The elevation and plan of a cone are shown in Fig. 4.3. Draw (i) the given elevation (ii) the plan (iii) the first auxiliary elevation on plane X'-Y'.

This problem involves drawing orthographic projections and an auxiliary view of a cone. Given Dimensions: Cone height (H) = 50 mm Cone base diameter (D) = 50 mm Auxiliary plane X'-Y' is inclined at 45^ to the horizontal. --- Part (i): Draw full size the given elevation The given elevation is a front view of the cone. Since the cone's axis is vertical, its elevation is an isosceles triangle. Step 1: Draw a horizontal line AB of length 50 mm. This represents the base of the cone. Step 2: Locate the midpoint M of AB. Step 3: From M, draw a vertical line upwards, MV, of length 50 mm. Point V is the apex of the cone. Step 4: Connect V to A and V to B. This completes the triangular elevation. --- Part (ii): Draw full size the plan The plan view is a top view of the cone. Since the cone's axis is vertical, its plan is a circle. Step 1: Below the elevation, draw a horizontal center line. Step 2: Draw a vertical center line, aligning it with the apex V and the midpoint M of the base in the elevation. The intersection of these center lines is the center of the cone's base in the plan view. Step 3: Draw a circle with a diameter of 50 mm (radius 25 mm) centered at the intersection of the center lines. Step 4: Divide the circumference of the circle into 12 equal parts (each 30^). Label these points (e.g., 0 to 11). These points represent the bases of the generator lines of the cone. Step 5: Project these 12 points vertically upwards to the base AB of the elevation. These projected points on AB correspond to the bases of the generator lines in the elevation. --- Part (iii): Draw full size the first auxiliary elevation on plane X'-Y' The diagram shows the X'-Y' line in the elevation view, inclined at 45^ to the horizontal. This indicates that the auxiliary plane is inclined at 45^ to the horizontal plane. For an auxiliary elevation projected from the original elevation, the view is taken perpendicular to X'-Y', and the depths are transferred from the plan view. Step 1: Draw the reference line X'-Y': Draw a line X'-Y' at an angle of 45^ to the horizontal. Position it conveniently, for example, passing through the lowest point of the base of the cone in the elevation (e.g., point A). Step 2: Project points from the elevation: Draw projection lines perpendicular to X'-Y' from the apex V and from each of the 12 points on the base AB of the cone in the elevation. Step 3: Transfer depths from the plan: The distances from X'-Y' in the auxiliary view represent the depths* of the points, which are measured from the center line of the plan view. For the apex V: The apex is at the center of the plan, so its depth is 0. Therefore, the apex in the auxiliary elevation will lie on the X'-Y' line. For the base points (0 to 11): Measure the perpendicular distance of each point from the center line of the plan view. Let R be the radius (25 mm). Points on the center line (e.g., 0 and 6): Depth = 0. These points will lie on the X'-Y' line. Points at 30^ from center line (e.g., 1 and 11, 5 and 7): Depth = R (30^) = 25 × 0.5 = 12.5 mm. Points at 60^ from center line (e.g., 2 and 10, 4 and 8): Depth = R (60^) = 25 × 0.866 = 21.65 mm. Points at 90^ from center line (e.g., 3 and 9): Depth = R (90^) = 25 × 1 = 25 mm. Plot these depths along the corresponding projection lines from X'-Y'. For example, for point 3, measure 25 mm perpendicular to X'-Y' along its projection line. Step 4: Complete the auxiliary elevation: Connect the plotted points for the base with a smooth curve to form an ellipse. This is the elliptical base of the cone in the auxiliary view. Connect the apex (which lies on X'-Y') to the extreme points of the elliptical base to complete the auxiliary elevation of the cone. The final drawing will show: An isosceles triangle for the elevation. A circle for the plan, divided into 12 segments. An auxiliary view showing the cone as if tilted, with an elliptical base and the apex on the X'-Y' line, projected perpendicular to the 45^ inclined X'-Y' line. That's 2 down. 3 left today — send the next one.