Airfoil geometry

Thread Starter

Hello

Joined Dec 18, 2008
82
Hi,


I've constantly tried to come to a solution to the problem attached but haven't managed to get the answer.


Any help whatsoever would be very much appreciated,


Regards
 

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Kermit2

Joined Feb 5, 2010
4,159
Don't make it hard. The radius of the circle on the outer edge of your arc goes from 1 to 0 as the cosine of the function of the angle of the arc.

Rcos30=r simple pimple.
 

Thread Starter

Hello

Joined Dec 18, 2008
82
I can't make what you're trying to say...anyhow, it isn't that straight forward, you have to show r equals the expression.
 

Kermit2

Joined Feb 5, 2010
4,159
I tried this and it works. if the answer is right you should be able to twist things around to show this as a cosine function.

I still can't see where the offset distance between your two arc centers affects the answer to finding 'r'. It's the side opposite the angles that is fixed; and that's the sine function.

???



Edit: Okay, I think I'm understanding your question now.

:)

I took a little while for me to 'get it'. The function defining little r will approach zero before the cosine of the angle does by the amount it is offset by the difference between the defined angle(30) and theta. So the cosine function is being reduced by another cosine function which should be the difference of the angles.

Am I getting warmer? :)
 
Last edited:

Georacer

Joined Nov 25, 2009
5,182
The above approach cannot be used.

I would call the distance you name L as D, because L already exists in Hello's drawing.

Your error is that you assume that \( \overline{R1} \) and \( \overline{R2} \) end in the same point.
But in Hello's drawing it is clear that \( \overline{R1}\) ends in the outer perimeter and \( \overline{R2}\) ends in the inner (until they meet, that is).

Your set of equations is valid only in the point where the two circles meet.

P.S. I couldn't use the LaTex command \overrightarrow to denote the vector format.
 
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