yeah, what he said ^^^
I will say that is the best (if not, slightly overstated
Thanks.
Well, I have to say that it is my feeling that for a given MANUAL trannied car you will not have an accurate estimation of rwhp for a 1500 hp car if you first determine the % drivetrain loss for the vehicle at say 300hp.
What is drivetrain loss? I can think of four things that contribute, though there are probably more:
Well, a torque applied at one end of a static object will apply an identical torque on the other end. However, we're not dealing with static objects, but accelerating objects. The acceleration in and of itself "consumes" some of that force meaning that the force applied at the other end of the object will now be less the torque required to accelerate that object's mass into rotational velocity. I think it's somewhat difficult to determine the force required because while we could estimate the rotational inertia of a given drivetrain, the rest of the equation would depend on the rate of acceleration of each component of the drivetrain, each of which would depend on the rate of acceleration of said vehicle. They would each be proportional and dependant on the gear ratios/tire radius involved. This would, therefore, not necessarily be a constant percentage of the power applied, but would depend on where you are in the powerband, which gear you're in, and . As a general rule, more force equals more acceleration, so this requirement as Mike mentioned should increase with the drivetrains rate of acceleration. Inertial dynos, therefore, would show an increase in measured torque loss that increases with the amount of torque produced. However, I'm uncertain whether or not this could accurately be represented with a constant percentage of the rate of acceleration. On a dyno that held a constant velocity, however, the loss of force due to this phenomenon would be 0.
Another loss in the drivetrain would be the loss of torque due to friction between the gears in the transmission, and the friction between the ring/pinion gears in the rear-end housing. This too, would increase proportionally with the amount of torque applied, and would be accurately represented by using a constant percentage of that torque. That's because the kinetic friction coefficient is that constant, and the only other information required to determine the friction involved is the normal force which would be proportional to the torque applied to the system, and which would account for the angles of the gears and the gear ratios used as gear ratios directly multiply the normal force.
There would be a some amount of loss in the hysterisis due to deflection in the tire, too. This loss of force is the reason that some tires allow a car to get better gas mileage than the previous tires did. With most street tires, the force loss should not be noticeably different when more force is applied. The biggest factors in tire deflection include the weight of the vehicle on the tire, and the speed of the tire. Neither of these need change in a higher HP application. However, non-standard tires like bias ply tires used for racing will deflect significantly due to the torque applied and as a result the force loss here should increase as the torque applied to the tires does. However, I would expect this loss to be relatively small by comparison to the other factors already mentioned because the tire only deflects once as the power is applied and then begins to recover its normal shape slowly as torque decreases through the powerband, and then more so when the driver gets out of the throttle. However, this is really only one long event for each molecule whereas for the other factors discussed (speed and weight) each rotation of the tire causes a deflection and recovery.
The last loss of force that occurs to me is the force lost in the friction of the bearings of the moving componenets in the drivetrain. Though it's a relatively insignificant force-loss overall, some of it will also increase with the force applied, but some of it will not. That's because the force applied also adds weight to some of these bearings like the ones in the transmission, u-joint, and rear-end. As load is applied to the gears, they in turn add load to the shaft that they are on, which adds weight to the rollers on their bearings. Other bearings do not take the added weight because the force is not applied directly to them, only a rotational force. When that happens, no extra weight/load is applied to the bearing, and only the twisting force (torque) is applied to the component itself.
I'm sure this is getting to be a bit of a novel now, but it's interesting to me because I believe one side of the argument, though I've never actually thought through the forces and losses involved. This is not an argument so much as an open discussion. Feel free to add any input or correct my observation.
Well, I have to say that it is my feeling that for a given MANUAL trannied car you will not have an accurate estimation of rwhp for a 1500 hp car if you first determine the % drivetrain loss for the vehicle at say 300hp.
Yeah, I can see your point..still not sold
However, I DO realize I am the minority on this, and the majority goes by that, so obviously on to something. Call me hard headed...
Torque does not measure work. Think of it as a rotational force or more technically "a moment," which is "a force applied about a fixed axis." Just as no work is done when a force is applied to a non-moving object, no work is done with a torque is applied to a non-rotating object.Torque is nothing but a value of work
horsepower is nothing but a value of how quickly that work is done, at some measurable constant rate of speed. Neither takes into account any sort of a rate-of-change or acceleration.
A perfect example of this is how, all else equal, if you were to pit a 347 and a 351 against each other with identical horsepower and torque figures, the 347 will walk all over the 351 every time, because of its lighter rotating assembly and higher rate of acceleration. The HP and torque numbers may have been the same, but they don't include a consideration of inertia and acceleration.
All your other points of gear loading, bearing loading, tire deflection, etc, I agree with entirely. A few things I could add is losses from slinging fluid and losses from shear forces in bearing lubricant. Also, don't forget driving the accessories, as most engine dynos aren't done with PS, AC, smog, etc hooked up. A lot of times, they don't even have a belt driven water pump hooked up. Obviously, all of those are in place on a chassis dyno.
OK, another example of proportional losses is parasitic drag in a supercharger. The 14-71 blower that sits on top of a top fuel engine consumes somewhere around 800-1000 hp when the engine is running full steam. Obviously, even in the hottest street driven roots-blower application, you're coming no where near that kind of parasitic loss. Using that same logic, is it not unreasonable to think that if a drivetrain behind a 300hp car consumes 45hp (15% loss) that a drivetrain behind a 7000hp car might consume 1050hp (15%)?
Like I said before, the "%" method is not perfect, but estimating something between 12% and 15% (more for auto, less for manual) is actually fairly accurate across the board.
Please don't take this as any type of personal attack.
Torque does not measure work. Think of it as a rotational force or more technically "a moment," which is "a force applied about a fixed axis." Just as no work is done when a force is applied to a non-moving object, no work is done with a torque is applied to a non-rotating object.
Everything you say is true. However, you must take into account how you measure the hp&tq. This is getting into the "nitty gritty" details, and I'm not certain that we're talking about a significant difference. However, there's no doubt that on an intertial dyno the engine/drivetrain you are taking the torque measurement from is under acceleration, and while the rate of acceleration of the dyno's roller is what is used to calculate the torque applied to it, it cannot account for the rotational mass of the drivetrain that is applying that torque. Thus it cannot measure or calculate the force that is "lost" in the acceleration of that drivetrain.
I agree that if you have the same heads/cam/intake on a 347 that's on a 351, you'll have significantly more impressive results from the 347. However, I believe that these differences will be readily noticeable on a dyno. The rotating mass of the 351W is significantly greater. Thus, I believe the 347 will have a greater output. Nearly 20 lbs of weight in the rotating assembly will do that. However, if a 347 & 351 outputs are identical, I would expect them to perform nearly identically in the same vehicle except that the 347 is probably somewhere around 75-100 lbs lighter, and will therefore out perform the 351 equipped vehicle in the same way that it would another identically engined vehicle that was 75-100lbs heavier.
The thing is, once you measure the tq and hp, you've accounted for the rotating assembly's inertia.
Great points. Thanks!
Apples and Oranges. The 14-71 blower has to move 7000hp worth of air, which is going to be more than 10 times as much air as your average performance car. However, with regard to drivetrains, the 1500 hp car in question is potentially moving the same, albeit built, drivetrain as the 150 hp car.
I think for manuals you'd get the closest approximation, though my 'feeling' is that it still wouldn't be accurate. For autos, I think it's very tough to get any accurate approximation on a dyno. I've seen nearly 2 identically engined high HP cars that I modeled my engine after - exact same turbo, and equivalent c.i., heads, intakes. The manual trannied car (anybody no who Wolverine is?) made 1016rwhp, the auto trannied car made approx. 850 rwhp. Both setups were maxed out and the auto trannied car went 154mph. The manual car only went 140. These cars should have made about the same power at the crank based on the max output of the turbo. Clearly, there was nothing wrong with the auto trannied car at the track. Still, lets say they made 1200hp at the crank (a 15% loss with the manual), did the auto really eat 30%?
Maybe so... Then again, it was posting monster ETs and Traps (8.97@154).
I have heard that auto transmissions and torque converters built for big power and heavy track use post extremely misleading numbers on chassis dynos. I have actually HEARD people say 20-30% before. Also, was that done on the same dyno under the same conditions?
Not at all, man. I enjoy having these conversations with you, because I know you know what the hell you're talking about and that it always leads to a good technical discussion.
hehe...This is true for an inertia dyno, since all the computer can account for is the fixed inertia of the moving dyno parts. I prefer to think in terms of a brake dyno, in which the engine is held at a constant speed through loading and torque is measured, and the effects of inertia can be ignored. But then, chassis dynos aren't brake-type and then we DO have to consider drivetrain inertia. Hmm. I may be arguing myself in a circle here.
Now I understand where you're coming from. I'd have to agree with you, and hadn't considered it from that perspective. On a Brake dyno, I suppose that two engines with identical numbers may perform differently when installed in a vehicle.Now, this is where a brake type dyno would be "blind" to the increased rotating heft of the 351. If the two engines had identical output on a brake dyno, the 351 would likely show less power on an inertia dyno because it would be slower to accelerate.
LOLYou'rewelcomesir.
I have heard that auto transmissions and torque converters built for big power and heavy track use post extremely misleading numbers on chassis dynos. I have actually HEARD people say 20-30% before. Also, was that done on the same dyno under the same conditions?
I would also be inclined to think that the AT car's dominance at the track has a lot to do with the fact that high powered turbo cars really shine when backed with an AT. I assume that's a trans brake car?
Ok, ill admit i got a little lost in the shuffle on that convo and didnt read it all but let me ask this question...
You say the loss would be the same on a 1500hp manual transmission car as a 300 hp manual car... thats fine, but what transmission are you using? Say you use a TKO in the 300hp car, well thats not gonna survive in the 1500hp car. You would need a liberty or like a GF5 series g-force to hold that kind of power. The bigger transmission is going to soak up more power on the smaller motor. So its kind of hard to compare the two.
I think this may do more to support the people against the % theory but it tells me that the more power you make the beefier the transmission you will need which will soak up moar powa.
Im my case the 15% rule was pretty damn close. my motor made 440hp with a dual plane and a carb. On the chassis dyno it made 390hp and two days later it made 370 with more humidity. I'd consider this pretty close to 15%, and personally i dont trust chassis dyno numbers at all.
Hot Rod magazine had an article about how shops manipulate dyno numbers that was really good.
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