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Topic: How to calculate air resistance of an octagonal loop?  (Read 4370 times)

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Offline Boxxxed

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How to calculate air resistance of an octagonal loop?
« on: February 17, 2015, 04:00:13 PM »

Say I have an octagon that is constructed of some tubes. The diameter of this octagon is 26 meters (perpendicular from one parallel side to the other). The radius of these tubes is about 2.5 inches and the loop is "slicing" into the wind, like throwing a Frisbee but with no spinning motion.

The wind speed is 80 km/h or 22m/s

How do I calculate the drag force?

I did a messy calculation and got 621 lbs. I don't have a background in physics so any input would be appreciated.

Offline mjc123

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Re: How to calculate air resistance of an octagonal loop?
« Reply #1 on: February 18, 2015, 04:48:02 AM »
If you're mixing meters, inches, km/h, m/s and lbs, I'm not surprised the calculation is messy! Try using consistent units.

Offline Enthalpy

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Re: How to calculate air resistance of an octagonal loop?
« Reply #2 on: February 18, 2015, 06:27:43 AM »
Since the tube is thin, I consider the octagon never stands perfectly flat in the wind, so the rear side gets the full wind force as well.

For the inclined sides, I take the same drag (=component force parallel to the wind) as a cylinder of the same area as seen from the wind's direction. That overestimates the drag a bit.

Then the drag is the same as a cylinder of 2*26m length and 63.5mm diameter. At Re~105 the drag coefficient is just above 1, so the drag is 1000N.

Drag predictions are rather inaccurate, so take margins for real. In addition, objects often vibrate in the wind, and cylinders more so, which amplifies the mechanical loads.

Offline Boxxxed

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Re: How to calculate air resistance of an octagonal loop?
« Reply #3 on: February 18, 2015, 10:37:04 AM »

Then the drag is the same as a cylinder of 2*26m length and 63.5mm diameter. At Re~105 the drag coefficient is just above 1, so the drag is 1000N.

There are 2 sides perpendicular to the wind and so have the full effect of wind resistance. There are two sides parallel to the wind direction and so produce little drag and there are 4 sides at 45 degrees to the wind and I would assume produce half the drag. In other words those 4 tubes produce the same drag as 2 pipes perpendicular to the wind.

So I end up with the equivalent of about 43m of tube with a full drag effect. Your estimate at 2*26 is good since the loop is never flat.

Also slight correction, 63.5 mm is the radius not diameter. However, I made a spreadsheet to calculate air resistance quickly while modifying loop dimensions and I still get 1381 N for 63.5mm as the radius. Changing the radius to 2.5 inches gives me 2700 N which is the original 621 lbs I had.

Thanks for the input

https://drive.google.com/file/d/0B_LRZe8FgK92VEE0OGFBckRkckk/view?usp=sharing






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