John Cuthber Posted September 8, 2013 Share Posted September 8, 2013 First, by x axis i meant axis perpendicular to direction of motion, and second, im talking about a 2 dimensional plate, not a 3 dimensional one It it about a 30% difference in the cosine of 0 and 45 degrees, so it is correct. And according to my theory, they would have Reynolds numbers that are roughly equal Good start, now try the other shapes. Link to comment Share on other sites More sharing options...
Endercreeper01 Posted September 8, 2013 Author Share Posted September 8, 2013 Good start, now try the other shapes. The shapes all have roughly the same Reynolds numbers. The sphere and half sphere both have the same average angle and roughly the same Reynolds number, and so coefficients that are very similar. The same goes for the streamlined half- and full bodies. I don't know about the long and short cylinder because they must have very different Re, and I don't know the angle for the cone. Link to comment Share on other sites More sharing options...
doG Posted September 8, 2013 Share Posted September 8, 2013 The shapes all have roughly the same Reynolds numbers. The sphere and half sphere both have the same average angle and roughly the same Reynolds number, and so coefficients that are very similar. The same goes for the streamlined half- and full bodies. I don't know about the long and short cylinder because they must have very different Re, and I don't know the angle for the cone. Ummmmmmmmmmmmm......NO! Reynolds numbers are only dependent on the flow, not the shape of any submerged bodies. Turbulence, which is represented by reynolds number, has an effect on drag and that's why the drag coefficient is not a constant. Link to comment Share on other sites More sharing options...
Endercreeper01 Posted September 8, 2013 Author Share Posted September 8, 2013 (edited) Ummmmmmmmmmmmm......NO! Reynolds numbers are only dependent on the flow, not the shape of any submerged bodies. Turbulence, which is represented by reynolds number, has an effect on drag and that's why the drag coefficient is not a constant. I know that, but according to my theory, they do. How can you be sure they don't? Its likely that they all have different Re Edited September 8, 2013 by Endercreeper01 Link to comment Share on other sites More sharing options...
doG Posted September 9, 2013 Share Posted September 9, 2013 I know that, but according to my theory, they do. How can you be sure they don't? Its likely that they all have different Re Only if you're making up some new definition for Re.....Feel free to imagine whatever you want. Link to comment Share on other sites More sharing options...
Endercreeper01 Posted September 9, 2013 Author Share Posted September 9, 2013 (edited) Only if you're making up some new definition for Re.....Feel free to imagine whatever you want. Re=Dvρ/μ according to my theory then they must have different reynolds numbers and also what would θ be for a half-sphere? I think it is 45. Is that correct? According to my theory, a half sphere and sphere have the same Cd. And also i found this link: https://www.google.com/url?sa=t&source=web&cd=2&ved=0CBIQFjAB&url=http%3A%2F%2Fwww.chem.mtu.edu%2F~fmorriso%2FDataCorrelationForSphereDrag2013.pdf&ei=V0QtUuChBNb94APqo4HICQ&usg=AFQjCNFux5E9ERsURhBQxTL8Zq5U-KqChw&sig2=Abqy5fRUhntiDepifZMMjQ it talks about Cd based on Re up to Re=107 so If i divide that by the cosine of 45, which is 21/2/2 then i get how to calculate the coefficient of drag for a plate up to Re=107. So the equation becomes 48/2^1/2 Re + 5.2Re/5)/21/2(1+(Re/5)1.52 + .822(Re/263000)-7.94/21/2(1+(Re/263000)8)+ 2Re0.8/463000*21/2 Edited September 9, 2013 by Endercreeper01 Link to comment Share on other sites More sharing options...
Endercreeper01 Posted September 10, 2013 Author Share Posted September 10, 2013 So what do you think? Link to comment Share on other sites More sharing options...
doG Posted September 10, 2013 Share Posted September 10, 2013 I think your understanding of fluid dynamics is insufficient for you to postulate theories on fluid dynamics... Link to comment Share on other sites More sharing options...
Endercreeper01 Posted September 10, 2013 Author Share Posted September 10, 2013 I think your understanding of fluid dynamics is insufficient for you to postulate theories on fluid dynamics... How is it insufficient? Link to comment Share on other sites More sharing options...
imatfaal Posted September 16, 2013 Share Posted September 16, 2013 ! Moderator Note Please note that theanonymouse has been banned as a sockpuppet of the member endercreeper01. I have now hidden the sockpuppets contribution as well as various bumps by the OP. Please be aware that sockpuppets are against the forums rules in both word and spirit; the use of a sock-puppet to further an argument is reprehensible. Do not respond to this modnote within the thread - report it if you feel it is unjust 1 Link to comment Share on other sites More sharing options...
Endercreeper01 Posted September 23, 2013 Author Share Posted September 23, 2013 I have something to add to my theory. It can also be described as Dcos(∫(over area)θ). But it is basically Dcos(θ_avg) where θ_avg is the average of θ Link to comment Share on other sites More sharing options...
Endercreeper01 Posted September 28, 2013 Author Share Posted September 28, 2013 I also want to change something else. My theory does not include Cdi. Cd=Cd0+Cdi, and this theory is only for how to calculate Cd0. In the future it mught, but for now it is just how to calculate Cd0 -1 Link to comment Share on other sites More sharing options...
Endercreeper01 Posted September 28, 2013 Author Share Posted September 28, 2013 The reason for not including Cdi is because Cdi is based on the coefficient of lift. I would need a new theory to calculate Cl, and I just started working on it. When I do come ou with a new theory about calculating Cl, then this theory will include Cdi -1 Link to comment Share on other sites More sharing options...
Endercreeper01 Posted October 5, 2013 Author Share Posted October 5, 2013 But actually, it would be for the full Cd because there would be D0 and Di just like Cd0 and Cdi, so therefore, it would be Cd =Dcos(θavg )=D0cos(θavg )+Dicos(θavg ) Link to comment Share on other sites More sharing options...
Endercreeper01 Posted October 15, 2013 Author Share Posted October 15, 2013 (edited) Actually it wouldn't be and it would just be for CD0. It's because the lift coefficient would not be Lcos(θavg ), where L is the lift coefficient for a 2D plate, and that's because you also have to consider the effects from the airfoils and angle of attack. Accoring to my new theory for coeficient of lift (WIP), one term in the equation would be Dcos(θavg ), but except the x axis would now be at the bottom instead of perpendicular to the velocity. Edited October 16, 2013 by Endercreeper01 Link to comment Share on other sites More sharing options...
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