Shobobo99
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| Joined: 30 Dec 2008 |
| Total Posts: 5754 |
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| 05 Aug 2011 02:21 PM |
| On a wedge part, how can you get the angle of the front surface based on the size of the part? |
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| 05 Aug 2011 02:22 PM |
Surfaces don't have angles...
⅟½⅓¾⅔⅕⅖⅗⅘⅙⅚⅛⅜⅝⅞ |
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Shobobo99
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| Joined: 30 Dec 2008 |
| Total Posts: 5754 |
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| 05 Aug 2011 02:23 PM |
| Okay, what I meant was the angle of the cut off side of the wedge part. |
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Shobobo99
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| Joined: 30 Dec 2008 |
| Total Posts: 5754 |
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| 05 Aug 2011 02:31 PM |
Look into sine, cosine and tangent.
⅟½⅓¾⅔⅕⅖⅗⅘⅙⅚⅛⅜⅝⅞ |
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Shobobo99
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| Joined: 30 Dec 2008 |
| Total Posts: 5754 |
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| 05 Aug 2011 02:35 PM |
| I know but I was hoping someone knew the formula :3 |
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Shobobo99
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| Joined: 30 Dec 2008 |
| Total Posts: 5754 |
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| 05 Aug 2011 02:56 PM |
... (a*a)+(b*b) = (c*c)
That's the pythagorean theorem which you learn in geometry.
Find the square root of c*c, and that's the lenth of the hypotnuse. |
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Shobobo99
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| Joined: 30 Dec 2008 |
| Total Posts: 5754 |
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| 05 Aug 2011 02:59 PM |
| I am not asking about the length of the hypotnuse of the WedgePart, I am asking how to calculate the angle of hypotnuse to the world axis. |
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Shobobo99
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| Joined: 30 Dec 2008 |
| Total Posts: 5754 |
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| 05 Aug 2011 03:26 PM |
welp. I figured it out.
math.acos(s.Z/(s.Z^2+s.Y^2)^0.5)
s being the size of the part. |
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Shobobo99
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| Joined: 30 Dec 2008 |
| Total Posts: 5754 |
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| 05 Aug 2011 03:45 PM |
| Wait.. nvm that only works if the angle is 45 degrees. HELP. ._. |
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