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HP vs Torque Discussion (No Jokes, No bantering. Just facts and reality)

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Old 02-14-2009, 09:10 PM
  #301  
mark kibort
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VR, I was using the two 300hp cars with their dynos, because i didnt have access to two 400hp engine dynos to use as an example.

I dont mean to be condesending, but its getting a little frustrating that my message is not clear.

It doesnt matter that the engines belong to a 928 or an M3. They both could be going in an e36 M3, one you seem to have some knowlege about. And reverting back to the question, if you have a choise of two 400hp or even 300hp engines, if they have different torque values, which would be better for a road course and the answer, logically, physically, emotionally, has to be , IT DEPENDS. IT depends on the shape of the hp curves. IF the two 400hp curves look like the 928 and M3 curves, then the lower torque engine would have the ADVANATAGE at ANY vehicle speed, of any turn, down any straigth.
This is because, HP at any speed, determines acceleration. Its not my law, its Newton's. Its also real life experince.

Real life experience, on the race track requires this basic knowledge, plus all the things you tout to have a solid grasp of. Hp /weights, set up, conditions, geariing, and the list goes on as we know.

So, the "greater torque producton" as you say, means realively nothing, unless you have access to the HP or torque curves with RPM. Without it, its a guess that you probably will be right on most of the time, but its not the rule. That is and has been my only point.

Respectfully yours

Mk




Originally Posted by Veloce Raptor
Mark, even after all the warnings, is this level of condescension REALLY necessary?


Again and for the 100th time, you are changing the subject. We ARE NOT and WERE NOT comparing a V8 928 and an I6 M3, were we?

We were comparing 2 identical 400 hp cars with different torque on road courses.

And despite all your condescension and sanctimony, every time you try to change that to suit your predetermined conclusion, it speaks volumes.

And yes, Mark, I do hold fast to my experience, and also to my ability to actually compare apples to apples. And the car with the greater torque production--all other things being equal--will do better on a road course. Period.
Old 02-14-2009, 09:18 PM
  #302  
mark kibort
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Thanks Bruce!

Actually, we are talking that there are cases where there are some HP curves of near equal shape, but with disparate torque values, like the CTSV vs M5. In those two curves, both cars would have pretty much equal torque to the driven wheels at any speed on the track. its becuase their HP curve shape was the same, even though one had 125ft-lbs more torque! Generally, VR is right. the greater torque engine usually produces a broader hp curve. But, as you know in cup cars, the close ratio gear box makes up for a lot of the disadvantages vs a boader HP competitor. Thats why they use them!

The M3 vs 928 example shows the lower torque engine making more rear wheel torque at ANY speed on the track. If I was to pick most any other lower torque rating car, that might not be the case. Point being, as you say, it depends!

mk'
Originally Posted by brucegre
I haven't been in here for a few pages, but I can't resist.

If we had two identical cars (same tire diameter, same rear end, same transmission, in the same cars) and we had two 400 hp cars, but with radically different torque curves then it still depends:

The shape of the torque curve is critical, but let's assume that the high torque motor makes power sooner than the lower torque motor. Somewhere in there, the low torque motor will catch up in horsepower, let's assume that it's similar to the CTS and M3 curves Mark was using.

So, both motors have about the same range of equal power, about 2000 rpm. Now, it depends on the track because with identical everything else, only one of those cars is going to be in the power band at any given point. So, which one is better? Damned if I know, but I didn't have anything else to do for the last five minutes.
Old 02-14-2009, 09:23 PM
  #303  
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Originally Posted by mark kibort
.... Generally, VR is right. the greater torque engine usually produces a broader hp curve.

..........................
mk'
now we are getting somewhere...without changing gears, etc., in the "ALL other things being equal" cars.
Old 02-14-2009, 10:02 PM
  #304  
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Simple explanations work best.
2 + 2 = SQRT (((9^2 + 1) + (7 * 6)) / ((4^2 *2) - 1))
Newton said F=MA. Simple is better. 2 + 2 = 4
Old 02-14-2009, 10:36 PM
  #305  
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Originally Posted by Bull
now we are getting somewhere...without changing gears, etc., in the "ALL other things being equal" cars.
/THREAD
Old 02-14-2009, 10:41 PM
  #306  
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Actually, I was talking about the two different powertrains in one cars. (all things being equal, meaning, car itself, weight, setup, etc) Only thing being the same with them on the power side of the equation, would be the gear spacing. In otherwords, as the charts show, at 60mph the two different powerplants would be at the beginning of their operational RPM range in what ever gear does that. And if gear spacing is the same, the gear reductions would be different for each car at that same speed. The engines would be puting out the same HP , but very differnent engine torque. However, the torque at the rear wheels would be near the same. For the sake of real life, I used the max torque and most favorable condition for the higher torque engine too, yet it failed to compete at the beginning and certainly at the end of its RPM range.

mk

Originally Posted by Bull
now we are getting somewhere...without changing gears, etc., in the "ALL other things being equal" cars.
Old 02-14-2009, 10:47 PM
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That's the crux of the entire debate. Some folks still are thinking about ENGINE torque ( rotational force) being the "F" in the "F=ma". However, Im talking about the "F" at the rear wheels, not the "F" at the engine, which is almost irrelevant when comparing two equal HP engines in the same car.

Its hard to boil down some physics in to a simple 2+2=4 formula. BUT, i think i did this in a post on the previous page. I showed that two equal HP engines with different torque can produce the same, greater or less torque to the rear wheels at any vehicle speed.

mk

Originally Posted by SundayDriver
Simple explanations work best.
2 + 2 = SQRT (((9^2 + 1) + (7 * 6)) / ((4^2 *2) - 1))
Newton said F=MA. Simple is better. 2 + 2 = 4
Old 02-14-2009, 10:58 PM
  #308  
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Default Final exam

VR or anyone,

Let's say, Dez wants to have you coach him to be a Speed WC Touring Racer.
He has a chance to buy two cars one with these two HP/torque dyno runs.
The cars are near identical BMWs with the same weight and balance.
The only difference is the engine and drive train that will be put in. (assume durabiltiy is the same as well. performance only!)

Both have the same peak HP of 288rwhp. One makes peakhp at 5,000rpm and the other at 8000rpm. One has a peak torque of 293ft-lbs at 4500rpm the other has a peak torque of 235ft-lbs at 5500rpm.

You can look at these curves and make your assement or use what I've provided above.

https://rennlist.com/forums/6282005-post246.html

Which car do you advice your student and paying customer to buy in order to be most succesful on any road course and why.
Old 02-14-2009, 11:17 PM
  #309  
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Originally Posted by mark kibort
You can approximate the usable rpm range, and then approximate the area under the HP curve by knowing the space between the gears. In general, most all street cars have a range of 70% to 80% drop of RPM between shifts. 73% is common for our cars and 83% for close ratio set ups like cup cars.

So, if you were to look at the HP available through all the gears, it would the shift point rpm, with a 73% drop to the next gear. This means if your engine revved out to 10,000, after a shift, the rpm would drop to 7300rpm. You would then use the HP level at that point all the way to the rpm shift point to determine the HP available for creating torque at the rear wheels

mk
So you mean a drop to eg 73%, not a drop of 73%. That makes more sense. Seems to me like there is actual physics and then there is actual driving plus mitigating things like road surface, tyre temps, tyre condition, pressure, boost/grip, cam timing, spark timing, yada yada yada. All these and more have an influence on eg corner exit speed. So while I am happy to read and 'discuss' this theory, it's probably too hard to measure quantifiably.
For those people that can't help getting personal perhaps tune out of the thread.
Old 02-14-2009, 11:31 PM
  #310  
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Originally Posted by mark kibort
I dont mean to be condesending, but its getting a little frustrating that my message is not clear.
^^^^^^^

Irony.
Old 02-14-2009, 11:41 PM
  #311  
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2 + 2 = SQRT (((9^2 + 1) + (7 * 6)) / ((4^2 *2) - 1))
2 + 2 = SQRT(((9^2 + 1) + (7 * 6)) / ((4^2 *2) - 1))


4 = SQRT(((81 + 1) + (42)) / ((16 *2) - 1))


4 = SQRT((82 + 42) / (32 - 1))


4 = SQRT(124 / 31)


4 = SQRT(4)

4 != 2

Someone please tell me I missed something.

-td
Old 02-14-2009, 11:47 PM
  #312  
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Yeah, I corrected that in an earlier post. I was thinking 73% of the RPM after the shift, but ment to write, 23% RPM drop. I didnt even have to equalize in the gearing department. I just used the gear charts that came with the engines. It still worked out to favor the lower torque engine in that case. not the rule of course.

mk

Originally Posted by 333pg333
So you mean a drop to eg 73%, not a drop of 73%. That makes more sense. Seems to me like there is actual physics and then there is actual driving plus mitigating things like road surface, tyre temps, tyre condition, pressure, boost/grip, cam timing, spark timing, yada yada yada. All these and more have an influence on eg corner exit speed. So while I am happy to read and 'discuss' this theory, it's probably too hard to measure quantifiably.
For those people that can't help getting personal perhaps tune out of the thread.
Old 02-15-2009, 12:12 AM
  #313  
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Yes I saw that you corrected it after I just finished reading through this whole tome. I somehow feel like I've died just a little after getting through that. Reminds me of another thread in the 951 forums about 'Standalone'. Jeez some guys just can't help getting personal AND not answering simple questions. Forgive me if I've missed some precedence however it never ceases to amaze me that guys get into a thread, drop bombs all over the place and just blow the thing up. As opposed to either answering a basic question (whether it's addressing the topic or showing their self proclaimed credentials), or just not posting. Not too hard from what I can see.
I'm not saying that I agree entirely with what you're saying but it's probably mostly as it's over my head physics. I always thought hp was a product of tq anyway, but that would be too basic for this thread.
Old 02-15-2009, 01:30 AM
  #314  
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Originally Posted by Veloce Raptor
Mark has to be involved getting it there, A Whyne. It's can't just be tweedle dumb and tweedle dumber.
Ahhh, Its a case of a very Dumb Driving Roach having too much Torque . Torque that turns to crap .... nothing like whyning the revs up .....Revs win all the time ...

Last edited by A.Wayne; 02-15-2009 at 02:17 AM.
Old 02-15-2009, 05:36 AM
  #315  
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Originally Posted by mark kibort
HP vs. Torque. How are they relevant? What do we look at to determine performance in any track, or racing application? That is the discussion.
We had a good one going before it spun out of control, just at the point where some had brought up some great points and facts and it was all getting disseminated.

The best part and turning point of the discussion was when the BMW M5 vs. Caddy CTSV dyno runs were posted. what was so good about these dyno runs was that they were equal in HP and in the shape of the HP curve. What this meant, was both cars would accelerate, at any point on a track, at almost exactly the same rate. (if both cars were able to swap engines at any time).
Some had thought the high torque of the CTSV would allow some better acceleration out of a particular turn on a track somewhere, vs. the BMW, but remember, the BMW normally has a 414hp rating and the CTSV has a 550hp rating. It just so happened that someone found a 430rwhp dyno run for both, meaning the BMW M5 was pumped up a bit.

Anyway, lets discuss.

Hp determines acceleration at any vehicle speed as shown in the Newtonian identity: acceleration = power/(mass x velocity).

what this means is, when any two same cars are being compared, if they have the same HP at any same vehicle speed, they will have the exact same accelerative forces. In reality, cars hp curves are not identical. Some are peaky and some are flat. The amount of time spent at the different HP levels, dictates a car's performance. I like to use Hp-seconds, as area under the curve and averages (both indicate the same) don't really show the full answer, as you spend different levels of time on a hp curve at the different HP values.

If we could use infinitely variable transmissions, your engine would run right up to max HP and stay there while you accelerate. since we dont have these today, we try and get gear boxes that have gear ratios that are close together. We also try an optimize our final drive ratios so that for a given track, we hit most of the ends of straights by hitting redline of our engines for the lowest lap time.

Many talk of "max torque" values, or "Flat" torque curves. After we discuss this here, you will see that these terms are not only misleading, but can misdirect assessment of a engines potential performance and also mislead a racer into shifting at inappropriate times, thus hurting overall lap times an performance.

Let the discussion begin, unless you had enough from the flamed out discussion earlier.

I've attached the BMW M3 -Modified at 430rwhp and the Caddy CTSV -stock at 430rwhp, curves to work from. As you can clearly see, the CTSV has 120+ft-lbs more engine torque, so how does this play out in the real world? Does it matter?

Post questions, thoughts, reasons, experience and keep it clean!

Mark


Going back to this post. Assume that both cars are geared such that gearing is irrelevant. What I mean by that is, each car redlines in each gear at the same speed. To simplify, let's say that's 8K for the Beemer and 6K for the Chevy - er - Caddy. Also assume 27% drop between gears.

Based on the above, I would derive the ideal RPM ranges for each as follows:

BMW: 5,840 - 8K RPM
Caddy: 4,380 - 6K RPM

Using the lower HP curves for each, I then find the point on the HP curve where we will be after shifting to the next higher gear:

BMW: 5,840 RPM, 330HP
Caddy: 4,380 RPM, 325 HP

This bears out your theory in a perfectly ideal situation. However, let's say that we hit that taller gear 500 RPM lower in the band, for any number of reasons. For example, held up in traffic or other situation where the gear change is too costly to be worth doing. Forget about sequential whiz-bang video game transmissions for a second, consider H-pattern and manual clutch, which I think most of us have. The cost of downshifting to use 500RPM to redline, then upshifting again may not be worth paying in some situations.

BMW: 5,340 RPM, 290 HP
Caddy: 3,880 RPM, 320 HP

Whoops, the BMW is down 10% on power in this situation. Not good. In a perfect world, on paper or a whiteboard, these two cars may seem to be equal. However, in the real world where nobody hits their shift point perfectly every single time, where things happen that upset your rhythm, the car with the broader and flatter torque curve (and resulting flatter HP curve) will be more forgiving and will benefit the less-than-perfect driver(just about everyone).

I've seen you go both ways in your videos, Mark. I've seen you hit 1st gear leaving turn 11 at Laguna. I've also seen you stay in a higher gear when on paper, you could have hit a lower gear for ~500RPM and chose not to.


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