The Solutions of Three Chorographic Problems, by a Member of the Philosophical Society of Oxford

The Solutions of three Chorographic Problems, by a Member of the Philosophical Society of Oxford. The three following Problems may occur at Sea, in finding the distance and position of Rocks, Sands, &c. from the shore; or in surveying the Sea Coast; when only two objects whose distance from each other is known, can be seen at one station: but especially they may be useful to one, that would make a Map of a Countrey by a Series of Triangles derived from one or more measured Bases; which is the most exact way of finding the bearing and distance of places from each other, and thence their true Longitude and Latitude; and may consequently occur to one that would in that manner measure a Degree on the Earth. The First Problem (Fig. 1. and 2.) There are two objects B and C, whose distance BC is known, and there are two stations at A & E, where the objects B, C being visible, & the stations one from another, the Angles BAC, BAE, AEB, AEC, are known by Observation, (which may be made with an ordinary Surveying Semicircle, or Crossstaff, or if the objects be beyond the view of the naked Eye, with a Telescopic Quadrant) to find the distances or lines AB, AC, AE, EC. Construction. In each of the triangles BAE, CAE, two angles at A, E, being known, the third is also known: then take any line αε at pleasure, on which constitute the triangles βαε, αεγ respectively equiangular to the triangles BAE, AEC; join βγ. Then upon BC constitute the triangles BCA, BCE equiangular to the correspond- dent triangles $\beta \gamma \alpha$, $\beta \gamma \epsilon$, join $AE$, and the thing is manifestly done. **The Calculation.** Assuming $\alpha$ of any number of parts, in the triangles $\alpha \beta \epsilon$, $\alpha \gamma \epsilon$, the angles being given, the sides $\alpha \beta$, $\alpha \gamma$, $\epsilon \beta$, $\epsilon \gamma$ may be found by Trigonometry: then in the Triangle $\beta \alpha \gamma$, having the angle $\beta \alpha \gamma$, and the legs $\alpha \beta$, $\alpha \gamma$, we may find $\beta \gamma$. Then $\beta \gamma$, $BC :: \beta \alpha$. $BA :: \beta \epsilon$, $BE :: \gamma \alpha$, $CA :: \gamma \epsilon$, $CE$. **The second Problem (Fig. 3 & 4.)** Three objects $B$, $C$, $D$, are given, or (which is the same) the sides, and consequently angles of the triangle $BCD$ are given; also there are 2 points or stations $A$, $E$, such, that at $A$ may be seen the three points $B$, $C$, $E$, but not $D$; and at the station, $E$, may be seen $A$, $C$, $D$, but not $B$, that is the angles $BAC$, $BAE$, $AEC$, $AED$, (and consequently $EAC$, $AEC$,) are known by observation: to find the lines $AB$, $AC$, $AE$, $EC$, $ED$. **Construction.** Take any line $\alpha \epsilon$ at pleasure, and at its extremitys make the angles $\epsilon \alpha \gamma$, $\epsilon \alpha \beta$, $\alpha \epsilon \gamma$, $\alpha \epsilon \delta$, equal to the correspondent observed angles $EAC$, $EAB$, $AEC$, $AED$. Produce $\beta \alpha$, $\delta \epsilon$, til they meet in $\phi$, join $\phi \gamma$; then upon $CB$ describe (according to 33. 3. Eucl) a Segment of a circle, that may contain an angle $= \gamma \phi \beta$; and upon $CD$ describe a Segment of a circle capable of an angle $= \gamma \phi \delta$; suppose $F$ the common section of these 2 circles; join $FB$, $FC$, $FD$; then from the point $C$, draw forth the lines $CA$, $CE$, so that the angle $FCA$ may be $= \phi \gamma \alpha$, and $FCE = \phi \gamma \epsilon$; to $A$, $E$, the common sections of $CA$, $CE$, with $FB$, $FD$, will be the points required, from whence the rest is easily deduced. **The Calculation.** Assuming $\alpha$ of any number, in the triangles $\alpha \gamma \epsilon$, $\alpha \phi \epsilon$, all the angles being given, with the side $\alpha$ al- sum'd, the sides \( \alpha \gamma, \epsilon \gamma, \alpha \phi, \epsilon \phi \), will be known; then in the triangle \( \gamma \alpha \phi \), the angle \( \gamma \alpha \phi \) with the legs \( \alpha \gamma, \alpha \phi \) being known, the angles \( \alpha \phi \gamma, \alpha \gamma \phi \) with the side \( \phi \gamma \) will be known: then as for the rest of the work in the other figure, the triangle \( BCD \) having all its sides and angles known, and the angles \( BFC, BFD \), being equal to the found \( \beta \phi \gamma, \beta \phi \delta \); how to find \( FB, FC, FD \) by Calculation (and also Protraction) has been shewn by Mr Collins (in Phil. Transact n. 69 p. 2093.) as to all its cases, which may therefore supercede my shewing any other way. But here it must be noted, that if the sum of the observed angles, \( BAE, AED \), is 180 degrees: then \( AB \) and \( ED \) cannot meet, because they are parallel, and consequently the given solution cannot take place; for which reason I here subjoin another. Another Solution. Upon \( BC \) (Fig. 5.) describe a segment \( BAC \) of a circle, so that the angle of the segment, may be equal to the observed \( \angle \beta \alpha \gamma \), (which as above quoted is shewn 33. 3. Euclid) and upon \( CD \) describe a segment \( CED \) of a circle, capable of an angle equal to the observed \( CED \); from \( C \) draw the diameters of these circles \( CG, CH \); then upon \( CG \) describe a segment of a circle \( GFC \), capable of an angle equal to the observed \( \angle AEC \); likewise upon \( CH \) describe a circles segment \( CFH \), capable of an angle equal to the observed \( CAE \): suppose \( F \) the common section of the two last circles \( HFC, GFC \), join \( FH \), cutting the circle \( HEC \) in \( E \), join also \( FG \), cutting the circle \( GAC \) in \( A \): I say that \( A, E \), are the points required. Dem: For the \( \angle BAC \) is \( = \beta \alpha \gamma \) by construction of the segment, also the angles \( CEH, CAG \), are right, because each exists in a semicircle: therefore a circle being described upon \( CF \) as a diameter, will pass thro \( E, A \); Therefore fore the angle \( \angle CAE = \angle CFE = \angle CFH \) (by construction) to the observed angle \( \gamma \). In like manner the \( \angle CEA = \angle CFA = \angle CFG \): observed angle \( \gamma \). If the stations \( A, E \) fall in a right line with the point \( C \), the lines \( GA, HE \) being parallel, cannot meet: but in this case the problem is indeterminate and capable of infinite Solutions. For as before upon \( CG \) describe a Segment of a circle capable of the observed \( \angle \gamma \), and upon \( CH \), describe a Segment capable of the observed \( \gamma \): then thro' \( C \), draw a line any way cutting the circles in \( A, E \), these points will answer the question. The Third Problem. Four points \( B, C, D, F \), (Fig. 6.) or the 4 sides of a quadrilateral, with the angles comprehended are given; also there are 2 stations \( A \) and \( E \) such, that at \( A \), only \( B, C, E \) are visible, and at \( E \) only \( A, D, F \), that is, the angles \( BAC, BAE, AED, DEF \) are given: to find the places of the two points \( A, E \); and consequently, the lengths of the lines \( AB, AC, AE, ED, EF \). Construction. Upon \( BC \) (by 33. 3.Eucl.) describe a segment of a circle, that may contain an angle equal to the observed angle \( BAC \), then from \( C \) draw the chord \( CM \), or a line cutting the circle in \( M \), so that the angle \( BCM \) may be equal to the supplement of the observed angle \( BAE \), i.e. its residue to 180 degrees. In like manner on \( DF \) describe a segment of a circle, capable of an angle equal to the observed \( DEF \), and from \( D \) draw the chord \( DN \), so that the angle \( FDN \) may be equal to the supplement of the observed angle \( AEF \), join \( MN \), cutting the 2 circles in \( A, E \): I say \( A, E \), are the two points required. Dem: Join \( AB, AC, ED, EF \), then is the \( \angle MAB = \angle BCM \) (by 21. 3. Eucl.) = supplement of the observed \( \angle BAE \) by construction, therefore the constructed \( \angle BAE \) \( \angle BAE \) is equal to that which was observed. Also the \( \angle BAC \) of the segment, is by construction of the Segment, equal to the observ'd \( \angle BAC \). In like manner the constructed angles \( AEF \) and \( DEF \), are equal to the correspondent observed angles \( AEF, DEF \), therefore \( A, E \) are the points required. The Calculation. In the Triangle \( BCM \), the \( \angle BCM \) (= supplement of \( BAE \)) and \( \angle BMC \) (= \( BAC \)) are given, with the side \( BC \); thence \( MC \) may be found; in like manner \( DN \) in the \( \triangle DNF \) may be found. But the \( \angle MCD \) (= \( BCD - BCM \)) is known with its legs \( MC, CD \), therefore its base \( MD \), and \( \angle MDC \) may be known. Therefore the \( \angle MDN \) (= \( CDF - CMD - FDN \)) is known, with its legs \( MD, DN \); thence \( MN \) with the angles \( DMN, DNM \), will be known. Then the \( \angle CAM \) (= \( \angle DMC + DMN \)) is known, with the \( \angle MAC \) (= \( MAB + BAC \)) and \( MC \) before found; therefore \( MA \) and \( AC \) will be known. In like manner in the triangle \( EDN \), the angles \( E, N \), with the side \( DN \) being known, the sides \( EN, ED \) will be known; therefore \( AE \) (= \( MN - MA - EN \)) is known. Also in the triangle \( ABC \), the \( \angle A \) with its sides \( BC, CA \) being known, the side \( AB \) will be known, with the \( \angle BCA \); so in the triangle \( EFD \), the \( \angle E \) with the sides, \( ED, DF \) being known, \( EF \) will be found, with the \( \angle EDF \). Lastly in the triangle \( ACD \), the \( \angle ACD \) (= \( BCD - BCA \)) with its legs \( AC, CD \) being known, the side \( AD \) will be known, and in like manner \( EC \) in the triangle \( EDC \). Note that, in this problem, as also in the first and second, if the two stations fall in a right line with either of the given objects: the locus of \( A \), or \( E \), being a circle, the particular point of \( A \), or \( E \), cannot be determined from the things given. As to the other cases of this third problem, wherein \( A \); and \( E \), may shift places, i.e. only \( D, F, E \) may be visible. sible at $A$, and only $A, B, C$ at $E$; or wherein $B, D, E$, may be visible at $A$, and only $C, F, A$, at $E$; or wherein $A$ may be of one side of the quadrilateral, and $E$ on the other; or one of the stations within the quadrilateral, and the other without it: I shall for brevity's sake omit the figures, and diversity of the Signs $\pm$ and $-$ in the calculation, and presume that the Surveyor will easily direct himself in those cases, by what has been said. The Solution of this third problem is general, and serves also for both the precedent. For suppose $C, D$ the same point in the last figure, and it gives the solution of the second problem: but if $B, C$, be suppos'd the same points with $D, F$, by proceeding as in the last, you may directly solve the first problem. --- **A Letter from William Molyneux Esq.; to one of the Secretarys of the R. S. concerning the Circulation of the blood as seen, by the help of a Microscope, in the Lacerta Aquatica.** _Dublin Octob. 27. 1685._ _Sir,_ OUR Society lately received transcripts of two of Dr Gardens Letters, the first dated form Aberdeen July 17. 1685. to Dr Middleton; the other Sept. 4. 1685. to Dr Plot. To both these Letters I have something to say. In the first he gives an Account of the Visible Circulation of the blood in the Water-Newt or Lacerta Aquatica; truely I am heartily glad, that this Learned and Ingenious Dr has hit upon this Experiment; tis now above two years and an half, since I first Discovered this surprising