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 Surface Area of an Archimedes Screw Post Reply Engineering Forum
 Posted by: cascon ® 06/27/2007, 23:16:11 Author Profile eMail author Edit I need help in finding the surface area of a Archimedes screw.?I need to calculate the material requirements for applying a protective coating to this screw conveyor. The length is 38'6", width of the conveyor is 3'6" and the torque tube is 2' in diameter. There are 20 flights in this conveyor from end to end with 2' of of tube on the top end to facilitate the turning motor. I need to separate the area for the flights and tube as they will get different materials. If anybody can help me I would appreciate it very much. I just do not know what kind of formula to use. I do not know the pitch of the blade, but this screw was engineered for a waist water treatment plant, so I think it would have been engineered for optimal efficiency.

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 : Surface Area of an Archimedes Screw : Surface Area of an Archimedes Screw -- cascon Post Reply Top of thread Engineering Forum
 Posted by: haruspex ® 12/08/2009, 17:18:10 Author Profile eMail author Edit If the helix is mounted on a shaft of length L, zero radius, and has outer radius B then the surface area is [B*sqrt(B*B+L*L)/2 - L*L*arcsinh(B/L)] If the shaft radius is A then just subtract out the same formula replacing B with A. This is obtained by considering the length of the curve at the outer radius. This is simply, by Pythagoras sqrt(B*B+L*L) To get the area, replace B by x and integrate wrt x from A to B. To assist with the integration, substitute x = L*sinh(phi)

 Re: Surface Area of an Archimedes Screw : Surface Area of an Archimedes Screw -- cascon Post Reply Top of thread Engineering Forum
 Posted by: jboggs ® 06/28/2007, 19:32:00 Author Profile eMail author Edit (1) No matter what method you eventually use, the screw pitch will be part of the calculation so it must be either determined or assumed. (2) If I were doing it, I would recognize the fact that the surface area of a series of discs of the screw diameter and material thickness spaced about a pitch length apart plus the surface area of the central longitudinal tube, would be a very close approximation. No matter what your answer is you're going to order more coating than you calculate just for safety and to account for application variations, right? So a close approximation is good enough. (And a LOT simpler than precisely calculating the area of a helical surface!) No matter what you calculate, the actual usage will be different - guaranteed!

 Re: Re: Surface Area of an Archimedes Screw : Re: Surface Area of an Archimedes Screw -- jboggs Post Reply Top of thread Engineering Forum
 Posted by: cascon ® 06/29/2007, 09:23:54 Author Profile eMail author Edit Thanks for your reply. I pretty much came to the same conclusion yesterday after talking with several people concerning this problem. I knew the math was way over my head and I did not have a chance in hell to figure this problem out on my own with any precision, but I had a mechanical engineer scratching his head over this one and that surprised me. Now its just one of those things I'd like to know because I dig math but calculus is about 20 years in my past. I guess now days the trick is to dump the variables into a CAD program and hit the tools area button and let the computer figure it out. But your right on this one, "a close approximation is good enough".Once again, thanks for your help.