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Question for the math gurus

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Old August 8th, 2016, 07:07 PM
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Question for the math gurus

I have a bore that has a 1.920 diameter. This bore has a shaft that is .210 running across it in the middle. What % of the bore is blocked by the shaft?

Is it possible to flow this bore with and without the shaft installed to see what the restriction figure is? Not sure how a flow bench works for heads but could that be adapted to a single bore like this? If it matters the bore is about 3 inches deep. I know that will not matter for the math figures.
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Old August 8th, 2016, 07:48 PM
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its here...ya just gotta make it work....i didnt have time to look..but theres just about anything ya want here

http://www.wallaceracing.com/Calculators.htm

Last edited by marxjunk; August 8th, 2016 at 08:34 PM.
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Old August 8th, 2016, 08:18 PM
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Area of bore will be (pi/4 x 1.92^2). The shaft will appear to be a rectangle looking down the bore, so that's (0.210 x 1.92). I calculate roughly 14% area reduction.

You could flow test it if it was critical. There should be some formulas out there to calculate drag from a cylinder in cross flow, just not sure when I'll have some time to dig them out.
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Old August 8th, 2016, 09:02 PM
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I get a bore of 2.895 sq inches, and a shaft, assuming circular, of .0346 sq inches, which is a reduction of some 1.2%
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Old August 9th, 2016, 04:16 AM
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Originally Posted by Koda
I get a bore of 2.895 sq inches, and a shaft, assuming circular, of .0346 sq inches, which is a reduction of some 1.2%
The shaft is running 90 degrees to the bore across the middle. Like a throttle butterfly shaft in a throttle plate bore.
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Old August 9th, 2016, 07:06 AM
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Originally Posted by jensenracing77
The shaft is running 90 degrees to the bore across the middle. Like a throttle butterfly shaft in a throttle plate bore.
Ok, I thought it was coaxial like a metering rod. I understood across the middle to be a description of the diameter of the shaft.

Bore is still 2.895 sq inches. Luke is correct on the figuring on the cross sectional area of the shaft, which is 0.4032, and his % reduction is also correct.
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