How to make an Oloid

oloid made from marble, 7cm

After making several Oloids in the way shown at the previous page, I was asked to describe how to make an Oloid, and I recognized the way I did it was far from perfect.
So I studied some more and found an other way to get the shape.

The oloid is a shape formed by two perpendicular circles with the middle of the one at the edge of the other. The surface is the connection between the circles when rolling the oloid.
The tangents can be found with the formula cos(alpha)*cos(beta)=cos(alpha)+cos(beta)

drawing

The procedure I follow is based on the fact that the cuttingpoint of two tangents to a circle have a distance of tan((alpha - beta) / 2 ) to the points on the circle.

The making is an process where each step produces new tangents. I calculated the distance from the tangent to the lines that are needed to produce the next tangent.
All measurements are made from the points that lay on the circles

Dependent on the material and size of the oloid you may need more steps. The last steps are sanding and polishing.
drawing

The first tangents are at 90 degree at each circle. Get a square block with dimensions length width about 1 : 2.2 After removing the corners it looks like this. The edge has an angle of 60 degrees.
The Oloid has a proportion of 1 : 1.5
As the diagonal of the block is twice radius of the circle, the block has a proportion of 1 : 1.5 sqrt(2)
The first lines to draw are the centers and ends of the circles. The distance between the lines is sqr(2)/2 the width of the block.
The line is drawn from the edge to sqrt(3)r, that is from the 90 degree tangent to tan(15 degree) =~ 0.27r
Draw the next faces and cut them
On the outer circle the distance from the lines to the 180 and 90 degree tangent is r*(1-sqrt(2)) =~ 0,41r
From the above formula can be found that the tangent is at 65.5 deg. at the other circle, so the outer line is at tan(12.25)r =~ 0.2 r from the middle of the circle.
The inner line is at tan(2.75)r =~ 0.05 r above the 60 degree line.

Next step
On the outer circle the new lines are at tan(11.25)r =~ 0.20 r from the previous tangents. The inner faces go up to approximaly the same point, as the tangent touches at 61.3 degrees. The outer tangent goes to 74 deg. so the most outer line goes to tan(8) r =~ 0.14 r from the middle of the circle, and the inner line of this face to tan(4.25)r =~ 0.07 r from the 65,5 degree line

The applet I wrote shows the steps in building the Oloid. On the JAVA console of your browser you can see the values of each buiding step.
JAVA 3D examples of the building process

If you have any questions or remarks, please mail me. Arie Brederode

The first tangents are at 90 degree at each circle. Get a square block with dimensions length width about 1 : 2.2	After removing the corners it looks like this. The edge has an angle of 60 degrees.
		The Oloid has a proportion of 1 : 1.5 As the diagonal of the block is twice radius of the circle, the block has a proportion of 1 : 1.5 sqrt(2) The first lines to draw are the centers and ends of the circles. The distance between the lines is sqr(2)/2 the width of the block. The line is drawn from the edge to sqrt(3)r, that is from the 90 degree tangent to tan(15 degree) =~ 0.27r
Draw the next faces and cut them
		On the outer circle the distance from the lines to the 180 and 90 degree tangent is r*(1-sqrt(2)) =~ 0,41r From the above formula can be found that the tangent is at 65.5 deg. at the other circle, so the outer line is at tan(12.25)r =~ 0.2 r from the middle of the circle. The inner line is at tan(2.75)r =~ 0.05 r above the 60 degree line.
Next step
		On the outer circle the new lines are at tan(11.25)r =~ 0.20 r from the previous tangents. The inner faces go up to approximaly the same point, as the tangent touches at 61.3 degrees. The outer tangent goes to 74 deg. so the most outer line goes to tan(8) r =~ 0.14 r from the middle of the circle, and the inner line of this face to tan(4.25)r =~ 0.07 r from the 65,5 degree line