# Find the projection ?

In the drawings, the image of the solids are constructed using the method of projection.But for that one picture is not enough, you need at least two projections.With them and determined point in space.Therefore, it is necessary to know how to find the projection point.

## projection point

To do this, you need to consider the space of dihedral angle located within a point (A).It uses horizontal and vertical P1 P2 plane of projection.Point (a) is projected onto the projection plane orthogonally.As for projecting perpendicular rays, they are united in the projected plane, perpendicular to the plane of projection.Thus, when combining the horizontal and frontal P1 P2 by rotation planes P2 / P1 axis receive planar drawing.

then perpendicular to the axis is shown with a line disposed thereon points of view.It turns a complex drawing.Thanks to draw a line on it and a vertical link, you can easily determine the position of a point relative to the projection planes.

To make it easier to understand how to find the projection, it is necessary to consider the right-angled triangle.His short side is the leg, and the long - the hypotenuse.If you run on the hypotenuse leg projection, it is divided into two segments.To determine their size, it is necessary to perform calculation of the initial data set.Consider in this triangle, the methods of calculating the basic projection.

Typically, this problem indicate the length of the hypotenuse leg N and D the length of whose projection and want to find.To do this, we learn how to find the projection of the leg.

consider finding a way to leg length (A).Given that the geometric mean of the projection and the leg length of the hypotenuse is equal to the value of the desired contact leg: N = √ (D * Nd).

## How to find the length of the projection

root of the product can be found by squaring the value of the desired length of leg (N), and then divided by the length of the hypotenuse: Nd = (N / √ D) ² = N² / D. When referring to the source datavalues only the legs D and N, the length of the projection should be found using the Pythagorean theorem.

We find the length of the hypotenuse D. To do this, use the values of the legs √ (N² + T²), and then substitute the value obtained in the following formula of finding the projection: Nd = N² / √ (N² + T²).

When the source data specified data about the length of the projection leg RD, as well as data on the magnitude of the hypotenuse D, should calculate the length of the second leg ND projection using a simple subtraction formula: ND = D - RD.

## projection

speed consider how to find the projection speed.In order to represent a given vector description of the motion, it should be placed in the projection on the coordinate axes.There are one-coordinate axis (beam), two axes (plane) and three axes (space).In finding the necessary projection of the ends of the vector perpendicular to the axis of the drop.

In order to understand the values of the projection, you must learn how to find the vector projection.

## projection vector

When a body moves perpendicular to the axis, the projection will be provided in the form of points, and have a value of zero.If the movement is carried out parallel to the coordinate axis, the projection will coincide with the vector unit.In the case when the body moves such that the speed vector directed at an angle φ relative to the axis (x), the projection onto this axis will be a segment of: V (x) = V • cos (φ), where V - a model of the velocity vector.when the direction of the velocity vector and the same coordinate axis, the projection is positive, and vice versa.Take

following coordinate equation: x = x (t), y = y (t), z = z (t).In this case, the function speed to be projected on the three axes and will have the following form: V (x) = dx / dt = x '(t), V (y) = dy / dt = y' (t), V (z)= dz / dt = z '(t).It follows that it is necessary to take for finding the speed derivative.The very same velocity vector expressed by the equation of this form: V = V (x) • i + V (y) • j + V (z) • k.Here, i, j, k are the unit vectors of the coordinate axes x, y, z respectively.Thus, the modulus of the velocity is calculated as follows: V = √ (V (x) ^ 2 + V (y) ^ 2 + V (z) ^ 2).