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diagram, is in a in the VP, and a 1 in the HP, and therefore they are MECHANICAL AND ENGINEERING DRAWING 37 the required projections, a being an elevation or vertical projection of A, and a its plan or horizontal projection. If it were required to find from its projections the position of the original point A with respect to the YP and HP, then perpendiculars to those planes let fall from its projections a and a would intersect in A, giving it as the position of the original point. Knowing how to obtain the projections of a point, we shall now be able to find the projections of a straight line. 1st. Let the line AB (Fig. 70) be perpendicular to the VP. Here AB being perpendicular to the YP, will be parallel to the HP; therefore, from its position with respect to the YP, its projec- tion on that plane will become a point a, as the eye being directly opposite the end of it, the visual ray or projector proceeding from the eye will travel along the line itself, coinciding with it, and penetrate the YP in a, then a is the "elevation" of the line AB. To find its " plan " or projection on the HP, let fall projectors perpendicular to the HP from both ends of AB, and the points a b,\ where these pro- jectors penetrate the HP, will be projections of the ends A and B of the line AB, and if a b be joined, then ab will be the plan or hori- zontal projection of the original line. 2nd. Let CD (Fig. 70) be the given line, and let it be perpendicular to the HP, and its projections required. In this case CD is parallel to the YP, and its projection on that plane will be obtained by letting fall from C, D its ends, projectors to the YP, and the points c d, where these fall on that plane, will be the