## To draw a rectangular hyperbola from part of a cone

Figure 12.4 shows the method of drawing the hyperbola, which is a true view on the line AB drawn parallel to the vertical centre line of the cone.

1 Project point B to the circumference of the base in the plan view, to give the points B1 and B2.

2 Mark points B1 and B2 in the end elevation.

3 Project point A onto the end elevation. Point A lies on the centre line in the plan view.

4 Take any horizontal section XX between A and B and draw a circle of diameter D in the plan view.

5 Project the line of section XX onto the end elevation. B1 I B2

Hyperbola Fig. 12.4

6 Mark the chord-width W in the plan, on the end elevation. These points in the end elevation form part of the hyperbola.

7 Repeat with further horizontal sections between A and B, to complete the hyperbola.

The ellipse, parabola, and hyperbola are also the loci of points which move in fixed ratios from a line (the directrix) and a point (the focus). The ratio is known as the eccentricity.

Eccentricity =

distance from focus perpendicular distance from directrix

The eccentricity for the ellipse is less than one.

The eccentricity for the parabola is one.

The eccentricity for the hyperbola is greater than one.

Figure 12.5 shows an ellipse of eccentricity 3/5, a parabola of eccentricity 1, and a hyperbola of eccentricity 5/3. The distances from the focus are all radial, and the distances from the directrix are perpendicular, as shown by the illustration.

To assist in the construction of the ellipse in Fig. 12.5, the following method may be used to ensure that the two dimensions from the focus and directrix are in the same ratio. Draw triangle PA1 so that side A1 and side P1 are in the ratio of 3 units to 5 units. Extend both sides as shown. From any points B, C, D, etc., draw vertical lines to meet the horizontal at 2, 3, 4, etc.; by similar triangles, vertical lines and their corresponding horizontal lines will be in the same ratio. A similar construction for the hyperbola is shown in Fig. 12.6.

Commence the construction for the ellipse by drawing a line parallel to the directrix at a perpendicular distance of P3 (Fig. 12.6 (a)). Draw radius C3 from point F1 to Hyperbola

Parabola

Hyperbola

Parabola 5 units

1 2 3 4 5 6 Perpendicular distance . from directrix

5 units

1 2 3 4 5 6 Perpendicular distance . from directrix

3 units

1 2 3 4 5 Perpendicular distance from directrix

Fig. 12.6 (a) Ellipse construction (b) Hyperbola construction intersect this line. The point of intersection lies on the ellipse. Similarly, for the hyperbola (Fig. 12.6 (b)) draw a line parallel to the directrix at a perpendicular distance of Q2. Draw radius S2, and the hyperbola passes through the point of intersection. No scale is required for the parabola, as the perpendicular distances and the radii are the same magnitude.

Repeat the procedure in each case to obtain the required curves.

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### Responses

• carlo
How to draw rectangular hyperbola?
9 years ago
• leonida piazza
How to draw a hyperbola in engineering drawing?
9 years ago
• jan-erik
How to draw hyperbola given eccentricity?
9 years ago
• lucas
How to draw hyperbola in engineering graphics?
8 years ago
• pansy grubb
How to draw rectangular hyperbola in drawing?
8 years ago
• doda pinto
How to draw centre lines on an engineering drawing?
8 years ago
• david
How to draw hyperbola as section of a cone?
3 years ago
• maik
When make rectangular hyperbola from cone section?
3 years ago
• eric
How to draw a hyparabola?
3 years ago
• maggie
How to draw a hyperbola in technical drawing?
3 years ago
• Ami McLean
How rectangular hyperbola form from cones?
3 years ago
• Sara Kortig
How to obtain a rectangular hyperbola from a cone?
2 years ago
• rezene
How to draw rectangular hyperbola in engineering graphics?
2 years ago
• costantino
How to construct a rectangular hyperbola in engineering drawing?
2 years ago
• BLAKE
How to generate a hyperbola from a cone?
3 months ago