Pipe bend circumference

[cjcurry]

Right maths wizards, I have a problem for you. I need to bend a pipe. It needs two radius bends following on from one another, one to make an angle, the other to return it to a path parallel to its original course. That is, it starts traveling vertically, bends, then returns to vertical. The constraints are the height in which this happens (800mm) and the offset difference between the vertical paths (360mm). So for each bend the height is 400mm and the width is 180mm. So one can calculate the chord of the each required bend fairly easily as the hypotenuse of that right angled triangle, 438.63mm. But I'm stumped from there. You could, of course, do any old bend and have a straight angled run in-between, but I need the smoothest bend possible for the fluid. What I'm after is the length of the arc, ie the length of pipe between points A, B & C. The diagram below should make it clearer!

Thanks!

 

[blamocur]

Since this looks like a practical question, as opposed to homework, here goes:
[math]R = \frac{d^2+h^2}{2d}[/math][math]L = 2r \arcsin \frac{h}{R}[/math]
In your case [imath]d=180[/imath] and [imath]h=400[/imath], so [imath]R=534[/imath] and [imath]L=904[/imath]

While I never make any mistakes ( ), I would still double check this with a piece of rope or something.

 

[blamocur]

Since this looks like a practical question, as opposed to homework, here goes:
[math]R = \frac{d^2+h^2}{2d}[/math][math]L = 2r \arcsin \frac{h}{R}[/math]
In your case [imath]d=180[/imath] and [imath]h=400[/imath], so [imath]R=534[/imath] and [imath]L=904[/imath]

While I never make any mistakes ( ), I would still double check this with a piece of rope or something.

I/ve noticed one case of using [imath]r[/imath] instead of [imath]R[/imath] but missed another :
[math]L = 2R \arcsin \frac{h}{R}[/math]

 

[cjcurry]

Fantastic, thanks blamocur!