# Thread: Horizontal angle of rotated vertical bend

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## Horizontal angle of rotated vertical bend

Does anyone have a routine to calculate the horizontal angle of a bend (11 1/4, 22 1/2, or 45) given the required vertical angle (or vice-versa)?
I've spent a lot of time trying to figure it out on my own but my math skills just aren't strong enough.

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## Re: Horizontal angle of rotated vertical bend

I must admit, I don't quite understand what angles are you mentioning. Can you attach sketch of your question?

M.R.

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## Re: Horizontal angle of rotated vertical bend

Assuming the pipe in the attachment is a vertical bend (i.e. water main, gas line), if you rotate it along the axis formed be the horizontal section, the section moving diagonally up will flatten and you will create a horizontal bend. In those cases where it's advantageous to achieve a non-uniform vertical bend (something other than 11-1/4, 22-1/2, or 45 degrees), you could rotate the bend down but you would need to know what the resulting angle would be in plan view.

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## Re: Horizontal angle of rotated vertical bend

If I understood correctly, you don't know how to obtain angle of projection on plan view of bend that was rotated along horizontal axis down by alpha degree starting from vertical position of 90 degree... If you look from side where axis of rotation is viewed in special position as point, you will see bend as line that when rotates makes circle... If you rotate from vertical position by alpha, your horizontal projection would be L*cos(alpha), vertical L*sin(alpha), where L is projection length of bend in our special projection view, and if bend is bended at 45 degree as shown in attached file, after some calculations, I've found :

Code:
```(defun c:hpa-ra () (c:hprojang-rotang) )
(defun c:hprojang-rotang ( / alpha gamma gammar)
(setq alpha (getangle "\nInput angle in decimal degrees for rotation of bend along main pipe axis : "))
(setq gammar (- PI (atan (cos (- (/ PI 2) alpha)))))
(setq gamma (* (/ 180 PI) gammar))
(princ "Horizontal projection angle is : ")(princ gamma)(princ " degrees")
(princ)
)
(princ "\nType \"hpa-ra\" for shortcut")
(princ)```
This only works for bend of 45 degrees...

This works for all curvatures of bend pipes :
Code:
```(defun c:hpa-ra () (c:hprojang-rotang) )
(defun c:hprojang-rotang ( / alpha beta v-h gamma gammar)
(setq beta (getangle "\nInput angle in decimal degrees for curvature of bend of pipes : "))
(setq v-h (/ (sin beta) (cos beta)))
(setq alpha (getangle "\nInput angle in decimal degrees for rotation of bend along main pipe axis : "))
(setq gammar (- PI (atan (* v-h (cos (- (/ PI 2) alpha))))))
(setq gamma (* (/ 180 PI) gammar))
(princ "Horizontal projection angle is : ")(princ gamma)(princ " degrees")
(princ)
)
(princ "\nType \"hpa-ra\" for shortcut")
(princ)```
M.R.

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## Re: Horizontal angle of rotated vertical bend

Well, I think you're on the right track but I need something that will work for 11-1/4 and 22-1/2 degree elbows as well. Also, the results, while correct, are confusing.
Thanks for the effort!

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## Re: Horizontal angle of rotated vertical bend

I did it for all elbow angles...

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## Re: Horizontal angle of rotated vertical bend

Oh. Sorry...
Thank you very much. You've saved me a lot of time. Not from trying to figure this out; I couldn't. You've saved me from having to manually create a 3D part, manually rotating it, and measuring the new horizontal angle. All through trial and error.

Thanks again!

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