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October 12, 2005, 09:08 |
C4 airfoil data
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#1 |
Guest
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Does someone know the C4 airfoil data or the eqautions to calculate its coordinates? thanks !
dragon |
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October 12, 2005, 09:49 |
Re: C4 airfoil data
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#2 |
Guest
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I've found RC(4) airfoil on this page: http://www.nasg.com/afdb/show-airfoil-e.phtml?id=996 Click on contour link and y'll find cordinates.
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October 12, 2005, 10:30 |
Re: C4 airfoil data
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#3 |
Guest
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thanks,dear Bounecer.
But I guess RC(4) is not the one I want. I read in the paper that the airfoil's name is C4. I dont know if it is NACA,but I guess not. This is the paper I read: "Non-Newtonian Flow over the trailing edge of an airfoil" Journal name:Experimental thermal and fluid science 1996;12:244-249 there is a picture of C4, but it does not look like RC(4). thanks again. dragon |
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October 12, 2005, 12:53 |
Re: C4 airfoil data
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#4 |
Guest
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I'm studing a 2D, incompressible, laminar, viscous flow about airfoil. I've used "structured C grid generation", with finite volumes method and second order upwind numerical scheem and SIMPLER algorithm. After great time and efforts, my code program diverge after i reach a precision of 10e-3.
In my work i'am using only NACA airfoil, but if i found what you look for, i'll contact you. Good chance friend |
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November 2, 2005, 07:07 |
HN273airfoil data
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#5 |
Guest
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please send all information about airfoil HN273. thank you.
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March 22, 2011, 12:49 |
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#6 |
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Geth
Join Date: Feb 2011
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Please click one of the Quick Reply icons in the posts above to activate Quick Reply.
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October 8, 2013, 15:22 |
matlab code
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#7 |
New Member
Akshay Khadse
Join Date: Sep 2013
Posts: 11
Rep Power: 13 |
Here is the code to generate C4 thickness distribution and cambered airfoil
clc clear all t = 0.1;%thickness, t and x are in % of chord length theta = 18.5*pi/180; %camber angle/2 mu = pi/6;%it is actually dy_c/dx, for simplicity i used it equal to 30 degrees. x = 0:0.02:1; y = (t/0.2)*(0.3048*x.^0.5-0.0914*x-0.8614*x.^2+2.1236*x.^3-2.9163*x.^4+1.9744*x.^5-0.5231*x.^6);%equation for symmetric airfoil C4 y_c = sqrt(((0.5/sin(theta/2)).^2-(x-0.5).^2))-0.5/tan(theta/2);%equation of camber line %for upper s/f wrapped co-ordinates around camber line y_c x_u = transpose(115*(x-y*sin(mu))); y_u = transpose(115*(y_c+y*cos(mu))); %for lower s/f wrapped co-ordinates around camber line y_c x_L = transpose(115*(x-y*cos(mu))); y_L = transpose(115*(y_c-y*sin(mu))); plot(x_u,y_u,x_L,y_L); ylim([-20 50]); regards, Ammofreak Last edited by Ammofreak; October 10, 2013 at 08:01. |
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