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Required a circle in which a polygon of given sides can be inscribed.
This problem gives an equation which is proportionate in degree to the number of sides of the polygon. To solve it by the method just expounded we describe any circle ABCD (Fig. 3) and lay off in this circle the given sides AB, BC, CD, DE, EF of the polygon, which for the sake of simplicity I here suppose to be pentagonal. If the extremity of the last side falls on A, the problem is solved. But since it is very improbable that this should happen at the first trial we lay off on the straight line PR
(Fig. 4) the radius PA of the circle, and erect on it at the point A the perpendicular AF equal to the chord AF of the arc AF which represents the error in the supposition made regarding the length of the radius PA. Since this error is an excess, it will be necessary to describe a circle having a larger radius and to
Fig. 4.
perform the same operation as before, and so on, trying circles of various sizes. Thus, the circle having the radius PA gives the error F jAj which, since it falls on the hither side of the point Aj,
should be accounted negative. It will consequently be necessary
in Fig. 4 in applying the ordinate AjF j to the abscissa PAj to draw that ordinate below the axis. In this manner we shall
obtain several points F , F j, . . . , which will lie on a curve of which the intersection R with the axis PA will give the true
radius PR of the circle satisfying the problem, and we shall find this intersection by successively causing the points of the curve
lying on the two sides of the axis as F , F j, . . . to approach nearer and nearer to one another.
From a point, the position of which is unknown, three objects are observed, the distances of which from one another are known.
The three angles formed by the rays of light from these three objects to the eye of the observer are also known. Required the position of the observer with respect to the three objects.
If the three objects be joined by three straight lines, it is plain that these three lines will form with the visual rays from the eye of the observer a triangular pyramid of which the base and the three face angles forming the solid angle at the vertex are given. And since the observer is supposed to be stationed at the vertex, the question is accordingly reduced to determining the dimensions of this pyramid.
Since the position of a point in space is completely deter- mined by its three distances from three given points, it is clear that the problem will be resolved, if the distances of the point at which the observer is stationed from each of the three ob- jects can be determined. Taking these three distances as the unknown quantities we shall have three equations of the second degree, which after elimination will give a resultant equation of the eighth degree; but taking only one of these distances and the relations of the two others to it for the unknown quantities, the final equation will be only of the fourth degree. We can ac- cordingly rigorously solve this problem by the known methods; but the direct solution, which is complicated and inconvenient in practice, may be replaced by the following which is reached by the curve of errors.
Let the three successive angles APB, BPC, CPD (Fig. 5) be constructed, having the vertex P and respectively equal to the angles observed between the first object and the second, the second and the third, the third and the first; and let the straight line PA be taken at random to represent the distance from the observer to the first object. Since the distance of that object to
Fig. 5.
the second is supposed to be known, let it be denoted by AB, and let it be laid off on the line AB. We shall in this way obtain the distance BP of the second object to the observer. In like manner, let BC, the distance of the second object to the third, be laid off on BC, and we shall have the distance PC of that object to the observer. If, now, the distance of the third object to the first be laid off on the line CD, we shall obtain PD as the distance of the first object to the observer. Consequently, if the distance first assumed is exact, the two lines PA and PD will necessarily coincide. Making, therefore, on the line PA, prolonged if neces- sary, the segment PE = PD, if the point E does not fall upon the point A, the difference will be the error of the first assump- tion PA. Having drawn the straight line PR (Fig. 6) we lay off upon it from the fixed point P , the abscissa PA, and apply to it at right angles the ordinate EA; we shall have the point E of the curve of errors ERS. Taking other distances for PA, and making the same construction, we shall obtain other errors which can be similarly applied to the line PR, and which will give other points in the same curve.
Fig. 6.
We can thus trace this curve through several points, and the point R where it cuts the axis PR will give the distance PR, of which the error is zero, and which will consequently represent the exact distance of the observer from the first object. This distance being known, the others may be obtained by the same construction.
It is well to remark that the construction we have been considering gives for each point A of the line PA, two points
B and Bj of the line PB; for, since the distance AB is given, to find the point B it is only necessary to describe from the
point A as centre and with radius AB an arc of a circle cut- ting the straight line PB at the two points B and Bj,—both of which points satisfy the conditions of the problem. In the same
manner, each of these last-mentioned points will give two more upon the straight line PC, and each of the last will give two more on the straight line PD. Whence it follows that every point A taken upon the straight line PA will in general give eight upon the straight line PD, all of which must be separately and succes- sively considered to obtain all the possible solutions. I have said,
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