Any help would be greatly appreciated.

Thanks in advance

Willy

- Thread starter wilster98
- Start date

Any help would be greatly appreciated.

Thanks in advance

Willy

- Joined
- Oct 6, 2005

- Messages
- 10,505

Code:

```
D/2
_______ _______
\ | | /
\ | | /
\ | T | x°/
\ | | /
\ | | / 90°-x°
\| |/
```

Let T be the thickness.

Then angle xº is arctan(D/[2·T]).

Use a calculator set to degree mode.

For example, if the diameters differ by 3/8" and the thickness is 9/16", then

D = 0.3750

T = 0.5625 so 2·T = 1.1250

arctan(0.3750/1.1250) = 18.43°

If you get 0.32175, instead, then the calculator is set to radian mode. You can multiply radians by 57.2958, to convert to degrees.

0.32175 · 57.2958 = 18.43

If you don't have a calculator with arctan, then google: arctan(0.3750/1.1250) in degrees

- Joined
- May 13, 2021

- Messages
- 1

Let D be the difference between the two diameters.Code:`D/2 _______ _______ \ | | / \ | | / \ | T | x°/ \ | | / \ | | / 90°-x° \| |/`

Let T be the thickness.

Then angle xº is arctan(D/[2·T]).

Use a calculator set to degree mode.

For example, if the diameters differ by 3/8" and the thickness is 9/16", then

D = 0.3750

T = 0.5625 so 2·T = 1.1250

arctan(0.3750/1.1250) = 18.43°

If you get 0.32175, instead, then the calculator is set to radian mode. You can multiply radians by 57.2958, to convert to degrees.

0.32175 · 57.2958 = 18.43

If you don't have a calculator with arctan, then google: arctan(0.3750/1.1250) in degrees

is that d/2 before so if your original diameter was .75 then you divided it by to create D so its actually the ARCtanx(Radius/(2xThickness))

- Joined
- Oct 6, 2005

- Messages
- 10,505

Hello PointBlank. No. I'd defined symbol D at the very beginning to represent the… is that [D/2] before so if your original diameter was .75 then you divided it by [2] to create D …

In my diagram, the bottom diameter is represented by the horizontal distance between the two right triangles. The combined length of the two horizontal legs of those triangles represents the difference between the two diameters. Due to symmetry, the two triangles are identical; therefore, the horizontal legs have equal length. In other words, the measure of each horizontal leg may be expressed as half the diameters' difference (D/2).

When you say, "original diameter", you're referring to the bottom diameter. Correct?

If your bottom diameter is 0.75 units, what is your top diameter?

PS: As the tangent of angle 'x' may be expressed as the ratio of sides 'Opposite/Adjacent', I could have also written the arctangent using the same ratio. That is, I could have written the ratio as [D/2]/[T] instead of D/[2T] because those are the same ratio:

arctangent([Opposite]/[Adjacent])

arctangent([D/2]/[T])