggs's system, when $s$ is equal to radius, and be denominated $v$, its value in any other case will be expressed by $\frac{v s}{m}$. Hence, to obtain the log. of the area $gc$, the quantity with which we are immediately concerned, we must find the log. value of $\frac{r m^3}{s \gamma}$, and divide it by $2$; we must then take out the log. cosine of the arc whose log. sine is equal to the quotient; and, having multiplied it by $2$, we must add the product to the constant log. of $v$ ($3,1015$), and the log. tangent of the sup- posed latitude, rejecting thrice the log. radius. But, if $\frac{r m^3}{s \gamma}$ be greater than radius, which must necessarily be the case when the azimuth is greater than a right angle, we must then consider $\frac{r m^3}{s \gamma} - m^2$ as the square of the tangent of the arc whose secant is $\frac{r m^3}{s \gamma}$, observing in other respects the directions before given. The quantities $r$ and $s$ are both employed in the first computation, from the result of which we also obtain $\gamma$; and, consequently, this operation will not be attended with much trouble. The above instructions, it is manifest, are given upon the supposition of $r$ and $s$ having the same sign; but, if the declin. and latit. should not be of a similar denomination, then will our expression become $m^2 + \frac{r m^3}{s \gamma} \cdot \frac{v s}{m^3}$, and we must consider $m^2 + \frac{r m^3}{s \gamma}$ as the square of the secant whose corresponding tangent is $\frac{r m^3}{s \gamma}$. With this exception, the process will be the same as when the tangents $r$ and $s$ are both affirmative. The preceding formula naturally suggests to us another method of finding the log. area $gc$; and, as some perhaps may think this more eligible than either of the former, I shall take the liberty of explaining it. When the latit. is given, the area $GC$, it is obvious, must invariably preserve the same magnitude at all distances from the meridian; and, consequently, the area $gc$, which is proportional to it, must likewise remain constant. If, therefore, we can ascertain this area when the hour-angle is supposed to vanish, we may employ it when the sun is at any distance from noon. Let us now conceive the declin. to be equal to nothing; then will our expression for the area $gc$ become $\frac{\lambda}{\lambda'}$; and, consequently, (since the tangents of the azimuth and hour-angle vanish in the ratio of their sines, or of the sines of the opposite sides in the triangle alluded to before,) we shall have the area $GC = \frac{\lambda m}{\lambda' l}$, when the sun is upon the meridian. But this area is always the same when the latitude is given, whatever be the sun's declination, and therefore may always be represented by $\frac{\lambda m}{\lambda' l} = \frac{v s}{m} \times \frac{m}{l}$ $= \frac{v m}{\lambda}$; and the area $gc$ will be generally expressed by $\frac{v m}{\lambda} \times \frac{\cos. \text{ of merid. altit.}}{\cos. \text{ of declin.}}$, when the hour-angle does not exceed the limits which have been recommended. Hence, if we add together the constant log. 3,1015, the log. radius, and the log. cosine of the merid. altitude, and subtract from their sum the log. cosines of the latit. and declin. we shall obtain the log. value of $gc$. It will be necessary, perhaps, to meet an objection which some may be inclined to urge against the method of deducing the hour-angle in terms of the cosine, when this angle is very small. But it should be recollected, that with the angle itself we have no immediate concern, the accuracy of our conclusion depending entirely upon the accuracy with which the area corresponding to any particular increment of time can be determined. Now this area, whatever be the sun's distance from the meridian, will be nearly proportional to the increment of the latit. and, consequently, its magnitude is totally unconnected with that of the hour-angle. A given error in the quantity which expresses this area will equally affect our conclusion, whether the angle be 2, or whether it be 20 degrees. But let us inquire what effect will actually be produced, by admitting an error of half an unit in each of the log. cosines whose difference is equal to the area $gc$; and, of course, in some instances, an error of an unit in the area itself, upon any particular supposition of latit. and declin. We have only to ascertain the ratio which this area bears to unity; for the same ratio will the correction of the latit. bear to the error in our result. If the latit. for instance, be $50^\circ$, and the declin. $10^\circ$, on the same side of the equator with the latit. then, radius being unity, $\delta$ (the increment of the hour-angle) will equal $\frac{L}{\lambda T}$ $$= \frac{L}{\cos. 50^\circ \times \frac{s}{\Sigma}} = (g \text{ being the sine of the hour-angle, and } \alpha$$ the cos. of the altit.) $\frac{L}{\cos. 50^\circ} \times \frac{\sqrt{\alpha^2 - g^2 \delta^2}}{g \delta} = 11$ minutes nearly, when $\delta$ is $5^\circ$; and we have seen that, at any other distance from the meridian, the incremental area will be of the same magnitude. Hence, subtracting the log. cosine of $5^\circ$ from that of $5^\circ 11' 0''$, we get the difference equal to $1238$; and, consequently, the error in our approximation will be to the error in the assumed latit. as $1 : 1238$, when the log. cosines are carried to seven places of decimals. But, when the zenith distance of the sun, at his greatest altit. is very small, and there is moreover a considerable uncertainty with respect to the latit. this error will probably become of more importance, and we may find it necessary to guard against it. Now it is manifest that, by diminishing the multiple which the area $gb$ is of the area $gc$, exactly in the same proportion we shall diminish this error; and I shall afterwards explain in what manner it may be accomplished. The same expedient is also calculated to prevent another species of inac- accuracy, to which, under similar circumstances, our process is liable. If we are desirous of knowing how much our conclusion is affected by substituting the tangent of the hour-angle for the tangent of the azimuth, it may be done with the greatest facility. We have seen, that the area $GC$ (fig. 1.) is to $GB$ as one minute to the error in the assumed latitude. But our method supposes that $gc$ (fig. 2.) is to $gb$ as one minute to the correction required; and we must therefore estimate the difference betwixt these ratios, in order to ascertain how far the approximation is inaccurate. Let $EG = m \times FG$; and then, since $GC$ may be considered as $\frac{GD + FC}{2} \times FG$, and $GB = \frac{GD + EB}{2} \times EG = \frac{GD + EB}{2} \times m \times FG$, the former ratio will equal $\frac{GD + EB}{GD + FC} \times m = (according to the notation employed above) m \cdot \frac{zT - mT}{zT - T} = m - m \cdot \frac{m - 1}{2} \times \frac{T}{T}$ nearly = the minutes contained in the computed error of latitude. Their difference will equal $m \cdot \frac{m - 1}{2} \cdot \frac{T}{T} - \frac{t}{t} = m \cdot \frac{m - 1}{2} \cdot \dot{z} \cdot \frac{1 + T^2}{t} - \dot{z} \cdot \frac{1 + t^2}{t} = m \cdot \frac{m - 1}{2} \cdot \frac{T^2 - t^2}{t} \cdot \dot{z} = m \cdot \frac{m - 1}{2} \cdot \frac{T^2 - t^2}{\lambda T t} \cdot \dot{L} (\dot{L} being substituted for its equal $\dot{z}$) = the minutes contained in the error of our approximation. It will be more convenient, however, to express this difference independently of the azimuth. Now, preserving the notation before adopted, we have $T = \frac{s}{\Sigma} = \frac{g \delta}{a \times l_a - d} = \frac{l_y \delta \lambda}{l_a - d} = \frac{l_y \delta}{l_y \delta - d \lambda}$; therefore, $\frac{T}{t} = \frac{\gamma \delta}{l_y \delta - d \lambda}$, and $\frac{T}{t} = \frac{t}{T} \left(= \frac{T^2 - t^2}{T t}\right) = \frac{\gamma \delta}{l_y \delta - d \lambda} = \frac{l_y \delta - d \lambda}{\gamma \delta}$ Mr. Lax's Method of finding the Latitude of a Place, \[ \frac{\gamma^2 \delta^2 - l^2 \gamma^2 \delta^2 + 2 dl \gamma \delta \lambda - d^2 \lambda^2}{\gamma \delta \times l \gamma \delta - d \lambda} = \frac{\gamma^2 \delta^2 \lambda^2 + 2 dl \gamma \delta \lambda - d^2 \lambda^2}{\gamma \delta \times l \gamma \delta - d \lambda}. \] being rejected, as incomparably less than the sum of the other terms,) \( \lambda \cdot \frac{\gamma \delta \lambda + 2 dl}{l \gamma \delta - d \lambda} = \lambda \cdot \frac{\gamma + 2 rs}{s \gamma - r}; \) and, consequently, we have \( m \cdot \frac{m-1}{2} \cdot \frac{T^2 - t^2}{\lambda T t} \cdot \hat{L} = m \cdot \frac{m-1}{2} \cdot \frac{29}{100000} \cdot \frac{\gamma + 2 rs}{s \gamma - r} \) minutes. Hence it appears that, in the latitude of Cambridge, when the sun's declination is \( 2^\circ \) north, and his distance from the meridian \( 5^\circ \), this error will be equal to \( m \cdot \frac{m-1}{2} \cdot \frac{1}{4312} \) minutes of a degree. Let the assumed differ from the true latitude ten minutes, i.e. let \( m = 10 \), then will the error amount to half a second; and, if we suppose \( m = 30 \), the error will not exceed four seconds. If, instead of varying the assumed latitude one minute, we vary it \( n \) minutes, in calculating the area \( gc \), the above expression will be transformed into \( m \cdot \frac{m-1}{2} \cdot \frac{29 \cdot n}{100000} \cdot \frac{\gamma + 2 rs}{s \gamma - r} \), the increment of the latitude being, in this case, \( n \hat{L} \). The error will consequently vary as \( m \cdot \frac{m-1}{2} \cdot n \), when the declination, latitude, and time of observation are given; and, if we suppose the real to differ from the assumed latitude \( p \) minutes, this last expression will become \( \frac{p}{n} \cdot \frac{p-1}{2} \cdot n = p \cdot \frac{p-n}{2n} \), which varies as \( \frac{p-n}{n} \), when \( p \) remains constant; and, of course, the error may be diminished in any proportion by diminishing this quantity \( \frac{p-n}{n} \). Now, the increment of the latitude is always \( = \lambda T \dot{x} \); and, therefore, since \( \dot{x} \) is determined by the process explained above, we have only to ascertain the value of \( T \), in order to approximate nearly to \( p \). But the sine of the azimuth is $\frac{s}{\alpha}$, which is known, and consequently the tangent may likewise be obtained. Having thus discovered a near value of $p$, or of the error in our first assumed latitude, we are enabled so far to correct it in our second hypothesis, and in a much greater degree to reduce the error in our conclusion. As it will sometimes, though not often, be necessary to have recourse to this expedient, two short tables might be added, in order to facilitate the operation. The first might contain the log. cosines to every degree from the 15th to the 90th of the quadrant; and be so contrived, as likewise to exhibit the log. sine of any arc, as far as the 75th degree, expressed in minutes and seconds of time; that, by subtracting the log. cosine of the altitude from the log. sine of the hour-angle, (the cosine of declination being considered as equal to radius,) we might obtain the log. sine of the azimuth. This should be made one of the arguments of the second table; the other being the cosine of the latitude to every five degrees of the first sixty of the quadrant; and the table should give us the value of $15 \times \lambda T$, which, multiplied into the minutes contained in the error of time, would determine with sufficient exactness the quantity $p$. We might, indeed, with the same facility, compute the fourth table according to the mean value of the cosine of declination, if it could be supposed that such a degree of precision would ever be required. Perhaps it would be advisable for the mariner, in order to avoid all distinction of cases, to calculate, in every instance, his incremental area with the lat. varied ten minutes, instead of one; and then he would always be secure of a result sufficiently correct for his purpose. In the case supposed above, if the real were to differ from the assumed latitude 30 minutes, the error in his conclusion would not exceed the fourth part of a second. We must observe, that if \( n \) (the approximate value of \( p \), thus deduced) be the minutes by which the first assumed lat. is varied in our second hypothesis, we ought to multiply the correction, after it is taken from the 2d table, by this quantity; but, as \( n \) is always supposed to be a whole number, the additional trouble arising from this process can never be an object of the smallest consideration. The 2d table, however, may be as conveniently applied to the case, where the lat. is varied ten minutes, as where it is varied only one. The logarithms which form the argument are the same in both cases, except that the index in the former is less by unity than in the latter. Another species of inaccuracy originates in our conceiving the fluxion of the lat. to vary as \( T\dot{x} \), instead of \( \lambda T\dot{x} \); and it may be useful to ascertain how much our conclusion is affected by this circumstance. We suppose, in fact, that the increments of the lat. corresponding to the finite times \( gf \) and \( ge \) are to each other as the areas \( gc \) and \( gb \) drawn into the cosine of the first lat.; whereas, in strict propriety, they are to each other as the sum of all the elements of these areas drawn into the cosines of the latitudes belonging to each, i.e. (if \( ge = m \times gf \)) we consider these increments as being in the ratio of \( gc \times \lambda : gb \times \lambda \), instead of \( gc \times \frac{2\lambda + \dot{\lambda}}{2} : gb \times \frac{2\lambda + m\dot{\lambda}}{2} \). Hence \[ \frac{gb}{gc} \cdot \frac{2\lambda + m\dot{\lambda}}{2} = \frac{gb}{gc} = m \cdot \frac{m - 1}{2} \cdot \frac{\dot{\lambda}}{\lambda} = m \cdot \frac{m - 1}{2} \cdot s \hat{L} (\text{rad.} = 1) \] \( = m \cdot \frac{m-1}{z} \cdot \frac{2g s}{100000} \) the minutes contained in the error proceeding from this cause. When the real latitude, for instance, is \(52^\circ\), and the assumed \(51^\circ 50'\), this error will scarcely amount to a single second. But, if either the cosine of the latitude should be so small, or the difference betwixt the supposed and true latitude so great, as to render this error of any importance, we may prevent it by the same means that were recommended in the preceding case. In computing the incremental area \(gc\), we must correct our first hypothesis respecting the latitude with the assistance of the third and fourth tables; and, for the reasons assigned above, the cause of this inaccuracy will be removed. There is still another part of our process, which will sometimes be the source of a small error. We are directed to distribute the whole increment of the time betwixt the two observations, by equalizing the areas corresponding to each in the "figura tangentium;" whereas they can only be considered as perfectly equal in the original curve whose ordinates are the tangents of the azimuth. It is supposed, in reality, that \(gc\) bears to \(GC\) the same ratio which \(gc\) bears to \(GC\); or (which is nearly the same thing) that \(gd\) is to \(GD\) as \(gd\) to \(GD\): we must therefore estimate the difference betwixt these ratios, and we shall easily deduce the amount of this error. Let us retain the notation employed above, and moreover let \(\Gamma\) represent the cosine, and \(\tau\) the tangent of the hour-angle at the observation farthest from noon; \(\Delta\) the difference betwixt the hour-angles, and \(\delta\) the excess of \(\gamma\) above \(\Gamma\): then, from what has already been demonstrated, we have \[ \frac{gd}{GD} = \frac{l\gamma - d\lambda}{\gamma \delta}, \quad \text{and} \quad \frac{gd}{GD} = \frac{l\Gamma - d\lambda}{\Gamma \delta}; \] consequently, \( \frac{g_d}{GD} - \frac{gd}{GD} = \frac{d\lambda}{r\gamma\delta} \) (\( d \) being neglected) will express the part of \( BG \), or \( BG \), (when they are supposed to be very small,) by which \( bg \) exceeds \( bg \). Let this be multiplied into \( \frac{GD}{gd} \), and it will equal \( \frac{dl\dot{\gamma}}{r.l\gamma\delta-d\lambda} = \frac{r\dot{\gamma}}{r.s\gamma-r} \) the part of \( bg \) itself by which it is greater than \( bg \). But our method of equalizing the areas will take from \( bg \) a portion of this excess bearing to the whole the ratio of \( t : t + \tau \); and, consequently, \( bg \) will be made too small by \( \frac{r\dot{\gamma}}{r.s\gamma-r} \times \frac{t}{t+\tau} = \frac{rt\dot{\Lambda}}{2.s\gamma-r} \) nearly. Hence \( \frac{mrt\dot{\Lambda}}{2.s\gamma-r} \) will express the minutes contained in the error arising from this cause. The same conclusion may be deduced from the formula \( \frac{\dot{\gamma}}{\gamma} = 1 - \frac{r}{s\gamma} \cdot \frac{\dot{\lambda}}{\lambda} = \text{the area } gb, \text{ if the increment of the time be equal to } ge. \) For the same reasons, \( \frac{\dot{r}}{r} = 1 - \frac{r}{s\Gamma} \cdot \frac{\dot{\lambda}}{\lambda} = \text{the area } gb, \text{ if the increment of the time be represented by } ge. \) Hence \( gb - gb = \frac{\dot{\gamma}}{\gamma} - \frac{\dot{r}}{r} = \frac{r\dot{\gamma}}{s\gamma r} \times \frac{\dot{\lambda}}{\lambda}, \) and \( \frac{gb-gb}{gb} \times \frac{t}{t+\tau} = \frac{r\dot{\gamma}}{r.s\gamma-r} \times \frac{t}{t+\tau} = \text{that part of the whole area } gb \text{ by which it is made to exceed its proper magnitude; and } \frac{mrt\dot{\Lambda}}{2.s\gamma-r} = \text{the minutes contained in the error which we are investigating. If, for instance, the latitude be } 52^\circ 12', \text{ and the declination } 2^\circ, \text{ both of the same kind, and one of the altitudes be taken at the distance of } 5^\circ, \text{ the other at the distance of } 10^\circ, \text{ from the meridian, this error will amount to } \frac{1}{10000} \text{th part of the whole difference betwixt the real and the assumed } \) latitude. Hence, if this difference be ten minutes, the error in the approximation will not be more than $\frac{1}{7}$th part of a second; and, if the difference be one degree, the error will only be six times greater than in the former case. It will therefore always be inconsiderable, except when the latitude and declination are nearly equal, and on the same side of the equator. But, if this cause should be likely to produce an effect of some importance, we may prevent it by computing an incremental area with the least altitude likewise, and the latitude increased or diminished by a minute, and then taking a mean betwixt them both for the magnitude of the area $gc$. Thus we shall generally obtain a conclusion sufficiently exact; but, if we are desirous of rendering it perfectly so in every instance, we must divide the difference betwixt the real and apparent intervals, in such a manner that the areas assigned to the two observations may be to each other in the same ratio as their correspondent incremental areas. This division will be readily dispatched by making the difference betwixt the logarithms of the two first terms, in the proportion (taken from the 1st table) equal to the difference betwixt the logarithms of the two last, which are given when either of the formulæ $m^2 - \frac{r m^3}{s \gamma} \cdot \frac{v s}{m^3}$, or $\frac{v m}{\lambda} \times \frac{\cos. \text{ of merid. alt. } (\mu)}{\cos. \text{ of declin. } (\delta)}$, is employed; but must be found, if the first method of computing the area $gc$ be adopted. Suppose that $a$ and $b$, representing the logs. of the areas $gb$ and $gb$ in the table, appear from inspection to exceed each other nearly as much as the logs. of the areas $gc$ and $gc$; let $c$ and $d$ be the two next terms in succession, and let $m$ and $n$ be the logs. of the quantities expressing the incremental areas; then will Mr. Lax's Method of finding the Latitude of a Place, \[a \sim b \sim c \sim d : a \sim b \sim m \sim n :: a \sim c ;\] the quantity to be added to, or subtracted from, \(a\), to make it of the required magnitude. We have then only to subtract \(m\) from this corrected value of \(a\), to proceed with the remainder to the second table, and take out the corresponding error in the assumed latitude. This precaution, however, can very seldom be necessary; and, even when it is deemed advisable to adopt it, the division may be performed with so little regard to exactness as to render the process easy and expeditious. But we are not to conclude that, because this inaccuracy, and also the two first species that were considered, are likely to be introduced, when the latitude and declination are nearly equal and of the same kind, they will therefore unavoidably exist in these circumstances. On the contrary, they may always be prevented, when the weather is favourable, by making the observations within a smaller distance from the meridian. We have only to wait till the increase of the sun's altitude becomes so slow as not to produce a visible separation of the limbs in two or three seconds, and then we may be assured that the azimuth is small, and consequently that none of these errors will be considerable. I have hitherto supposed that this method is only to be adopted, when the sun, at each observation, is within fifteen degrees of the meridian; or (to speak more accurately) when both the azimuth and the hour-angle are so small that we may consider their tangents as bearing a given ratio to each other; and, indisputably, these limits should never be transgressed, when it can possibly be avoided; for we have seen (page 79th) that, whatever be the method employed, the smaller the hour-angle, the greater is the exactness with which the lat. is determined. Sometimes, however, it will be impossible to make both, or perhaps either of our observations within the distance which I have recommended; but, even in these cases, our rule may be conveniently applied. It has already been demonstrated that we can never be subject to any material error in consequence of the inequality of the areas $gb$ and $gb$, except when the zenith-distance of the sun, at his meridian altitude, is very small; and, for this case, an effectual remedy has been provided. We need not, therefore, make any farther remarks upon this species of inaccuracy. But perhaps it will be imagined, that because we still continue to suppose the areas of the "figura tangentium" to represent the increments of the latitude, a considerable error will be introduced. We can easily prove, however, that in consequence of the increment of the time being so much diminished by increasing the distance from noon, this error will seldom be of moment enough to claim our attention. Let $\tau$ be the tangent of the azimuth, and $\tau'$ the tangent of the hour-angle at the observation farthest from noon; then it follows, from what has been demonstrated before, that the ratio of $gb$ to $gc$ will exceed the ratio of $GB$ to $GC$, and, consequently, the ratio of $gb$ to $gc$, (supposed to be of a proper magnitude,) by $m \cdot \frac{m-1}{2} \cdot \frac{r+2rs}{s\tau-r} \cdot L$. Hence it is evident that, by diminishing one of these ratios and increasing the other, as we are directed in the present case, till they become equal, we augment the latter by a portion of this difference expressed by $\frac{t}{t+\tau}$, and, on this account, it will be made too great by $m \cdot \frac{m-1}{2} \cdot \frac{t}{t+\tau} \cdot \frac{r+2rs}{s\tau-r} \cdot L$. But the ratio of $gb$ to $gc$ will itself exceed the ratio of $GB$ to $GC$, its proper value, by $m \cdot \frac{m-1}{2} \cdot \frac{\gamma + 2rs}{s\gamma - r} \cdot \dot{L}$; and, by the method of equalizing the ratios, it is suffered to retain a portion of this difference expressed by $\frac{\tau}{t+\tau}$; and, therefore, it will be made too great by $m \cdot \frac{m-1}{2} \cdot \frac{\tau}{t+r} \cdot \frac{\gamma + 2rs}{s\gamma - r} \cdot \dot{L}$. The sum of these quantities, $m \cdot \frac{m-1}{2} \cdot \frac{29}{100000} \cdot \frac{1}{t+\tau} \cdot \frac{tr+2rst}{s\Gamma - r} + \frac{\tau y + 2rs\tau}{s\gamma - r}$ will determine, in minutes of a degree, the amount of the whole error, when the incremental area $gc$ represents one minute of latitude. But if, in our second hypothesis, the latitude be varied $n$ minutes, instead of one, and $p$ be the correction required, the above expression becomes $p \cdot \frac{p-n}{2n} \cdot \frac{29}{100000} \cdot \frac{1}{t+\tau} \cdot \frac{tr+2rst}{s\Gamma - r} + \frac{\tau y + 2rs\tau}{s\gamma - r}$ minutes. Hence, in the lat. of Cambridge, when the declination is $2^\circ$ north, and the sun's distance from the meridian at one observation is $5^\circ$, and at the other $45^\circ$, $p$ being equal to $10$, and $n$ equal to unity, the error will amount to little more than half a second. But, whenever it is judged necessary to guard against this species of error, we need only diminish the value of $\frac{p-n}{n}$ in the last expression; and this will readily be effected by dividing the difference betwixt the real and computed intervals of time, in such a manner that the portions belonging to the two observations may be to each other inversely as the quantities taken out of the fourth table, according to the instructions before delivered. When the portion belonging to the greatest altitude is thus obtained, we can immediately deduce the correction to be made in the first assumed latitude, and then proceed to calculate the incremental areas. Let us suppose, for instance, the difference betwixt the intervals to be \( q \), and the quantities deduced from the 4th table to be \( a \) and \( b \); then will \( \frac{abq}{a + b} = \text{the minutes by which the former latitude is to be varied.} \) If, however, there be any considerable uncertainty respecting the latitude, I would recommend the following method of obtaining an approximate value of it, before we begin the process for investigating its real magnitude. This will effectually preclude, in almost every instance, the various errors that have been described. The increment of the altitude is equal to \( \lambda S \dot{\lambda} \), and, therefore, when the azimuth is so small that its sine varies nearly as the arc, the whole increase of the altitude, whilst the sun is moving to the meridian, will be \( \lambda \times \frac{S}{2} = \frac{S^2 \lambda \delta}{2a} \), the sine of the hour-angle being considered as equal to the arc itself. Now the first of the two tables which have just been explained, will enable us to find the value of \( \frac{S^2}{a} \), by subtracting the log. cosine of the altitude from twice the log. sine of the hour-angle; and a third table might be added, to furnish us with the whole expression \( \frac{S^2 \delta \lambda}{2a} \), (according to the mean value of \( \delta \)) one of its arguments being the quantity \( \frac{S^2}{a} \) already determined, and the other the supposed latitude. It might, perhaps, be advisable to add another column to the first of these tables, containing twice the logarithmic sine of the hour-angle, as this would in some measure abridge the operation. We should find it more convenient too, if the last table were to give us the complement of the arc whose value is \( \frac{S^2 \delta \lambda}{2a} \), rather than the arc itself; because it MDCCXCIX. would then only be necessary to subtract the observed altitude from this complement, and we should immediately deduce the zenith distance of the sun, when he had arrived at the meridian. This being ascertained, we should have no farther difficulty in finding the latitude to be adopted in the subsequent computation. The exactness of the conclusion which is derived from this process, will necessarily depend upon the degree of certainty with which the time is given when the observation is made. An error in time may arise, both from an irregularity in the going of the clock and a small inaccuracy in estimating the difference betwixt the longitude of the place where the altitude is now taken, and that where the time was last determined; but these causes, it is evident, can seldom be very considerable. We may generally, I think, be sure of the time within two or three minutes. The principle upon which this approximation depends will open to us a more compendious way of finding the area $gb$, and which may always be pursued with advantage, when one of the altitudes can be taken at a small distance from the meridian. In the higher latitudes, indeed, the hour-angle may amount to five or six degrees; but, when the latitude and declination approach towards an equality, and are of the same denomination, this angle must be restrained within narrower limits. If, instead of deducing the meridian altitude from the altitude observed, according to the directions before given, we consider it as being actually equal to the meridian altitude, and employ the latitude resulting from this hypothesis in computing the hour-angle, this angle must necessarily be found equal to nothing. We may, therefore, spare ourselves the trouble of performing this part of the operation, and need only calculate the time belonging to the second observation, which is supposed to be made when the sun is at a greater distance from the meridian. The error in the assumed latitude will, in this case, be equal to the whole increase of the altitude betwixt the time of observation and noon; and from this consideration we may be enabled, in any particular instance, to discover at what distance from the meridian this plan may be safely adopted. We can extend it also to any number of observations that happen to be made within the proper limits, by connecting them successively with one on the opposite side of the meridian, and thus determining the areas $gb$ corresponding to each. Having afterwards calculated an incremental area with any of the assumed latitudes varied one minute, we must subtract its logarithm from the logarithmic value of each of the areas $gb$, and we shall discover, by means of the second table, the correction to be applied to each of the supposed latitudes. But the method which has just been explained might always be safely adopted, within the limits prescribed to the former rule, if, in computing the incremental area, we were to employ the latitude obtained by means either of the 3d and 4th, or of the 3d and 5th tables. This precaution would entirely prevent the error which might arise from the inequality of the ratios of $GC$ to $GB$, and $gc$ to $gb$, when the difference betwixt the real and the assumed latitude was very considerable. The error which is occasioned by the disparity of the hour-angles, might be obviated by the same means that were practised in the first method. There is only one objection, indeed, to this rule's being exclusively adopted, when either of the altitudes is taken within the distance at first recommended. It would be necessary, in that case, to extend the table much farther than we are obliged to do, if we adhere to the former rule; but the labour is so much abridged by this plan, that I am doubtful Mr. Lax's Method of finding the Latitude of a Place, whether its advantages might not be more than sufficient to compensate this inconvenience. The table should at least be carried to such a length as would enable us to proceed in this manner, whenever an observation was made within five or six degrees of the meridian. I must not forget to observe, before I conclude the theory, that although I have directed the altitudes to be taken on different sides of the meridian, it is by no means requisite that we should invariably adhere to this precept. We have seen the reason, indeed, why it is expedient, in most instances, to prefer this method, as being generally calculated to produce a much greater degree of exactness in the result. This, however, is not always the case; for, if one of the observations be made beyond the distance originally prescribed, it is of little importance whether the second altitude be taken on the same side of the meridian, or not. But it will sometimes be impossible to make the observations on different sides of noon; and hence it becomes necessary to inquire in what manner the real latitude may be discovered in these circumstances. The clock gives us the interval betwixt the observations equal to $ae - ae$; and by computation we find $ag$ and $ag$, and thence we deduce $eg - eg$, the difference betwixt the errors in time. Having then assumed, without any regard to accuracy, two portions of time, corresponding to the two observations, whose difference is the same as the difference betwixt the errors before determined, and which are to each other in the inverse ratio of the hour-angles, we must increase or diminish them both equally, till we get the areas in the first table of the same magnitude, and then we may conclude that we have obtained the proper value of each. The directions which have been given for the prevention of errors in the former case, when the altitudes are taken on different sides of the meridian, are very easily accommodated to the present; and it would therefore be superfluous to bestow any farther consideration upon them. From a review of the inaccuracies to which this method, in particular cases, may be liable, it appears that none of them can ever be of sufficient importance to affect the mariner. If he only computes the time with each of the altitudes and the latitude by account, and an incremental area with the greatest altitude and the former latitude varied ten minutes, the correction will generally be deduced within much less than a second, and, in the most unfavourable circumstances, within a minute, of the truth. But the astronomer, in every instance, even when the latitude and declination are nearly equal and of the same kind, by adopting the precautions which have been recommended, may be assured of a result perfectly exact. If, however, he should entertain any doubts upon this point, he might easily compute a second value of the incremental area with the latitude already determined; and this, it is evident, would necessarily produce a conclusion not less accurate than if it were obtained from the direct method. The most satisfactory way of proving the utility of this rule, will be to suppose a particular latitude and declination; with these to compute the altitudes, when the sun is at two given distances from the meridian; and thence to deduce the latitude, by an application of our own principles. **EXAMPLE I.** Let the latitude be $54^\circ 27' 50''$ north, and the declin. when the first altit. is taken, $2^\circ 22' 32''$, and, when the second is taken, Mr. Lax's Method of finding the Latitude of a Place, 2° 21' 35", (of the same denomination as the latit.) one observation being made in the morning, when the sun is 5° 3' 22", and the other in the afternoon, when he is 10° 1' 10", distant from the meridian. The altitude in the former case will be found equal to 37° 44' 52"; and in the latter to 37° 15' 20". | Observation 1st. | Observation 2d. | |-----------------|-----------------| | Lat. 54° 15' 0" | Lat. 54° 16' 0" | | Log. of \(a\) = 9,7868838 | 9,7868838 | 9,7820217 | | \(-d\) = 8,6175181 | 8,6175181 | 8,6146155 | | \(-l\) = 9,9093281 | 9,9094190 | 9,9093281 | | \(-dl\) = 18,5268462 | 18,5269371 | 18,5239436 | | \(-\frac{am^3}{dl}\) = 21,2600376 | 21,2599467 | 21,2580781 | | \(-\frac{am^3}{dl}\) = 10,6300188 | 10,6299733 | 10,6290390 | | Log. tang. = 10,6177463 | 10,6176983 | 10,6167094 | | \(2 \times \log.\) tang. = 21,2354926 | 21,2353966 | 21,2334188 | | Log. of \(r\) = 8,6178915 | 8,6178915 | 8,6149839 | | \(-s\) = 10,1427296 | 10,1429961 | 10,1427296 | Computed log. of \(\gamma\) = 9,9961137 | 9,9962842 | 9,9911323 | True = 9,9983068 | | 9,9933253 | Area \(gb\) = 21931 | 9,9961137 | Area \(gb\) = 21930 | Hence, \(\frac{gb}{gc} = \frac{21931}{1705} = 12' 51"\), and latit. = 54° 27' 51". If we employ the formula \(\frac{\dot{\gamma}}{\gamma} = m^2 - \frac{rm^3}{s\gamma} \cdot \frac{vs}{m^3}\) in computing the area \(gc\), we shall obtain, very nearly, the same result. Log. of \( r = \frac{8,6178915}{s} = 10,1427296 \) \[ \begin{align*} - & \gamma = 9,9961139 \\ - & s\gamma = 20,1388435 \\ - & \frac{r^3}{s\gamma} = 18,4790480 \\ - & \left(\frac{r^3}{s\gamma}\right)^{\frac{1}{2}} = 9,2395240 \end{align*} \] Log. cosine \( = \frac{9,9933560}{2 \times \log. \cos.} = 19,9867120 \) Const. log. \( = 3,1015 \) Log. \( s = \frac{10,1427296}{gc} = 3,2309416 \) \[ \begin{align*} - & gb = 4,3410188 \end{align*} \] Difference \( = 1,1100772 \), producing in tab. 2d. \( 12' 52'' \) Lat. corrected \( = 54° 27' 52'' \) But there is no occasion to write down the logs. of \( r \), \( s \), and \( \gamma \), again: we may add together the logs. of \( s \) and \( \gamma \), and subtract their sum from the log. of \( r^3 \) at the same time, as they stand in the former operation, and this will render the latter process extremely short. Let us resume the preceding example, and compute the area \( gb \) by the second method, and the area \( gc \) by the formula \( m^2 = \frac{r^3}{s\gamma} \cdot \frac{v}{s} \). Mr. LAX's Method of finding the Latitude of a Place, Complement of alt. $52^\circ 15' 8''$ Observation 1st. Observation 2d. Declin. $2^\circ 22' 32''$ Lat. $54^\circ 37' 40''$. Lat. $54^\circ 37' 40''$. Lat. $54^\circ 37' 40''$. Log. of $a = 9,7820217$ $- d = 8,6146155$ $- l = 9,9113753$ $\frac{am^3}{dl} = 21,2560309$ $\frac{am^2}{dl} = 10,6280154$ Log. tang. = $10,6156259$ $2 \times \log. \text{tang.} = 21,2312518$ Log. of $\frac{rm^3}{s\gamma} = 18,4711836$ Log. $r = 8,6149839$ $- s = 10,1487823$ Log. cosine = $9,9934771$ Computed log. $\gamma = 10,0000000$ True = $9,9983068$ Area $gb = 16932$ $gb + gb = 33859$ Mean value of $gb = 16929$ Whose log. (from tab. 1st) = $4,229$ $gc = 3,237$ Difference = $1992$ producing from tab. 2d Lat. corrected = $54^\circ 27' 51''$ EXAMPLE II. Let the real latitude be $10^\circ 0' 0''$, and the two observations be made on different sides of noon; one when the sun's distance from the meridian is $5^\circ 0' 0''$, and the other when his distance is $10^\circ 0' 0''$; the declin. in the former case being $7^\circ 40' 40''$, and in the latter $7^\circ 39' 43''$, of the same kind as the latitude. Then will the greatest altitude be $84^\circ 32' 28''$, and the least $79^\circ 50' 48''$. | Observation 1st. | Observation 2d. | |-----------------|-----------------| | Lat. $10^\circ 10' 0''$ | Lat. $10^\circ 10' 0''$ | | Log. of $a = 9,9980259$ | - | | $d = 9,1258124$ | $9,1258124$ | | $l = 9,2467746$ | $9,2460695$ | | $\frac{am^3}{dl} = 21,6254389$ | $21,6261440$ | | $\frac{am^3}{dl} = 10,8127194$ | $10,8130720$ | | Log. tang. = $10,8075134$ | $10,8078745$ | | $2 \times$ log. tang. = $21,6150268$ | $21,6157490$ | | Log. of $r = 9,1297233$ | $9,1297233$ | | $s = 9,2536477$ | $9,2529200$ | | Computed log. of $\gamma = 9,9983978$ | $9,9983923$ | | True = $9,9983442$ | $9,9934035$ | | Area $gb = 536$ | $9,9983978$ | | Area $gc = 55$ | Area $gb = 520$ | $\frac{gb + gb}{2} = 528$. Hence, correction $= \frac{528}{55} = 9' 34''$, and lat. corrected $= 10^\circ 0' 26''$. The difference, however, betwixt the real and the corrected lat. would only have amounted to $19''$, if we had determined the area $gb$ by the 1st table, instead of taking a mean betwixt the two areas for its proper value. This error might have been expected from the near approach of the lat. to the declin. and ought therefore to have been guarded against. It proceeds from the three causes of inaccuracy which, I have shewn, must necessarily be combined in these circumstances. The remedies MDCCXCIX. Mr. Lax's Method of finding the Latitude of a Place, to be applied are well known. We must find an approximate value of \( p \) (page 89) by means of the 3d and 4th tables, and compute an incremental area with the least, as well as with the greatest altitude. If this plan had been pursued, the latitude would have been ascertained within less than a second of the truth. Or, if the weather had been favourable, we might have prevented the error as effectually, by making the observations when the hour-angles were much less, and approached nearer to an equality. N. B. The log. sines and tangents of the declin. and lat. to be employed in all the operations should be taken out at the same time. But, when the first method of computing the incremental area is adopted, we may avail ourselves, with considerable advantage, of the following expedient. Instead of taking the whole log. sine and tangent of the new lat. take the increments of each, or the difference betwixt their respective values in the two suppositions; and find the increment of the log. tangent corresponding to the increment of the log. secant \( \frac{a m^3}{d l} \). This may readily be effected, when the log. tangent is deduced from the log. secant in the former process, by taking the rate of increase belonging to each, and thence inferring by an easy proportion the proper increment of the log. tangent in the latter case. The difference betwixt this increment and the increment of \( r \) is the value of the area \( gc \). An instance of the method thus improved shall be given in the next example. EXAMPLE III. Let us suppose the true lat. to be $54^\circ 27' 50''$, and two observations to be made in the afternoon—one when the sun's distance from the meridian is $4^\circ 30' 0''$, the other when his distance is $45^\circ 10' 15''$—the declin. in the former case being $8^\circ 7' 35''$, and in the latter $8^\circ 5' 3''$, on the same side of the equator with the latitude. Then will the greatest altitude be found equal to $43^\circ 31' 19''$, and the least to $31^\circ 20' 25''$. | Observation 1st. | Observation 2d. | Observation 1st. | Observation 2d. | |-----------------|-----------------|-----------------|-----------------| | Lat $54^\circ 10' 0''$ | Lat. $54^\circ 10' 0''$ | Lat. $54^\circ 11' 0''$ | Lat. $54^\circ 11' 0''$ | | Log. of $a = 9.8379894$ | - | - | - | | $d = 9.1503179$ | - | - | - | | $l = 9.9088727$ | - | - | - | | $\frac{am^3}{dl} = 20.7787988$ | - | - | - | | $\frac{am^3}{dl} = 10.3893994$ | - | - | - | | Log. tang. = $10.3498734$ | - | - | - | | $2 \times$ log. tang. = $20.6997468$ | - | - | - | | Log. of $r = 9.1547009$ | - | - | - | | $s = 10.1413981$ | - | - | - | | Computed log. of $y = 9.9958458$ | - | - | - | | True = $9.9986590$ | - | - | - | | Area $gb = 28132$ | Area $gb = 26816$ | Area $gc = 1572$ | Area $gc = 1494$ | Now, by diminishing $gb$ and $gb$, till the ratio of $gc$ to $gc$ becomes equal to the ratio of $gb$ to $gb$, we should get $gb = 28126$, and $gc = 26735$. Hence $\frac{gb}{gc} = \frac{28126}{1572} = 17' 53''$ the correction, and the latitude = $54^\circ 27' 53''$. But the same conclusion would have been obtained indepen- dently of the second incremental area, which I have demonstrated above can never be necessary, except when the zenith distance at noon is very small. The error of three seconds arises from a different cause, and might have been entirely excluded with the assistance of the 3d and 4th tables. We should, in fact, deduce from these tables an approximation of $22'$; and, if we calculate the area $gc$ with the lat. $54^\circ 32' 0''$, it will be found equal to $34762$, and consequently $\frac{28172}{36762} = 48'',62$, which, multiplied by $22$, produces $17' 49'',64$ for the correction, and the lat. is found $= 54^\circ 27' 49'',64$. It would have been still better, however, to have obtained a near value of the lat. in the first instance, by means of the 3d and 5th tables. The work might have been dispatched in rather less time, and the conclusion would have been as rigidly accurate. It is obvious that, if the time when the first observation was made could have been ascertained within two or three seconds, the area $gb$ might have been immediately found by either of the two methods which have been explained, without the assistance of a second altitude, or of the first table; and the latitude determined with much greater facility, and with sufficient exactness. When the azimuth indeed, as in the present case, does not exceed four or five degrees, an error of a second in time will not produce an error of more than a second in the result; and, as the azimuth decreases, the error in latitude, arising from a given error in time, will be diminished in the same proportion. In general, if $\dot{\lambda}$ be the error in the assumed time, the error in the corrected latitude will be nearly equal to $15 T \lambda \dot{\lambda}$. Let us then suppose the time to be given at the first observation, and let us determine the area $gb$ by the second method, and the area $gc$ by the formula $\frac{vm}{\lambda} \cdot \frac{\cos \text{ of merid. alt. } (\mu)}{\cos \text{ of declin. } (\delta)}$. Comp. of alt. = $46^\circ 28' 41''$ Declin. = $8^\circ 7' 35''$ Lat. assumed = $54^\circ 36' 16''$ Log. of $v = 3.1015$ $\mu = 9.8604043$ $m \nu \mu = 22.9619043$ Computed log. of $\gamma = 10,0000000$ True = $9.9986591$ $\delta = 9.9956171$ $\lambda = 9.7628420$ Area $gb = 13409$ Log. of area $gb = 4.127$ Area $gc = 3.2034452$ $gc = 3.203$ Difference = $.924$, producing from tab. 2d. $8' 23'' \frac{1}{2}$. Lat. corrected = $54^\circ 27' 52'' \frac{1}{2}$. **EXAMPLE IV.** The latitude, which is supposed in the last example to be $54^\circ 27' 50''$, is that of Ravensworth, a village about five miles to the north of Richmond, in Yorkshire, where I resided some months in the summer of 1797. During that time, I neglected no opportunity of taking the sun’s meridian altitude; and, accordingly, the latitude which is there assumed is the mean result of a considerable number of observations. Moreover, on the 8th of September, I took, with particular care, four altitudes, Mr. Lax's Method of finding the Latitude of a Place, when the sun was near the meridian; and with these I will now calculate the latitude, in order to apply the 1st and 2d tables, and likewise to exemplify the method of combining several observations together. The two first altitudes were taken in the morning, one at $19' 24''$, the other at $17' 40''$ before twelve o'clock, and were respectively $40° 49' 6''$, and $40° 50' 51''$; the two last in the afternoon, one at $17' 53''$, the other at $19' 44''$ past twelve o'clock, and were respectively $40° 49' 50''$, and $40° 48' 18''$. Now the longitude, found by the eclipses of Jupiter's satellites, is nearly $1° 40'$ west of Greenwich; and hence the declin. appears to have been, at the two first observations, $5° 26' 25''$, and at the two last, $5° 25' 47''$; the interval betwixt the first and second, and betwixt the third and fourth, being so small that we may consider the declin. as remaining constant during each of these times. | Observation 1st. | Observ. 2d. | Observ. 3d. | Observ. 4th. | |-----------------|-------------|-------------|--------------| | Lat. $54° 29'$ | Lat. $54° 29'$ | Lat. $54° 29'$ | Lat. $54° 29'$ | | Log. of $a = 9.8153538$ | $9.8156096$ | $9.8154610$ | $9.8152367$ | | $d = 8.9768457$ | $8.9768457$ | $8.9760047$ | $8.9760047$ | | $l = 9.9105959$ | $9.9105959$ | $9.9105959$ | $9.9105959$ | | $\frac{am^2}{dl} = 20,9279122$ | $20,9281680$ | $20,9288604$ | $20,9286361$ | | $\frac{am^3}{dl} = 10,4639561$ | $10,4640840$ | $10,4644302$ | $10,4643180$ | | Log. tang. $= 10,4366767$ | $10,4368218$ | $10,4372141$ | $10,4370861$ | | $2 \times \log. \text{tang.} = 20,8733534$ | $20,8736436$ | $20,8744282$ | $20,8741722$ | | Log. of $r = 8.9788063$ | $8.9788063$ | $8.9779578$ | $8.9779578$ | | $s = 10,1464648$ | $10,1464648$ | $10,1464648$ | $10,1464648$ | | Comput. log. $\gamma = 9.9986245$ | $9.9989147$ | $9.9988508$ | $9.9985948$ | Log. $r m^3 = 18,8300352$ $\frac{r m^3}{s \gamma} = 9,4150276$ Log. cosine $= 9,9847972$ $2 \times l \cos. = 19,9695944$ Log. $\nu = 3,1015$ $s = 10,1464648$ Log. $gc = 3,218$ by Means of two Altitudes of the Sun. | Observation | Time from the argument of 1st table | Error in time | |-------------|-----------------------------------|---------------| | 1st | 18' 14", by clock | 19' 24" | | 4th | 18 26 | 19 44 | Near log. value of $gb$ from 1st table | Log. value of $gb$ | Error in time | |--------------------|---------------| | 3,296 | 36 40 | | 3,276 | 36 40 | | 2)6,572 | 2 28 | First log. value of $gb$ | Log. value of $gb$ | Error in time | |--------------------|---------------| | 3,286 | | | Observation | Time from the argument of 1st table | Error in time | |-------------|-----------------------------------|---------------| | 2d | 16' 12", by clock | 17' 40" | | 3d | 16 40 | 17 53 | Near log. value of $gb$ from 1st table | Log. value of $gb$ | Error in time | |--------------------|---------------| | 3,294 | 35 33 | | 3,264 | 32 52 | | 2)6,558 | 2 41 | Second log. value of $gb$ | Log. value of $gb$ | Error in time | |--------------------|---------------| | 3,279 | | | Mean | 3,282 | | Log. value of $gc$ | 3,218 | Difference ,064 producing in 2d tab. 1' 10" Lat. corrected = 54° 27' 50" We now take the same example again, and apply the second method of computing the area $gb$, and the formula $\frac{m \nu \mu}{\delta \lambda}$ in determining the area $gc$. Mr. Lax's Method of finding the Latitude of a Place, Comp. of alt. at 2d observ'n 49° 9' 9" Declin. 5 26 25 Lat. = 54 35 34 Comp. of alt. at 3d observ'n 49° 10' 10" Declin. 5 25 47 Lat. = 54 35 57 Observations 1st and 3d. Lat. 54° 35' 57" Log. of \(a\) = 9.8153538 \(d\) = 8.9768457 \(l\) = 9.9112212 \(\frac{am^3}{dl}\) = 20,9272869 \(\frac{am^3}{dl}\) = 10,4636434 Log. tang. = 10,4363220 \(2 \times \log.\) tang. = 20,8726440 Log. of \(r\) = 8.9788063 \(s\) = 10,1483229 Computed log. of \(\gamma\) = 9.9997732 True = 9.9984422 Area \(gb\) = 13310 Computed log. of \(r\) = 10,0000000 True = 9.9986765 Area \(gb\) = 13235 \(\frac{gb + gb}{2}\) = 13272 Log of \(\frac{gb + gb}{2}\) = 4,1229 \(gc\) = 3,2194 Difference = .9039 Hence, correction from tab. 2d = 8' 1" - 7' 48" Lat. corrected = 54° 27' 56" - 54° 27' 44" Mean corrected lat. = 54° 27' 50" REMARKS. 1. All the altitudes that are taken on the same side of noon, tend only to correct the error which may be supposed to exist in the greatest of these altitudes, and can have no effect in removing any inaccuracy to which the greatest altitude on the other side of the meridian may be subject. Hence we must take more than one altitude on each side of noon, if we are desirous of obtaining a very exact conclusion. 2. When some of the observations are made in the morning, and others in the afternoon, the smaller the hour-angle, in every instance, the more favourable it will be for our purpose. But, if we cannot procure an altitude on each side of the meridian, we ought to make one observation when the hour-angle is as large as possible, and with this all the rest should be separately combined. We must be cautious, however, not to let the sun be too near the horizon, lest the apparent altitude should be affected by the uncertainty of the refraction. 3. If the clock were to furnish us with the true time, we might combine together any two observations made within the proper limits, without applying to the first table, and deduce a very exact correction. Should there even be a small error in the supposed time, we might still proceed in the same manner, without being liable to any material inaccuracy, provided the difference betwixt the hour-angles was not very considerable. The error, indeed, occasioned by adopting this method of finding the two areas, and taking a mean betwixt them for the value of $gb$, may easily be determined in any particular case. If $t$ and $t'$ be the respective tangents of the smaller and greater hour-angles, and $\dot{t}$ their difference; $\dot{x}$ the error of the clock in MDCCXCIX. Mr. Lax's Method of finding the Latitude of a Place, minutes of time; and \( \dot{Z} \) the whole error in the time computed with the greater altitude; then will the error in the result be to the whole correction of the latitude \( (\dot{L}) \) :: \( \frac{\dot{t} \dot{z}}{2} : t \dot{z} : \) \( \frac{\dot{t} \dot{z}}{2} : \frac{t \dot{L}}{15 T \lambda} : \frac{\dot{t} \dot{z}}{2} : \frac{l \gamma \delta - d \lambda}{15 \lambda \gamma \delta} \times \dot{L} :: 7.5 \times t \dot{z} : s - \frac{r}{\gamma} \dot{L} \). This error will consequently be equal to \( \frac{7.5 \dot{t} \dot{z}}{s - \frac{r}{\gamma}} \); and hence it appears that, if we had pursued this method in the last example, and there had been an error of a minute in the time given by the clock, there would not have been an error of a single second in the conclusion. 4. If the time were determined by equal altitudes, and one of them were to be employed in computing the area \( gb \), it is manifest that we should entirely exclude the error which has just been considered. It would be necessary, however, in order to correct by the second observation any inaccuracy that may have occurred in reading off at the first, to move the index, and then bring it apparently to the same position again, before we proceeded to take the second altitude. EXAMPLE V. On the 30th of January, 1799, the following altitudes of the sun's lower limb were taken in Trinity College, Cambridge; the height of the barometer being 29.6 inches, and of the thermometer 31 degrees. by Means of two Altitudes of the Sun. | Observed double alt. by the sextant | Time by the clock | Time from noon by the sun | Log. cos. of the hour-angle | Lat. from each pair of corresponding altitudes | |-------------------------------------|------------------|--------------------------|----------------------------|------------------------------------------------| | 39° 35' 10" | 11h 21' 0" | 23' 40" | 9.9976800 | 52° 12' 39" | | 40° 7 | 20 51 | 20 49 | 8206 | 46 | | 44° 55 | 26 49 | 17 51 | 8681 | 45 | | 48° 5 | 29 13 | 15 27 | 9012 | 44 | | 50° 26 | 31 1 | 13 39 | 9229 | 40 | | 52° 30 | 33 0 | 11 40 | 9437 | 41 | | 53° 40 | 34 35 | 10 5 | 9579 | 41 | | 55° 19 | 36 46 | 7 54 | 9742 | 43 | | 56° 50 | 38 56 | 5 44 | 9864 | 38 | | 57° 20 | 41 4 | 3 36 | 9946 | 41 | | 57° 55 | 43 54 | 0 46 | 9999 | 41 | | 57° 50 | 46 40 | - 2 0 | 9983 | 41 | | 57° 28 | 48 12 | 3 32 | 9948 | 41 | | 57° 0 | 49 59 | 5 19 | 9883 | 38 | | 56° 0 | 51 31 | 6 51 | 9806 | 43 | | 55° 15 | 53 0 | 8 20 | 9713 | 41 | | 53° 26 | 55 2 | 10 22 | 9556 | 41 | | 51° 5 | 57 44 | 13 4 | 9295 | 40 | | 48° 38 | 59 46 | 15 6 | 9057 | 44 | | 45° 17 | 12 2 | 9 17 29 | 8735 | 45 | | 41° 50 | 4 24 | 19 44 | 8388 | 46 | | 35° 10 | 8 20 | 23 40 | 7680 | 39 | 22) 1102 556 Error of adjust. 39° 50' 31" 2) 39° 51' 19" Refra. — parall. 19° 55' 39" Semi-diameter 19° 53' 4" Assumed merid. alt. 20° 9' 22" Mean declin. 17° 34' 13" 37° 43' 35" Assumed lat. 52° 16' 25" Correction from tab. zd. 3° 43' True lat. 52° 12' 42" not less accurate than the mean result from 22 meridian altitudes. Q 2 Mr. Lax's Method of finding the Latitude of a Place, This process will require very little explanation. The first column contains the observed double altitudes, as they were read off from the sextant; the second contains the corresponding times given by the clock; the third contains the times from noon determined, with sufficient exactness, by taking half the interval betwixt the first and last observations, (in which the altitudes are equal,) and supposing this to be the time elapsed betwixt the first observation and the sun's reaching the meridian. The fourth column is formed of the log. cosines of the hour-angles taken immediately from the argument of the first table; and the last column exhibits the latitudes deduced from every two corresponding altitudes, and is intended to shew the agreement betwixt these results. It is not necessary that we should employ the tables in this operation: for we may take the log. cosines of the hour-angles from Taylor's Logarithms, after the time is reduced into degrees, minutes, and seconds; and, by dividing the area $gb = 7537$, by the area $gc = 203$, (the natural number belonging to the log. $3.3081$,) we shall obtain the correction required. | o° 12" | o° 24" | o° 36" | o° 48" | 1' 0" | 1' 12" | 1' 24" | 1' 36" | 1' 48" | 2' 0" | 2' 12" | |--------|--------|--------|--------|------|-------|-------|-------|-------|------|-------| | 9.99 | | | | | | | | | | | | 16° | 8941 | | | | | | | | | | | - | 8923 | | | | | | | | | | | - | 905 | | | | | | | | | | | - | 887 | | | | | | | | | | | - | 869 | | | | | | | | | | | - | 850 | | | | | | | | | | | - | 832 | | | | | | | | | | | - | 813 | | | | | | | | | | | - | 795 | | | | | | | | | | | - | 776 | | | | | | | | | | | - | 757 | | | | | | | | | | | - | 737 | | | | | | | | | | | - | 718 | | | | | | | | | | | - | 699 | | | | | | | | | | | - | 679 | | | | | | | | | | | - | 659 | | | | | | | | | | | - | 639 | | | | | | | | | | | - | 619 | | | | | | | | | | | - | 599 | | | | | | | | | | | - | 578 | | | | | | | | | | | - | 558 | | | | | | | | | | | - | 537 | | | | | | | | | | | - | 516 | | | | | | | | | | | - | 495 | | | | | | | | | | | - | 474 | | | | | | | | | | | - | 453 | | | | | | | | | | | - | 431 | | | | | | | | | | | - | 410 | | | | | | | | | | | - | 388 | | | | | | | | | | | - | 366 | | | | | | | | | | | - | 344 | | | | | | | | | | | - | 322 | | | | | | | | | | | - | 300 | | | | | | | | | | | - | 277 | | | | | | | | | | | - | 255 | | | | | | | | | | | - | 232 | | | | | | | | | | | - | 209 | | | | | | | | | | | - | 186 | | | | | | | | | | | - | 163 | | | | | | | | | | | - | 139 | | | | | | | | | | | - | 116 | | | | | | | | | | | - | 992 | | | | | | | | | | | - | 668 | | | | | | | | | | | - | 444 | | | | | | | | | | | - | 200 | | | | | | | | | | | - | 996 | | | | | | | | | | | - | 972 | | | | | | | | | | | - | 947 | | | | | | | | | | | - | 922 | | | | | | | | | | | - | 910 | | | | | | | | | | | - | 885 | | | | | | | | | | | - | 860 | | | | | | | | | | | - | 835 | | | | | | | | | | | - | 809 | | | | | | | | | | | - | 784 | | | | | | | | | | | - | 758 | | | | | | | | | | | - | 732 | | | | | | | | | | | - | 706 | | | | | | | | | | | - | 680 | | | | | | | | | | | - | 654 | | | | | | | | | | The numbers in each cell likely represent some form of data or measurements, but without context, it's challenging to determine their significance. | 3' | 3' 12" | 3' 24" | 3' 36" | 3' 48" | 4' | 4' 12" | |----|--------|--------|--------|--------|---|--------| | 3° 3 | 3° 3 | 3° 3 | 3° 3 | 3° 3 | 3° 3 | 3° 3 | | 3' | 3' 12" | 3' 24" | 3' 36" | 3' 48" | 4' | 4' 12" | | 3° 3 | 3° 3 | 3° 3 | 3° 3 | 3° 3 | 3° 3 | 3° 3 | | 3' | 3' 12" | 3' 24" | 3' 36" | 3' 48" | 4' | 4' 12" | | 3° 3 | 3° 3 | 3° 3 | 3° 3 | 3° 3 | 3° 3 | 3° 3 | | 3' | 3' 12" | 3' 24" | 3' 36" | 3' 48" | 4' | 4' 12" | | 3° 3 | 3° 3 | 3° 3 | 3° 3 | 3° 3 | 3° 3 | 3° 3 | This table appears to contain numerical data arranged in a grid format, likely representing some form of trigonometric or astronomical values. The numbers are aligned in columns and rows, suggesting a systematic organization. Each cell contains a specific value, possibly fractions of degrees or radians given the format "x° y' z'". The table extends across many pages, indicating a comprehensive dataset. by Means of two Altitudes of the Sun. **Tab. II.** | h | Log. of g or α. | |---|-----------------| | 0 | 89 45 | | 1 | 89 30 | | 2 | 89 15 | | 3 | 89 0 | | 4 | 88 45 | | 5 | 88 30 | | 6 | 88 15 | | 7 | 88 0 | | 8 | 87 45 | | 9 | 87 30 | | 10| 87 15 | | 11| 87 0 | | 12| 86 45 | | 13| 86 30 | | 14| 86 15 | | 15| 86 0 | | 16| 85 45 | | 17| 85 30 | | 18| 85 15 | | 19| 85 0 | | 20| 84 45 | | 21| 84 30 | | 22| 84 15 | | 23| 84 0 | | 24| 83 45 | | 25| 83 30 | | 26| 83 15 | | 27| 83 0 | | 28| 82 45 | | 29| 82 30 | | 30| 82 15 | | 31| 82 0 | | 32| 81 45 | | 33| 81 30 | | 34| 81 15 | **Tab. III.** | Log. of g². | |-------------| | 7,64 | | 7,94 | | 8,12 | | 8,24 | | 8,34 | | 8,42 | | 8,48 | | 8,54 | | 8,59 | | 8,64 | | 8,68 | | 8,72 | | 8,75 | | 8,78 | | 8,81 | | 8,84 | | 8,87 | | 8,90 | | 8,93 | | 8,96 | | 8,99 | | 9,02 | | 9,05 | | 9,09 | **Tab. IV.** | 10° | 20° | 30° | 35° | 40° | 45° | 50° | 55° | 60° | |-----|-----|-----|-----|-----|-----|-----|-----|-----| | 8,64| ,64 | ,61 | ,57 | ,54 | ,50 | ,46 | ,42 | ,38 | | 8,68| ,70 | ,67 | ,64 | ,60 | ,56 | ,52 | ,48 | ,44 | | 8,71| ,77 | ,74 | ,71 | ,67 | ,63 | ,59 | ,55 | ,51 | | 8,75| ,84 | ,80 | ,76 | ,72 | ,68 | ,64 | ,60 | ,56 | | 8,78| ,90 | ,86 | ,82 | ,78 | ,74 | ,70 | ,66 | ,62 | | 8,81| ,97 | ,93 | ,89 | ,85 | ,81 | ,77 | ,73 | ,69 | | 8,84| 1, 0| ,98 | ,94 | ,90 | ,86 | ,82 | ,78 | ,74 | | 8,87| 1, 1| ,96 | ,92 | ,88 | ,84 | ,80 | ,76 | ,72 | | 8,89| 1, 2| ,94 | ,90 | ,86 | ,82 | ,78 | ,74 | ,70 | | 8,92| 1, 3| ,92 | ,88 | ,84 | ,80 | ,76 | ,72 | ,68 | | 8,94| 1, 4| ,90 | ,86 | ,82 | ,78 | ,74 | ,70 | ,66 | | 8,98| 1, 5| ,88 | ,84 | ,80 | ,76 | ,72 | ,68 | ,64 | | 9,02| 1, 6| ,86 | ,82 | ,78 | ,74 | ,70 | ,66 | ,62 | | 9,05| 1, 7| ,84 | ,80 | ,76 | ,72 | ,68 | ,64 | ,60 | | 9,09| 1, 8| ,82 | ,78 | ,74 | ,70 | ,66 | ,62 | ,58 | Mr. LAX's Method of finding the Latitude of a Place, | Tab. V. | |--------| | | 10° | 20° | 30° | 35° | 40° | 45° | 50° | 55° | 60° | |--------|-------|-------|-------|-------|-------|-------|-------|-------|-------| | 6,463 | 89° | 89° | 89° | 89° | 89° | 89° | 89° | 89° | 89° | | | 59' 30" | 59' 32" | 59' 34" | 59' 36" | 59' 38" | 59' 40" | 59' 42" | 59' 44" | 59' 46" | | 6,705 | | | | | | | | | | | | | | | | | | | | | | 6,941 | 58 30 | 58 35 | 58 42 | 58 46 | 58 51 | 58 56 | | | | | | | | | | | | | | | | 7,066 | | | | | | | | | | | | | | | | | | | | | | 7,163 | 57 30 | 57 38 | 57 49 | 57 57 | | | | | | | | | | | | | | | | | | 7,242 | | | | | | | | | | | | | | | | | | | | | | 7,309 | 56 30 | 56 42 | 56 58 | | | | | | | | | | | | | | | | | | | 7,367 | | | | | | | | | | | | | | | | | | | | | | 7,418 | 55 30 | 55 46 | | | | | | | | | | | | | | | | | | | | 7,464 | | | | | | | | | | | | | | | | | | | | | | 7,505 | 54 30 | 54 49 | | | | | | | | | | | | | | | | | | | | 7,543 | | | | | | | | | | | | | | | | | | | | | | 7,578 | 53 30 | 53 50 | | | | | | | | | | | | | | | | | | | | 7,610 | | | | | | | | | | | | | | | | | | | | | | 7,640 | 52 30 | 52 57 | | | | | | | | EXPLANATION OF THE TABLES. Tab. I. This table contains two arguments. One is the distance of the sun from the meridian, which is expressed in hours, minutes, and seconds; and also in terms of the log. cosine of the hour-angle. This argument is placed in the first column. The second stands at the top of the table, and consists of the minutes and seconds contained in the error of time. The table itself exhibits the logarithm of the area $gb$ corresponding to any given error at any particular observation. But it was necessary to give the log. value of this area in both directions from the tangent $gd$, i.e. both when the computed is greater, and when it is less, than the real interval of time. These two values are included in the same column, but are separated from each other by a point. The first quantity expresses the log. area in the latter, the second expresses it in the former case. Thus, for instance, in the fifth column we have $136.108$, and we are to understand that the log. of the area $gb$ is equal to $136$, when the computed time is too small; and equal to $108$, when the computed time is too great. The argument at the top is carried to every twelve seconds of time, or every three minutes of a great circle; and should be continued as far as four or five minutes of time for the first twelve or fourteen degrees from the meridian; as far as three minutes for the next ten degrees; two minutes for the next ten degrees; and so on, (the time being diminished as the tangent of the hour-angle increases,) as far as sixty or sixty-five degrees, beyond which the table need not be extended. The argument in the first column should be carried to every eight seconds of time for the first twelve or fourteen degrees; and to every twenty seconds for the rest of the table. Tab. II. This table consists of two columns. The first contains the differences betwixt the logs. of the area $gb$ and $gc$, when the latit. is varied one minute in calculating the incremental area. These differences are negative as far as the fourteenth, because so far the area $gc$ exceeds the area $gb$. In the second column are exhibited the errors in the assumed latit. corresponding to these differences. Tab. III. This table is composed of four columns. The first contains the computed distance in time from noon; the second the altitude of the sun; the third the log. sine of the hour-angle, or the log. cosine of the altitude, expressed in the first column; and the fourth twice the logarithm in the former column, being intended to save us the trouble of multiplying by 2, when we are finding the logarithm of $\frac{g^2}{a}$. Tab. IV. In this table there are two arguments. One of them occupies the first column, and consists of the log. sines of the azimuth, which are found by subtracting the log. cosine of the altitude from the log. sine of the hour-angle, as they stand in the preceding table. The other argument is placed at the top of the columns, and is formed of the assumed latitude. The table is designed to exhibit the product of the tangent of the azimuth multiplied into the cosine of the latit. and into 15; so that, by taking out this quantity, and multiplying it into the minutes contained in the error of time, we may at once determine how much the assumed latit. is to be varied in computing the incremental area. Tab. V. There are two arguments in this table likewise. That which stands in the first column is the log. value of $\frac{g^2}{a}$ obtained from the second table by subtracting the log. cosine of the altitude from twice the log. sine of the hour-angle. The second argument, which is placed above the columns, is the supposed latitude. The table itself contains the complement of the meridian altitude at the place of observation.