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09292013, 07:24 AM  #133  
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Seat of the pants is fine for a DE but you'll find racers hunched over a laptop between sessions looking for tenths at any club race weekend, paying coaches for data analysis. Posters like this are so disappointing to read on this board. A long time member goes out of his way with meaningful, thoughtful and factual input and we get this flippant childish reply and no respect. You contribute less than nothing. Last edited by consolidated; 09292013 at 09:07 AM.. 

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09292013, 02:33 PM  #134 
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You continue to proffer information is is incorrect in many ways. I was going to refute all of it but, this is the crux of it all.
You don't understand the physical definition of work. Work occurs when force or torque is applied over a distance or a rotation. Torque as measured by an engine dynamometer is not a measure of an engine's ability to perform work because it does not provide for the distance measurement, period. Did it not occur to you that torque is a vector quantity that is not conserved in a system where as work is a scalar quantity and is conserved? Really? Neither force nor it's rotational analog can perform work, without being applied over either a distance or a rotation, respectively. Torque is not a measure of an engine's ability to proper a car forward. It's physical fact. Also, I never said anything about Power versus Speed, I was talking about Power, exclusively. Reread what I said because you didn't read it correctly. 
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09292013, 03:19 PM  #135  
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No, talking about force or torque "conservation" is bs, but I never did. However, you are again showing your utter miscomprehension of the words and their physical concepts. Work is not conserved, 100% incorrect again. I think I am going to give up on you. Conservation concepts are not well applied to a car because it is filled with parasitic losses resulting in heat. However, the principle I think you meant to cite is the WorkEnergy theorem which states the that the work done is equivalent the the change in enegry of an object. Force and torque are physical and mathematical analogs in some fundamental ways they are identical they can both perform work. Dyno's provide a complete vs. rpm measurement of torque and knowing the entire torque curve and rpms is IDENTICAL to knowing the entire power curve vs. rpm. Quote:
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Sorry wrong. You stated I should scale the x and y axes of my plots and then I would be looking at power and your statement was and is 100% incorrect.
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09292013, 04:31 PM  #136  
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12: 7000 (redline) 23: 6900 34: 6700 45: 6600 (won't reach here, drag limited in 4th at ~135 mph) 56: 6300 " Hope that helps. Too much (boring) work to do the other car you mentioned. Build a spreadsheet, run the numbers yourself, if it very insightful and certainly nontrivial. P.S. What are your calculations Jonjt? And of course, last but not least, let's get back on topic.
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09302013, 01:55 AM  #137  
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I suspected it wasn't as "easy" as I suggested in my previous post. Thanks for clarifying and elaborating. Back to topic 

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09302013, 11:33 AM  #138 
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Calculate appropriate shift points for an IC engine within a street car Background: Given: F=ma AndP=Fv Where a is acceleration, v is forward velocity of the car, F is tractive effort at the wheels, P is power and m is the mass of the car. These equations are known to hold true at all instances in time, when applied to vehicle dynamics. Substitute the first equation into the second: P=mav Realize that P, a and v are all functions of engine speed (assuming WOT) and enough traction to “hold” the tractive effort.P(rpm)=ma(rpm)v(rpm) Shift points should be designed to maximize a(rpm) at all times. Therefore, P(rpm) and v(rpm) are known (independent) values, where as a(rpm) is dependent (unknown). (P(rpm))⁄(v(rpm))=ma(rpm) A sanity check shows that via units analysis, the units of the left of the equation reduce to newtons. The units on the right side do as well. So, what does this equation say? It relates the instantaneous tractive effort at the wheels of a car to the instantaneous acceleration experienced, at an instant in time. So, how does this relate to engine performance? Analysis: Given: v1=velocity in lower gear v2=velocity in higher,adjacient gear (P(rpm))⁄(v1(rpm))=ma(rpm) and (P(rpm))⁄(v2(rpm))=ma(rpm) Let us say that we are driving an S55 equipped vehicle and at an instant in time, we are accelerating, the throttle is wide open and the engine is spinning at exactly 7000rpm. We have a choice, we can either wind this motor out to redline or, we can shift up, to the next gear and continue accelerating. We need to know what choice will net the greatest a(rpm) value. Note that the last, say ~600rpms of the lower gear produces overlapping v(rpm) values with say, the first ~400rpms of the higher gear, due to the fact that the ratio of the higher gear is higher. During this overlap “phase” we can switch gears or we can stay in the lower gear and still have the same forward vehicle speed. The difference will be that the engine speed will be higher in the lower gear and lower in the higher gear. So, what do we want? Let’s revisit the equations. Since we already stated that there will be a velocity overlap, we know that v1(rpm)=v2(rpm). Also, recall that tractive effort at the wheels is equal to (P(rpm))⁄(v(rpm)). Since the velocity will be the same at the end of the lower gear and in the beginning of the higher gear, this means that determining the appropriate shift point to maximize a(rpm) is a function of maximizing power available at the wheels to accelerate the car. If torque values begin to roll off at high rpms, the acceleration the car experiences will begin to decrease as the rpms continue to climb, in comparison to the acceleration the car experienced when the torque was higher, in the same gear. You can characterize this by saying that in gear, tractive effort is a multiple of engine torque or, you can say that dKE⁄dt≈v(rpm)(dv(rpm))⁄dt. Either way is technically correct. That seems undesirable. And, certainly it is. But, that doesn’t mean that you will see higher tractive effort by up shifting, even if more engine torque is available at lower engine rpm. Why? Because though higher engine torque might be available (depending on the exact ratio of the higher gear), there is less torque multiplication because of the higher ratio. In order for there to be more tractive effort, the available torque at the lower engine speed has to be high enough to compensate for the lower torque multiplication. In other words, because power (unlike torque) is conserved in a drive train (Pwheel=Pengine(Pwindage+Pfriction+Ppumping+Pinertial+Pmiscellane ous in all gears), you shift when the power available to move the car forward is higher in the higher gear than it is in the lower gear. Given the fact that the torque curve of the S55 starts to decrease before redline and that power starts to taper a bit before fuel cutoff, it is possible that an early shift may be necessary in order to maximize a(rpm). HOWEVER, given the fact that the dropoff is slight and begins late, I am not yet convinced that short shifting is going to be necessary, particularly because we don’t have a proper engine dyno graph, because we don’t have gear ratios and because that fall off isn’t that great. To illustrate, pretend that two identical F80 M3s appear at an instant in time T and are a given displacement from an ordinate, ∆x. Assume that both cars are driven by identical drivers and that traction is not an issue. Assume that the only difference between the two cars is that one car is at say, 7000rpms in second gear and the other car is at, say 5500rpms (just a guess) in third gear. At this instant in time, the car that can provide the greatest power for acceleration will accelerate the fastest. This is regardless of the amount of torque the engine is producing at these respective rpms. Though tractive effort is a multiple of engine torque, decreasing maximum a(rpm) does not mean that an upshift to a higher peak engine torque value will net higher tractive effort. This was my main point. It is certainly possible to characterize tractive effort and shift points as a function of engine torque. But why? It requires extra computation and, is largely beyond the purview of the average M3/M4 owner. The easiest method to determine shift points would be to say to the owner, “when power available to more the car forward in the lower gear is less than that available in the higher gear, shift.” That is why I said one should concern themselves with power, in this case. CVT transmissions do exactly this. They are designed to maximize power available at the wheels, even if peak engine torque is lower at this point that it is at a lower point in the RPM range. Last edited by Jonjt; 09302013 at 12:01 PM.. 
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09302013, 12:13 PM  #139  
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You claim power is conserved then provide an equation showing how power is not conserved (we all know it is not based on dynos). And you still think you can calculate shift points without knowledge of RPM DEPENDENT parasitic losses using crank power only? Sorry bud, I am not going to wade through the rest of your analysis to find any other major errors or contradictions. Calculating shift points exactly just isn't trivial for the average enthusiast. Luckily many of us can still do a reasonably good job by feel alone. Use your "method" to show me exactly how to calculate shift points for any car where the exact shape of the power or torque curves are known. Until then STFU, really, your foot keeps going deeper and deeper.
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09302013, 12:40 PM  #140  
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You're going to have to expound on that one. There is no contradiction. From engine power, you subtract all the parasitic losses, as you called them (Pwindage, Ppumping, Pinertia, Pfriction, Pmiscelleanous, etc) and you get wheel power. Even after all the losses, the goal is still to maximize power available at the wheels, for a given vehicle speed. And, since the wheel speed is the same at the end of the low gear as it is in the beginning of the high gear, the drive train losses will be equal. The engine losses will be higher but thankfully, the engine dyno graphs we have measure power after all the engine losses have been accounted for. It does not measure combustion power.
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I any case, I did already. You quoted it. Power is neither created no destroyed in the drive train. That is what I mean by conserved. It all goes somewhere and, you can keep track of it. Pin=Pout (though heat, friction, moving fluids, accelerating masses, etc) whereas torque can and is multiplied by gears; ie Tin≠Tout. There is no physical law that mandates this. Quote:
Edit: When you say rpm dependent parasitic losses, I hope you aren't talking about engine induction, exhausting, pumping, windage, friction and inertial, etc losses. Right? Quote:
So, the only vehicles I have are ones I've built. If you provide values for engine power and torque (from a proper engine dyno graph) and assumptions about parasitic losses (including correlations with shaft speeds, and what components they come from) we can do the same vehicle and compare notes. If you REALLY want, I can even run some external flow analyses on those vehicles and include speed dependent drag information in the force balances. However, I can't include anymore than a few operating points. I don't have the time to do a proper sweep across the operational speed range these cars see. Interpolation error will be nontrivial at higher speeds but, that's all I can manage right now. Last edited by Jonjt; 09302013 at 01:20 PM.. 

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09302013, 12:47 PM  #142  
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09302013, 02:15 PM  #143 
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Jonjt:
Any system that converts power into heat (classic drivetrain losses) is NONCONSERVATIVE. Just because you can account for the power losses does not mean it is conserved. The power at the crank is not available at the wheel and hence power IS NOT conserved. It appears you understand this but are using imprecise language. Perhaps what you mean is that energy, counting heat energy is conserved. However, conservation of energy starting with fuel energy is not really a useful way to approach vehicle performance... How is power easier to be used to calculate shift points if you still need parasitic power losses. You claimed a power approach is simpler and say it is not. Using a dyno with wheel power would be superior but dynos vary RADICALLY in the real world and are totally unreliable except for extremely careful A to B comparisons before and after a modification under identical conditions. Even then very important things like intake system performance are not captured. Regular Cd and Area numbers are plenty accurate. CFD is needed for an OEM to improve Cd not or enthusiasts to calculate drag nor top speed nor any other basic performance metric. I too have CFD at my disposal but certainly do not have the very extensive training required with it to prevent "garbage in, garbage out". I suspect neither do you. Take your method and apply it the BMW 523i. I think it is time to simply put up or shut up... Really.
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09302013, 02:18 PM  #144 
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Which is? I'd say butt out when you have no idea.
Just for clarity sake I can give you a clue. I enjoy the application of basic science/math/engineering to vehicles, both performance and manufacturing. I enjoy sharing some insights on these topics with the generally well informed and often times the more technically leaning members here. I also try very hard to dispel common myths about vehicle performance. Unfortuntely that often leads to debates.
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09302013, 02:39 PM  #145  
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My basis is that power cannot be created or destroyed, you can account for it in any closed system. And since the difference between an open system and a closed system is a matter of scope/reference frame, analyzing this particular system with Pin=Pout is only a matter of picking the right components to include in the energy balance. If I had said that flywheel power is not equal to wheel power and I had not cared to deal with the losses in everything in between then sure, mathematically it isn't conserved. But, out of habit perhaps, I just like to say Pin=Pout since it's universally and macroscopically true. And, while I have done an analysis including combustion power, it is indeed arduous. The energy input, in the case, is the power the engine produces at the flywheel. Forget all the combustion processes and mechanical engine losses. Since we have an engine dyno graph, who the hell cares about combustion power? Quote:
That sucks. Characterizing losses at a function of shaft speeds anywhere in the drive train is easier, IMO if you just use a power based loss calculation instead of trying to find the reduction in shaft torque. The conversion to torque is IMO easier to do at the wheels because you dont need to worry about the mechanics of friction, pumping, windage, etc losses in each drivetrain component. Furthermore, most drivers can more easily (IMO) comprehend the idea of power and it's dependence on engine speed than the complexities of torque, torque multiplication and how lossess effect net torque in the moment balances throughout the drive train. Damn straight. Quote:
I would say that induction performance can be characterized on a dyno, just not in absolute terms. Quote:
Although, for our purposes a Cd value and area would at least provide consistent values for drag in the force balance, between us. Since we speculating about shift points, the 100% accuracy isn't needed. I actually do (perform? haha) CFD on a daily basis. External non compressible flow is not my specialty but, I have some experience with it. In any case, this is why I said I wouldnt have time to do a thorough analysis. External aero cases take millions of cells and lots of time to solve. Don't have time for that...... Quote:
If not, I'll just pick something and go with it. Last edited by Jonjt; 09302013 at 03:37 PM.. 

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09302013, 03:58 PM  #146  
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Award for quote out of context of the year. Really... Quote:
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Drag isn't required for this calculation as it does not affect the force applied, it is the resisting force. Really, how many times, hate to be so blunt but put up or shut up...
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09302013, 04:22 PM  #147  
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If you recall, this discussion is centered around an estimated power curve and an estimated torque curve. None of this is accurate, this is not practice, not yet. You even pointed out that we don't have a proper factory crank dyno and that we are all just guessing. I brought up this power method to SIMPLIFY this estimation, since we can't actually produce an accurate result regardless of how comprehensive the methods are. I've responded twice asking for you to provide the same losses you are guessing at so I can do my calculation but, you have yet to produce anything. Comparative analyses are useless without data sharing. You can't validate anyone's technique without knowing what went into it. You even pointed this out by referencing the "garbage in, garbage out" nature of CAE. I can go pull some data from some source, make some loss estimations and just produce some numbers. It won't happen until I get home, though. Last edited by Jonjt; 09302013 at 04:58 PM.. 

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09302013, 04:34 PM  #148  
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Not sure if that was what you were asking though... 

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09302013, 04:38 PM  #149 
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09302013, 04:51 PM  #150 
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Just to save time looking through the thread, link?

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09302013, 05:35 PM  #151 
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09302013, 06:36 PM  #154  
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1. You have no idea what reasonably accurate losses are, and, 2. Your "simplified" power based calculation method (that is so simple and insightful) requires these values. Then I am at a total loss here (no pun intended) as to the superiority of your method, either in theory but especially in practice. I'll be happy to tell you all of the details that went into my calculation once you provide a post with a detailed account of your calculation of these F10 523i shift points. You could also run the E92 M3 MDCT and show redline shift points across the board. A better test case might be the F10 M5. There has been discussion here and at m5board.com (me included) about it's shift points. I'm pretty sure there was empirical evidence posted over there that the M5 needs to be shifted pre redline in many gears and my simulation matched that. Wow, I am getting bored with this...
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