# Building a half-arch of a specific height, while consistently using an angle divisible by 90

I am trying to construct a lift handle in the CAD Sketchup. The circled part of the image shows the arch I'm referring to.

The program, Sketchup, has a 'tool' called push/pull, and another tool called rotate. I have a video uploaded on streamable demonstrating this, but I'm not sure if it's within the site's rule to do so. Here's a brief clip partially demonstrating it in action: https://i.imgur.com/pgVRw9d.mp4

I use the push/pull tool, to raise the face of the object by 250 mm, and the rotate tool, to rotate it by 15 degrees. $$90/15 = 6$$. The arch ends up with an overall height of 1074 mm.

I repeat the task 2 more times, but with the angle of 7.5 and 30 degrees. With 7.5 degrees, the arch looks much curvier, but the height is 2032 mm. With 30 degrees, the arc sides are noticeably pointy, but its height is more acceptable, 591 mm.

Here is an image of all 3 arcs together:

The dimension of each side on the inner part of the arc, are all 250 mm.

My question is, what equation may I use, so I end up with an arch that looks curvy enough, in this case 15 degrees or lower, and has a height of 508 mm from the 'ground' (or x-axis)?

I tried the following, with no success:

Circumference of a circle $$C=2\pi r$$

The arc in this case represents 1/4th of a circle; so $$C_{arc}=2\pi r/4$$

\begin{align} 2\pi508/4 &= 797.96 \\ \dfrac{797.96}{90/15} &= 132.99 \end{align}

I push/pull the face of the object by 132.99, and rotate its top face by 15 degrees. I repeat this process 5 more times. The 6th arch rotated, is perfectly perpendicular to the x-axis. Unfortunately, the height is 571 mm, instead of 508 mm.

• If you're not aware, you can define a path and a shape, and use the "follow me" tool to extrude the shape along the path. That may be a more direct way to get what you want. A question though: I'm not clear where you're measuring, but is it possible you're calculating the centerline of the arch, but measuring to the top? – Mark Dec 17 '18 at 17:17
• I asked on the forum, and it's not feasible/practical to use the follow me tool, as the handle gets narrower as it gets closer to the base. Clearer picture of the height: i.imgur.com/v2fRxIV.jpg – gimmegimme Dec 17 '18 at 20:56
• I just put an answer below. However, it's not clear to me how you're tapering the handle using your approach either, so I don't understand why follow me doesn't work. – Mark Dec 17 '18 at 23:46
• Thank you for the answer, I plan on expanding or narrowing each pulled segment using the scale tool, as this clip briefly shows: i.imgur.com/wLndPvP.mp4 – gimmegimme Dec 18 '18 at 0:07
• In following image, I use the scale tool on the sides after drawing all 6 segments: i.imgur.com/KgZKM8M.jpg – gimmegimme Dec 18 '18 at 0:13

It looks like you need a = 118.1978.

I came up with this using the formulas on this site. A section through your handle looks like part of a regular polygon with 12, 24, or 48 sides.

You can use the calculator on that site, but you have to make a minor adjustment because your first segment is perpendicular to the surface (so if you placed a mirror image below it you would have two adjacent sides with no angle in between). You can account for this by rotating their illustration by 1/2 the angle of each segment, and using r instead of R in their computation, and then adding half the side length to the answer. In other words, their calculated r + a/2 = height.

Using this approach, I perfectly reproduced all four of your results.

To calculate a, given the desired height is a little more complicated because of the modification I described above. In general, you can do it by solving two equations simultaneously: a = 2 r tan(pi/n) and r + a/2 = height.

Plugging in your desired height = 508, and n = 24, I get the answer a = 118.1978.

EDIT: Here's a sketch that explains why r + a/2 = 508 mm:

• I'm confused about the part where you say 'their calculated r + a/2 = height'. i.imgur.com/DjPz0uy.jpg, i.imgur.com/DEPFIhU.jpg – gimmegimme Dec 18 '18 at 2:15
• What did you substitute for r in the equation: a = 2 r tan(pi/n) and r + a/2 = height? It seems r + a/2 = height, is to be re-written as 2(508 - r) = a – gimmegimme Dec 18 '18 at 3:14
• Yes, you can use a = 2(508 - r). Then put that into the first equation as 2(508 - r) = 2 r tan(pi/n) and solve for r. You should get the same answer, though I did it the other way so I ended up solving directly for a instead of r. – Mark Dec 18 '18 at 5:46
• I see what you're doing to get the taper - that makes sense. – Mark Dec 18 '18 at 5:47
• I solved for r in the problem: 2(508-r) = 2rtan(pi/24), and got 506.84205110150227816067041891348; I substituted this number for r, on both sides, and got 2.31589...[only 1st 5 numbers after decimal included. I don't know what to do with the 506.84 – gimmegimme Dec 19 '18 at 1:42