Bignose Posted October 24, 2013 Posted October 24, 2013 There is also the drag from the wind If you have a wind speed of vw and an initial velocity of v0, then the total velocity is v=(vw2+v02)1/2 So, this equation can't be right. Because squaring the terms effectively eliminates their sign. In other words your formula here is telling me that the same amount of drag will happen for an object moving 10 mph into a 5 mph wind as an object moving 10 mph with a 5 mph wind. That just isn't right. And really, this entire thread is missing comparisons with the current drag correlations in the literature and comparisons with measurements. All I keep seeing is equations being tossed out, but nothing being done to actually evaluate how good they are. Furthermore, it looks like almost all of them are just being tossed out, with little to no reference to how they were derived. Drag itself can be calculated from first principles upon solving the Navier-Stokes equations. A large number of the correlations published today stem from using a particular solution of the N-S eqns and showing how perturbations in the solution lead to the correlations. Is there anything like that here? 2
Endercreeper01 Posted October 24, 2013 Author Posted October 24, 2013 (edited) So, this equation can't be right. Because squaring the terms effectively eliminates their sign. In other words your formula here is telling me that the same amount of drag will happen for an object moving 10 mph into a 5 mph wind as an object moving 10 mph with a 5 mph wind. That just isn't right. And really, this entire thread is missing comparisons with the current drag correlations in the literature and comparisons with measurements. All I keep seeing is equations being tossed out, but nothing being done to actually evaluate how good they are. Furthermore, it looks like almost all of them are just being tossed out, with little to no reference to how they were derived. Drag itself can be calculated from first principles upon solving the Navier-Stokes equations. A large number of the correlations published today stem from using a particular solution of the N-S eqns and showing how perturbations in the solution lead to the correlations. Is there anything like that here? 1. I meant minus 2. I did show how they were derived Also, for the form drag, there is a region behind the object where there is low pressure. This adds to the drag. According to my theory, the coefficient of flow drag would actually be: Cf=2Dcosθavg-DcosΦavg Where Φavg is the same as θavg, but the axis is at the back rather then at the front. This is why I think this: First, we have the term Dcosθavg. You need to add something to it, and that is the effect from the pressure difference. The force from the pressure difference is: F=AΔp Where A is the surface area and Δp is the change in pressure between them. Now, you need to find the coefficient of drag for this. You get it by: Cd=2Fd/Aρv2=Fd/qA In this case, Fd is AΔp, so we can rewrite this as: AΔp/qA=Δp/q But, what is Δp? Δp is p2-p1, where p1 is the pressure in the low pressure region and p2 is the pressure in the front. Sine F=Ap, this means p would be qCd In this case, Cd is just Cf not including the effects from the low pressure area. This would mean it is Dcosθavg. Therefore: p=qDcosθavg And so: Δp=qDcosθavg - qDcosΦavg This means Δp/q is: qDcosθavg - qDcosΦavg/q=Dcosθavg-DcosΦavg But also with the form drag, if the object is concave instead of convex, then it is D/cosθavg instead of Dcosθavg, and the same with Φavg Remember my previous theory about the coefficient of drag? Now, I have a new theory on the coefficient of drag. My theory is for the coefficients that affect the coefficient of drag. First, let's start off with the coefficients themselves. The equation for Cd in terms of other coefficients is: Cd=Cf+Cw+Cs+Cl2/eπAR First, lets start with Cl2/eπAR, where e is the efficiency factor and AR is the value of the span squared over the area, s2/A. That is the equation for induced drag. Cl is the lift coefficient, and my theory does not cover that. A new theory I am still developing does, but I need to finish a separate theory in order to have an equation for that. Now, lets go on to the parasitic drag. Let's start with the form drag coefficient. The form drag coefficient is: Cf=Dcosθavg Where D is the drag coefficient it would have if it was a flat 2D plate and is a function of Reynold's Number. θavg is the average of the angles that each surface makes. The angles are relative to the axis perpendicular to the velocity. The reason that I think this is the equation is because the form drag goes around the object. Let's say you had two 2D plates joined together at one side and they were both tilted at an average angle of θavg. Then they would now have a form drag coefficient of Dcosθavg. This can be applied to all shapes. The form drag is moving around the object. But we do not count angles that are greater then 90 degrees in the average. The drag from that is covered in the interference drag. The skin friction drag coefficient Cs already has an equation for it that I did not come up with. The equation is: Cs=2dΘ/dx where Θ is the momentum thickness. Finally, let's talk about the wave drag. The wave drag is the drag from shock waves and only has an affect for supersonic or transonic velocities. The equation for that is: Cw=Av/vs Where A is the area of the shock wave, v is the velocity, and vs is the speed of sound. v/vs is a ratio called the Mach number. It is the ratio of the velocity to the speed of sound. I think this because the more area the shock wave covers, then the more dag it would make, and A=0 when v<vs, and this means there is no drag from the shock waves when there is none, which is correct. I put the Mach number into the equation because when you have a bigger Mach number, you have stronger shock waves, and so more drag. This makes the equation for drag coefficient: Cd=Dcosθavg+Av/vs+2dΘ/dx+Cl2/eπAR What do you think? Do you actually read the whole post? 3. This is for the coefficient of drag Edited October 24, 2013 by Endercreeper01
Bignose Posted October 24, 2013 Posted October 24, 2013 Minus or Plus doesn't fix anything. The fact that you are squaring the velocities inside the ()'s means you lose the sign of the velocities. How most people in the literature fixes this is something like (v1 - v2)^2. They square the difference between the two velocities... that way the sign of the two velocities do make a difference in the calculation. And yes, I read the post. All you quoted there are you just throwing out forms of the equations. For example.... Cf=2Dcosθavg-DcosΦavg why? Why cosines? Why 2 in front of the 1st term. How do you calculate θavg? and so on. and so on. Lastly, you really didn't address anything of my biggest complaint... which is all these equations are fine and dandy, but how do they actually compare to data? I can live with ad hoc equations if they work well. For a very long time, an awful lot of practical fluid mechanics was done using ad hoc equations. But those were accepted because they work. So far, you haven't demonstrated a single shred of evidence that your work. Just piles and piles of equations. So, please post a plot of measured data, some of the best correlations in the literature and then your correlation. Show us how well your equations work, please.
Endercreeper01 Posted October 24, 2013 Author Posted October 24, 2013 (edited) Minus or Plus doesn't fix anything. The fact that you are squaring the velocities inside the ()'s means you lose the sign of the velocities. How most people in the literature fixes this is something like (v1 - v2)^2. They square the difference between the two velocities... that way the sign of the two velocities do make a difference in the calculation. And yes, I read the post. All you quoted there are you just throwing out forms of the equations. For example.... Cf=2Dcosθavg-DcosΦavg why? Why cosines? Why 2 in front of the 1st term. How do you calculate θavg? and so on. and so on. Lastly, you really didn't address anything of my biggest complaint... which is all these equations are fine and dandy, but how do they actually compare to data? I can live with ad hoc equations if they work well. For a very long time, an awful lot of practical fluid mechanics was done using ad hoc equations. But those were accepted because they work. So far, you haven't demonstrated a single shred of evidence that your work. Just piles and piles of equations. So, please post a plot of measured data, some of the best correlations in the literature and then your correlation. Show us how well your equations work, please. First, read them again. I explained it in my posts. Now, lets address how they correlate with data. I am assuming that they all have roughly the same skin friction drag and Reynolds numbers that are close to each other. Let's start with the cubes. The cube has a drag coefficient of 1.05 and the angled has a drag coefficient of .8. The average angle of the angled cube is 45 and it is 0 for the cube. Because Φavg=θavg for both cubes, it becomes just Dcosθavg. For the cube, the average angle is 0, and the cos of 0 is 1. Then, it reduces to D. The cos of 45 is roughly .7, and the answer is .735. It is close. The other factors would be the skin friction drag and the difference in D between them. For the cone, then it is different because we don't know what θavg is, and the same for the streamlined bodies. For the sphere and half sphere, they have close drag coefficients, and that is because of how Φavg affects the drag, or it could be because of the difference of D And also, there are a few ways to calculate an average. One way is to sum over all angles and divide by the number of angles. Edited October 24, 2013 by Endercreeper01
Bignose Posted October 25, 2013 Posted October 25, 2013 So, you have 1 value that is kind of 'close'. For some unspecified Re number. From some uncited source. I guess I'm not really impressed. Any chance we're getting that plot I asked about (that you ignored twice now)? Also, your avergae angle idea as described cannot work. What is the average angle of an airfoil like these? http://history.nasa.gov/SP-4305/p98a.jpg The average should be exactly the same as a sphere, which it clearly isn't.
Endercreeper01 Posted October 25, 2013 Author Posted October 25, 2013 So, you have 1 value that is kind of 'close'. For some unspecified Re number. From some uncited source. I guess I'm not really impressed. Any chance we're getting that plot I asked about (that you ignored twice now)? What do you want a graph of? Also, your avergae angle idea as described cannot work. What is the average angle of an airfoil like these? http://history.nasa.gov/SP-4305/p98a.jpg The average should be exactly the same as a sphere, which it clearly isn't. Why should the average of the airfoil be that of a sphere?
Bignose Posted October 25, 2013 Posted October 25, 2013 What do you want a graph of? Seriously? You gripe at me to read your post twice and yet you can't be bothered to go back and read mine? Ok... I want a plot of measured data from the literature (there are tons!), a plot of some of the best correlations presented in the literature (again, tons of them out there!), and a plot of your correlation. Show us how your correlation matches the experimental data better than the existing correlations. Please include detailed citations. Why should the average of the airfoil be that of a sphere? Because you said to get the average angle to "sum over all angles and divide by the number of angles." Well, a sphere has one of every angle as does the airfoil. Therefore by your method the avg of the two are the same, hence you are effectively saying that an airfoil is the same as a sphere. Which it clearly isn't.
Endercreeper01 Posted October 25, 2013 Author Posted October 25, 2013 (edited) Seriously? You gripe at me to read your post twice and yet you can't be bothered to go back and read mine? Ok... I want a plot of measured data from the literature (there are tons!), a plot of some of the best correlations presented in the literature (again, tons of them out there!), and a plot of your correlation. Show us how your correlation matches the experimental data better than the existing correlations. Please include detailed citations. Because you said to get the average angle to "sum over all angles and divide by the number of angles." Well, a sphere has one of every angle as does the airfoil. Therefore by your method the avg of the two are the same, hence you are effectively saying that an airfoil is the same as a sphere. Which it clearly isn't. We don't sum over angles greater then 90, and you would take the limit as n--> infinity of the sum over n. Ok, I will show some graphs. This one is how the angle of attack a effects the drag coefficient: From http://www.aerospaceweb.org/question/airfoils/q0259c.shtml Pay attention to the wind tunnel part. There is an intercept, and that is the other effects of the drag. But, do you see what Cf is proportional to? It is proportional to sin(90-a) because the angle of attack is not relative to the axis perpendicular to the velocity, but parallel. The constant of proportionally is Cf0, or the coefficient of form drag it has when a=0. Another reason it is proportional to sin(90-a) is because at a=0, the sin(90-a)sin(90)=1, and so it becomes just Cf. It has to be proportional to a trigonometric function because you are dealing with angles, and it can't be cos a because if it were, it would get smaller instead of bigger. A 3rd reason is because the function appears to approach something. It appears to approach the constant of proportionality. Here is another graph from http://www.aerospaceweb.org/question/airfoils/q0150b.shtml for the angle of attack that uses the same equation: The equation for skin friction drag I can not take credit for because that is not what I came up with in my theory Edited October 25, 2013 by Endercreeper01
Bignose Posted October 25, 2013 Posted October 25, 2013 Yep. Those are graphs. But they don't show how well your correlation works. So, I don't see how in any way it answers what I am asking. One more time... I am asking for you to present just how well your correlations work. That is, show me just how useful your formulas are.
Endercreeper01 Posted October 25, 2013 Author Posted October 25, 2013 Yep. Those are graphs. But they don't show how well your correlation works. So, I don't see how in any way it answers what I am asking. One more time... I am asking for you to present just how well your correlations work. That is, show me just how useful your formulas are. So, you want a plot of my equation? Or do you want me to present it in another way? I am confused about what exactly you want me to tell you.
Bignose Posted October 26, 2013 Posted October 26, 2013 So, you want a plot of my equation? Yes. Plot experimental data, the best correlations in the literature, and your correlation. All on the same plot. You should show us that your predictions lie right on top of the experimental data points. This would be the very basic requirements -- demonstrating that your idea actually, you know, works. I want you to show us that it works.
Endercreeper01 Posted October 26, 2013 Author Posted October 26, 2013 Yes. Plot experimental data, the best correlations in the literature, and your correlation. All on the same plot. You should show us that your predictions lie right on top of the experimental data points. This would be the very basic requirements -- demonstrating that your idea actually, you know, works. I want you to show us that it works. Ok, I will show a plot. Let's go back to a previous plot. My theory predicts that Cf=Cf0sin(a) where a is the angle of attack. In this case, Cf0 is about 1.8 because the value of the function at 90 degrees is the form drag it has with no angle of attack, and the sin of 0 is 1, so then you get Cf0. Now, let's make a plot of this equation: Cf=1.8sin(a) The 2 graphs are exactly the same, except that the second one uses radiants.
Bignose Posted October 26, 2013 Posted October 26, 2013 (edited) 1) Why can't you plot these on top of each other? 2) Your CD = a sum of a bunch of terms, Cf among them. The top plot is CD, your plot is Cf. You can't directly compare them unless you show that all the other terms in your sum go directly to zero (which they shouldn't given the equations you've posted for them). 3) In this post you write: Cf=Cf0sin(a) But several posts back you write: Cf=2Dcosθavg-DcosΦavg Obviously these two are different. Why is this? 4) Lastly, I guess just a general comment in that your 'prediction' here really isn't new. Your graph is taken from a 1981 report. The sinusoid shape is pretty clear, and you just fitted your Cf0 parameter. (BTW isn't clear the 1.8 is actually the best fit. There are experimental data points above and below 1.8 at the peak. You need to provide your fitting method (e.g. least squares, minimum entropy, or other etc.); I suspect in this case it was the Mark I eyeball fitting method) Fitting parameters is fine (as I wrote above a great deal of fluid mechanics has been and still is calculated this way), it just isn't novel. 5) it is 'radians' not 'radiants' Edited October 26, 2013 by Bignose
Endercreeper01 Posted October 27, 2013 Author Posted October 27, 2013 1. I don't know how to plot them over each other 3. In this case, it is different. Cf0 is just the Cf it has with no angle of attack. The angle of attack effect goes to both terms because of the distributive property. This is why it is Cf0sin(a) 2. It must be roughly 0 because there is little intercept at a=180 4. Then it would be about 1.8. The method I used was fitting parameters and my theory. Because a is measured parallel instead of perpendicular to the velocity, it would be sin(a)
Bignose Posted October 27, 2013 Posted October 27, 2013 (edited) 1. I don't know what to tell you, then. Learning how to create high quality plots and graphs is an incredibly useful skill. I would do my utmost to learn it. Excel makes somewhat ugly graphs, but at least it makes graphs. I personally like matplotlib. But there are many, many other graphing programs out there. If you ever want to present your ideas like in a article or similar, people are going to want to see graphs. 2. You can't just hand-wave this. You need to demonstrate it. And, the rest of the terms aren't just zero at a=180*, but they must be zero across the entire range of a, per your plot. I just don't see this based on the formulas you've posted. For example, you take about the pressure difference in your formula. An airfoil doesn't make this go to zero, there is a wake behind an airfoil, especially at oblique angles of attack. 3. And speaking of those formulas, you can't just change it willy-nilly. You don't get to just change the functional form from a sum of two cosines to a single sine in your proposed equation because it fits this specific example. And you really don't get to just change the function form without demonstrating how the angle of attack turns into your average angle. Saying 'distributive property' here makes zero sense. Changing the formula to fit a specific case ruins the general nature of the formula. If someone tries to use your work, how are they supposed to know whether to use the sum of two cosines, the single sine, or maybe some other third form? 4. I specifically asked how you fitted the parameters. Did you use sum of squares? Maximum entropy? Some other method? Please report the error due to the fitting. Or if you just wild-ass-guessed it, that's fine, just acknowledge this. NEW 5. Even if you accept that you did match one set of data (which per the above comments I still have my doubts), there is a ton more data out there. You should be actively seeking other data sets to see how well your predictions match. Are you doing this? Just a general comment, that I want to preface by saying it is not personal in any way whatsoever: I am curious if you would share what is your end goal here? Are you trying to get some of this published? Because it is my opinion that there are some significant gaps in knowledge demonstrated. Again, this is not a personal attack, but simply an observation. And if you want some references to improve those knowledge gaps, I would be happy to share some. Edited October 27, 2013 by Bignose 1
Endercreeper01 Posted October 28, 2013 Author Posted October 28, 2013 Yes, can you share some? I am trying to make a theory. I might try to publish it
Endercreeper01 Posted October 30, 2013 Author Posted October 30, 2013 1. I know how to create graphs, but not how to put them on top of each other 2. It is likely that they are talking about the coefficient of form drag only 3. it gets changed because the average angle from the angle of attack is cos(90-a). This is because the angle of attack is measured parallel to the velocity, and the angle we use is measured perpendicular to the velocity. Therefore, we use cos(90-a). Because cos(90-x)=sin(x), we can rewrite it as sin(a) 4. I don't know how
Endercreeper01 Posted November 1, 2013 Author Posted November 1, 2013 What do you think? Do you like my theory?
Endercreeper01 Posted November 6, 2013 Author Posted November 6, 2013 And also, we need to change D to the coefficient of drag it would have if it was a rectangular prism with that characteristic length, since D depends on Re and Re depends on characteristic length
Endercreeper01 Posted November 7, 2013 Author Posted November 7, 2013 We could also set the value of D to the drag coefficient it would have if it was a cube with length 1 times the length of the object
Endercreeper01 Posted November 13, 2013 Author Posted November 13, 2013 Are there any effects my theory lacks?
Endercreeper01 Posted November 15, 2013 Author Posted November 15, 2013 Also, it has been hard to find good graphs for my theory.
hypervalent_iodine Posted November 15, 2013 Posted November 15, 2013 ! Moderator Note It's obvious people have given up responding to this topic, so unless you have something substantive to add to your hypothesis or you feel like properly addressing Bignose's points on the previous page (specifically 2 and 5, the latter of which was ignored), please stop bumping this topic. Also, 'I don't know how' isn't a good excuse. Learning how to make proper graphs is something you could achieve very easily by spending a few minutes on Google. 1
Bignose Posted November 15, 2013 Posted November 15, 2013 Also, it has been hard to find good graphs for my theory. I don't believe this for a second. The literature is full of published drag correlations. I have serious doubts that you've even bothered to look. It's your idea, no one else should be expected to do your work for you. Go and start with a fluid mechanics text, and start going through the references the book will have. I did promise you some references a few posts back, so let me fulfill that promise now: I'd recommend Fox and McDonald's Introduction to Fluid Mechanics. This books is an excellent intro, and will give you a good working knowledge of how fluid mechanics is typically used. Then, you need to tackle a graduate level text that will do a more in depth dive into the properties of the Navier-Stokes equations. I'd recommend L. Gary Leal's Laminar Flow and Convective Processes. Slattery's Advanced Transport Phenomena is a good supplement. Then, you need to dive deeper into the fluid flow at a surface. Schlichting's Boundary Layer Theory is the gold standard in this area of research. You also need a good understanding of turbulence. P. A. Davidson's Turbulence: An Introduction for Scientists and Engineers looks good; Davidson has been in the field for a very long time. Another good one is Pope's Turbulent Flows. Lastly, a delve into multiphase flow would help as well. A lot of the drag correlations have been measured in order to make the multiphase flow predictions more accurate. Crowe et al.'s Multiphase Flow with Droplets and Particles is good. Fan and Zhu's Principles of Gas-Solids Flows is also highly recommended as is Rhodes' Introduction To Particle Technology. From these texts, you will find an extraordinary wealth of references which describe the papers doing the experiments that measure drag. Especially, those multiphase flows. So, to recap: I don't believe your quoted word for second, because the references I posts about is literally 1/100th of all the good texts on fluid mechanics. Maybe 1/500th of all the poor texts. But even the poor texts discuss drag and will have citations you can look up to find more. If you are serious about this endeavor, you need to be putting in the time with the advanced texts. Though I guess really, the first point is that you need to learn how to make a graph. That is a really basic skill that cannot be ignored. 1
Endercreeper01 Posted November 15, 2013 Author Posted November 15, 2013 (edited) I mean I can't find a coefficient of drag vs Re graph for a cube and for the thing with the average angle. I know how to make a graph, but I don't know how to put graphs on top of each other. I know how to create graphs, but not how to put them on top of each otherAnd I mean through the internet, also. ! Moderator Note It's obvious people have given up responding to this topic, so unless you have something substantive to add to your hypothesis or you feel like properly addressing Bignose's points on the previous page (specifically 2 and 5, the latter of which was ignored), please stop bumping this topic. Also, 'I don't know how' isn't a good excuse. Learning how to make proper graphs is something you could achieve very easily by spending a few minutes on Google. I compared the two graphs. And as I said before, I do know how to make graphs.I wasn't purposely bumping Edited November 15, 2013 by Endercreeper01
Recommended Posts