Endercreeper01 Posted August 27, 2013 Posted August 27, 2013 I have made a way to calculate a coefficient of drag based on what it would be if it was just a plate. I will give the coefficient of drag of a plate a name. I will just refer to it as just D. So, lets start simple. Lets say that we had a 2D plate and it was tilted at an angle θ from the x axis (axis orthogonal to direction of velocity. Then the drag coefficient would be Dcosθ. Now lets say it was not 1, but 2 plates and they are connected somehow. Now you have 2 angles, θ1 and θ2. If θ1=θ2, then it is still Dcosθ, but if θ1#θ2, then you need to find the average angle, or (θ1+θ2)/2 and so it is Dcos(1/2(θ1+θ2)) In fact, you can do this with a shape with any number of sides, you just need to find the average angle θavg,and just find the cosine of that angle and then multiply by D. So we can just write it as Cd=Dcos((Σθ)/n), where n is the number of angles. So then is this correct? And also, what would the value of D be?
Enthalpy Posted August 27, 2013 Posted August 27, 2013 ...plate tilted at an angle θ... then the drag coefficient would be Dcosθ. Ouch. ...Two plates [at] θ1 and θ2 then you need to find the average angle, or (θ1+θ2)/2 and so it is Dcos(1/2(θ1+θ2)) Ouch again. So then is this correct? Alas, no. And also, what would the value of D be? The mean value of all angles would be zero for a sphere, and so would have been the drag according to this theory. There is no decent theory to evaluate a drag coefficent in subsonic flows. Drag coefficients are essentially experimental. One reason is that liquid flows are complicated and unsteady, one other is that body elements interfere. Well-trained aerodynamics designers can give their gut evaluation to 0.01 accuracy, which is better than software does. Things improve at supersonic flows, for which several theories can make sensible predictions, and even finite element software achieves meaningful results.
Endercreeper01 Posted August 27, 2013 Author Posted August 27, 2013 (edited) The mean value of all angles would be zero for a sphere, and so would have been the drag according to this theory. I meant the angles for only the parts that the air would hit against, and even if it was zero, then it would be D according to this theory because the cosine of 0 is 1. Ouch. Ouch again. Alas, no. How so?And this theory is just dependent on the coefficient of drag of a 2D plate, so it can work. Why shouldn't it? And basically, the air wouldn't be hitting against any angle 90 degrees or greater, so we would just exclude all angles greater then or equal to 90 degrees Edited August 27, 2013 by Endercreeper01
Endercreeper01 Posted August 27, 2013 Author Posted August 27, 2013 (edited) And what would it be then for supersonic speeds? Edited August 27, 2013 by Endercreeper01
J.C.MacSwell Posted August 28, 2013 Posted August 28, 2013 (edited) I meant the angles for only the parts that the air would hit against, and even if it was zero, then it would be D according to this theory because the cosine of 0 is 1. How so? And this theory is just dependent on the coefficient of drag of a 2D plate, so it can work. Why shouldn't it? And basically, the air wouldn't be hitting against any angle 90 degrees or greater, so we would just exclude all angles greater then or equal to 90 degrees Why do you think it should work? Edited August 28, 2013 by J.C.MacSwell
Endercreeper01 Posted August 28, 2013 Author Posted August 28, 2013 Why do you think it should work? Because of the logic of this. Cd is proportional to cos θ and the constant of proportionality would be the coefficient of drag for a plateSo why don't you think it works?
Enthalpy Posted August 28, 2013 Posted August 28, 2013 Many things go the wrong way in your theory attempt, like the cos which is wrong, or the trailing areas - the make much drag at subsonic flow. Rather than discussing dozens of points, I suggest that you first read a book about experimental aerodynamics. A good old book with experimental graphs and formulas, not the useless differential equations. Then you'll have the observation basis needed to make a theory - or avoid making one.
Endercreeper01 Posted August 28, 2013 Author Posted August 28, 2013 (edited) Many things go the wrong way in your theory attempt, like the cos which is wrong, or the trailing areas - the make much drag at subsonic flow. Rather than discussing dozens of points, I suggest that you first read a book about experimental aerodynamics. A good old book with experimental graphs and formulas, not the useless differential equations. Then you'll have the observation basis needed to make a theory - or avoid making one. Hmm... Tell me a good book on areodynamics? And discuss your points, i have all day. And how is the cos wrong? With a 2D plate, then it should be D cos θ, where D is the coefficient of drag for a plate and θ is the angle between the plate and the axis orthogonal to the direction of motion Edited August 28, 2013 by Endercreeper01
Endercreeper01 Posted August 30, 2013 Author Posted August 30, 2013 Just discuss your points, and ill see if they are good.
studiot Posted August 30, 2013 Posted August 30, 2013 Rather than discussing dozens of points, I suggest that you first read a book about experimental aerodynamics. A good old book with experimental graphs and formulas, not the useless differential equations. Then you'll have the observation basis needed to make a theory - or avoid making one. Hmm... Tell me a good book on areodynamics? The Mechanics of Flight A C Kermonde, Fellow of the Royal Aeronautical Society Pitman Volume1 should fit the bill nicely.
Endercreeper01 Posted August 30, 2013 Author Posted August 30, 2013 Ever heard of Reynold's number? Yes, but in this theory, you use the drag coefficient it would have if it was a flat 2D plate (based on Reynolds number) and you multiply that by the cosine of the average value of all angles less then 90 degrees.
doG Posted August 30, 2013 Posted August 30, 2013 Yes, but in this theory, you use the drag coefficient it would have if it was a flat 2D plate (based on Reynolds number) and you multiply that by the cosine of the average value of all angles less then 90 degrees. For all velocities?
Endercreeper01 Posted August 30, 2013 Author Posted August 30, 2013 (edited) For all velocities? Im not sure exactly, but i think it would be because the Cd for a plate would probally be different if it was supersonic. someone said there was a way to calculate Cd at supersonic speeds, so im not sure if this also works at supersonic speeds Edited August 30, 2013 by Endercreeper01
doG Posted August 31, 2013 Posted August 31, 2013 I think cd increases with turbulent flow, i.e. high reynold's numbers...
Endercreeper01 Posted August 31, 2013 Author Posted August 31, 2013 (edited) I think cd increases with turbulent flow, i.e. high reynold's numbers... Doesn't it decrease? And I also posted this: in this theory, you use the drag coefficient it would have if it was a flat 2D plate (based on Reynolds number). Edited August 31, 2013 by Endercreeper01
doG Posted September 1, 2013 Posted September 1, 2013 High fluid velocities lead to high reynold's numbers which indicates increased turbulence which increases drag. This simply means that cd is dependent on velocity and theories must account for that. It is not as simple as shape and angle of incidence.
J.C.MacSwell Posted September 1, 2013 Posted September 1, 2013 High fluid velocities lead to high reynold's numbers which indicates increased turbulence which increases drag. This simply means that cd is dependent on velocity and theories must account for that. It is not as simple as shape and angle of incidence. High velocities and larger objects lead to higher lead to higher Reynolds numbers- the boundary layer transitions to turbulent relatively earlier on the object. This boundary layer turbulence/mixing with the outer flow generally reduces the overall turbulence of the flow (swept downstream more readily and less build up of larger eddies) Though shape dependant this generally means that higher Reynolds numbers lead to lower coefficients of drag. At lower Reynolds numbers turbulence in the boundary layer are often induced for this reason, the earlier transition from laminar reducing drag (or increasing lift of a wing by delaying stall or assisting reattachment of the flow)
doG Posted September 1, 2013 Posted September 1, 2013 High velocities and larger objects lead to higher lead to higher Reynolds numbers- the boundary layer transitions to turbulent relatively earlier on the object. This boundary layer turbulence/mixing with the outer flow generally reduces the overall turbulence of the flow (swept downstream more readily and less build up of larger eddies) Though shape dependant this generally means that higher Reynolds numbers lead to lower coefficients of drag. At lower Reynolds numbers turbulence in the boundary layer are often induced for this reason, the earlier transition from laminar reducing drag (or increasing lift of a wing by delaying stall or assisting reattachment of the flow) My own personal experience with orifices in hydraulics systems indicate the opposite. The higher the flow through an orifice, including a pipe which is just a long orifice, is that higher velocities create chaotic flows as opposed to laminar flow and the result is a higher pressure differential from the increased friction.
J.C.MacSwell Posted September 1, 2013 Posted September 1, 2013 My own personal experience with orifices in hydraulics systems indicate the opposite. The higher the flow through an orifice, including a pipe which is just a long orifice, is that higher velocities create chaotic flows as opposed to laminar flow and the result is a higher pressure differential from the increased friction. I think that is the key. The friction is greater. But often with the drag of objects the form drag is significantly more important and makes up a higher portion of the drag. If it does not trigger less form drag (a straight pipe would not have any) you would just have the greater friction drag.
doG Posted September 1, 2013 Posted September 1, 2013 (edited) I think that is the key. The friction is greater. But often with the drag of objects the form drag is significantly more important and makes up a higher portion of the drag. If it does not trigger less form drag (a straight pipe would not have any) you would just have the greater friction drag. Actually a straight pipe does at significant flows. Flow is laminar at lower velocities but it becomes chaotic at greater velocities. An article from one of the magazines I get actually suggests using this principle for cleaning fluid conductors. Here's a reference also from World of Physics - Drag Coefficient... Edited September 1, 2013 by doG
J.C.MacSwell Posted September 1, 2013 Posted September 1, 2013 (edited) Actually a straight pipe does at significant flows. Flow is laminar at lower velocities but it becomes chaotic at greater velocities. An article from one of the magazines I get actually suggests using this principle for cleaning fluid conductors. Here's a reference also from World of Physics - Drag Coefficient... Does what? Turbulence does not equate to form drag. Form drag is due to pressure differential on the leading and trailing ends of a body in a flow. In a straight pipe, regardless of whether laminar or turbulent, all drag is affected as shear along the pipe wall. Obviously if a pipe has irregular buildup and needs cleaning as in the article you referenced on flushing procedures...then it's not really a straight pipe. Also note that your reference from the World of Physics - Drag Coefficient...includes a table where the drag coefficient of spheres (golf and tennis balls) increases at lower Reynolds numbers. I don't see where it helps your claim to the opposite. Edited September 1, 2013 by J.C.MacSwell
Endercreeper01 Posted September 2, 2013 Author Posted September 2, 2013 According to my graph i showed you before, the coefficient of drag decreases with a higher reynolds number
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