I need to create a template for a cone that has a 2" opening at the top, a
4" base and stands 6" tall. Can you tell me how to achieve this?

Sonny

 

Hi Sonny,

I recently answered a very similar question from Wendy. Her cone had a base opening of 4" and a top opening of 2" but it was 12" tall where your truncated cone has a height of 6". I am going to refer to the diagrams I drew for her to answer your question.

For your situation |BC| = 6" so |AC| = 12 " and

|AD|2 = |AC|2 + |CD|2 = 144 + 16 = 160

and hence |AD| = √160 = 12.65 inches.

Now slice the cone from bottom to top and roll it out flat. What you get is a sector of a circle of radius 12.65 inches. The length of the arc is the circumference of the base circle of the cone. This is a circle of radius 4 inches so the length of the arc is

arc length = circle circumference = 2 radius = 2 4 = 25.13 inches.

The relationship among the length of an arc, the radius of the circle and the angle at the centre of the circle is

arc length = radius angle

where the angle is measured in radians. Thus

25.13 = 12.65 angle

and hence the angle is 1.99 radians. There are radians in 180o and thus

1.99 radians = 180/ 1.99 = 114.020

Thus you need to draw a circle of radius 12.65 inches. Cut out a sector with a centre angle of 114.020 . Cut out a sector of radius 12.65/2 = 6.33 inches inches and roll up what's left to form your truncated cone.

Penny