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**if it had ~515**

*RWHP*Discuss...

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^ what is the deal with "I now drive a Cavalier" ? lol

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KillerVette said:^ what is the deal with "I now drive a Cavalier" ? lol

I sold my Trans Am, bought a house, and I now drive a Cavalier.

Should be around April that I'll find my way into a nice C5 Z06. Not much sense in buying one now with the first snow fall less than 2 full months away.

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Ok. The driver needs to get some exercise. Very cheap mod.Jay2002WS6 said:but a car that weighs 3150 - call it 3350 with driver

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I went through the same thing on the "other" forum when that Motor Trend MPH was posted. Ran the numbers through a few online HP calculators, and they averaged roughly around what Jay posted above.

Which of course conflicts with the stated HP, especially when you consider the 505 figure was derived using the new testing methodology.

What several people think - and I'm inclined to go along with this idea - is the (online) HP calculators don't correctly factor in the efficiency of the Z06 drivetrain, gearing and Cd (drag). In fact, some engineering type ran the acceleration numbers through a formula with an extended calculation for Cd (which included frontal area drag, etc.), and while I certainly wouldn't profess to understand them, they did show a MPH number closer to 125/126 with an assumed RWHP of ~450.

I think the MT car might have been a tad on the optimistic side of the HP range (didn't someone post that a some cars were showing closer to 515-ish?), but mainly the car is just making outstanding use of the power it has!

:cheers:

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seemed pretty close for my car. Listed RWHP as 381, where my measured was 371. The 11.7 @ 127 for the new one at 3350 lbs shows 474 RWHP. If it's a little high like mine was (about 2.6%), I'm guessing about 460 RWHP, or around 525 at the engine, if using 12% loss.

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Since HP is proportional to engine RPM, you can make good numbers two ways, high torque at low RPM or low torque at high RPM. The problem with low torque high RPM motors that make all their HP at the top end. For most of the RPM range they're not making very good HP. A high torque motor makes better power down low. And even if that motor has a lower peak HP, it may still produce a greater amount of power across a larger range of the power band.

Picture a graph of a Dyno run. It's the area under the HP curve dictates how effectively the motor will accelerate the car. Depending on the shape of the curves, two different engines with the same peak HP could have very different "areas" and they will perform differently because of that difference.

Just for comparison. I have trapped 126.9mph with 433rwhp in good air without powershifting.(etstreets and gears). The c6Zo6 should be able to trap 127 with tires/good air and a good driver. But I believe the 127 figure was inflated.(and the et). Just my opinion. I don't see anyone trapping that mph in the summer who is not familiar with the car. Hopefully Ranger will get some passes this fall. I bet he traps in the 125-126 range bone stock on a cold day.Jay2002WS6 said:if it had ~515RWHP

Discuss...

ok, with that said - major auto rag publications rarely test 1/4 results on an actual NHRA/IHRA drag facility. They find a nice, flat (hopefully), open piece of road and measure performance numbers with their own timing devices. They most likely take one "trap" speed at the 1320 mark and go with that as the figure for the article.

Now, a real drag strip is not going to give you a mph trap speed as the car crosses the 1320 mark - it's going to take two readings (one around 1260 out and another at 1320). Then it averages those times to give you the 1/4 trap speed on your timeslip.

This could create the "inflated" number so many are trying to figure out. Now if MT did the test at a real strip - then disregard what I just typed.

Imagine you have a one pound weight attached to the floor and try to lift it with one pound of force (or 20, or 100 pounds), you'll have applied force and exerted energy but no work would have been done. If you detach the weight and apply enough force to lift the weight one foot, then one ft. lb. of work would have been done. If that took you one minute to do then you would've been doing work at the rate of one ft. lb./min. If it took you one second to do the work it would've been done at the rate of 60 ft. lbs./min.

Now, in order to apply these measurements to a cars performance you need to address three variables: force, work and time.

Now, let's recall this from the post above: Remember James Watt determined that a horse could lift 550 pounds, one foot, in one second thereby performing work at the rate of 550 ft. lbs./ sec., or 33,000 ft. lbs./min. (550 X 60 Secs/min.= 33,000), for an eight hour shift. He then published his observations and stated that 33,000 ft. lbs./min. of work was equal to the power of one horse, or one horsepower. And we've been using that standard to describe one horsepower ever since.

Now, since we're talking about cars here we need to measure the units of force of rotating objects like crankshafts so I'll use terms that define a 'twisting' force, such as ft. lbs. of torque. A ft. lb. of torque is the twisting, or torsional, force needed to support a one pound weight on a weightless (imaginary if you will) horizontal bar, one foot from the fulcrum. "Fulcrum" meaning the support about which a lever turns.

You have to understand that you can't measure horsepower directly. It's a calculated value. A dyno for example doesn't measure horsepower, it measures torque which is expressed in ft. lbs. as above. From this you can calculate the horsepower by converting the twisting force of torque into the work units called horsepower.

Imagine that one pound weight again, one foot from the fulcrum on its imaginary bar. If you rotate that weight one full revolution against a one pound resistance you've moved it a total of 6.2832 feet (Pi X a 2 foot circle) and presto you've done 6.2832 ft. lbs. of work.

Watt said that 33,000 ft. lbs. of work/min. was equal to one horsepower. Now we know one horsepower equals 33,000 ft. lbs./min. so if we divide 33,000 by 6.2832 we get 5252 RPM and we are making one horsepower. Therefore if we multiply the torque by the RPM and then divide by 5252 we can determine the horsepower. If we move that weight at the rate of 2626 RPM instead (half speed) it's the same as 1/2 horsepower (16,500 ft. lbs/min.), and so forth. So, here's the formula for calculating horsepower from a torque measurement:

HP = Torque X RPM/5252

Now, as far as driving a car we all know torque is what we feel. Torque wins races, horsepower sells cars as the saying goes. Any car, in any gear will accelerate at a rate that matches its torque curve. In other words, a car will accelerate best at its torque peak in any gear, and will not accelerate as well below or above its torque peak. 400 ft. lbs. of torque will accelerate the car as hard at 2000 RPM as it would if it were making the same torque at 4000 RPM in the same gear. However, per the formula above the horsepower would be doubled at 4000 RPM. Therefore, horsepower doesn't really mean anything as far as the driver is concerned and the two numbers only mean anything at 5252 RPM where the horsepower and torque are always the same.

In contrast to the torque curve horsepower climbs quickly with RPM, especially when the torque value is also climbing. Horsepower will continue to climb however until past the torque peak and will continue to climb as the RPM does until the torque curve starts to fall faster than the engine RPM is climbing. So you can see that horsepower really has nothing to do with what you feel in the SOTP.

Example: Take your car to its torque peak in 1st gear and floor it. Notice how you feel it push you back in the seat? Now take it up to the power peak and floor it. Notice that the sensation is less?

So we've seen how important torque is and it's pretty obvious that torque is what you feel. So, why is horsepower important you ask?

Here are some examples of how horsepower can make a big difference in how fast a car accelerates even though perceived torque, that is what you feel, is telling you otherwise.

Peak h.p. @ RPM & Peak Torque @ RPM

Car A: 250 [email protected] RPM 340 ft. [email protected] RPM

Car B: 300 [email protected] RPM 340 ft. [email protected] RPM

Let's assume for now the cars have the same gearing and are the same weight.

First of all both cars will push you back in the seat about the same amount, at least at or near their peak torque in each gear. They'll both feel about the same but Car B will actually be quite a bit faster than Car A even though it doesn't pull any harder. Now let's see if we can figure out why Car B is faster. In order to do that we need to look at this formula:

Torque = HP X 5252/RPM

If we plug in some numbers we can see that Car A is making 328 ft. lbs. of torque at its horsepower peak of 250 h.p. @ 4000 RPM amd it can be inferred that it can't make any more than 263 lb. ft. of torque @ 5000 RPM otherwise it would be making more than 250 h.p. at that RPM and it would be rated at that. In reality, Car A is probably not making any more than 210 lb. ft. or so at 5000 RPM, and if you owned it you'd probably shift at about 4600-4700 RPM because more torque is available at the wheels in the next highest gear at that point.

On the other hand: Car B is making 300 lb. ft. @ 5000 RPM and is good up to its redline RPM which is probably somewhere around 5500 RPM for example.

Now, in a drag race assuming both drivers launch about the same time. Car A might have a slight advantage because its peak torque comes a little sooner in the RPM range, but that's debatable, since Car B has a wider, flatter curve (which is pretty much by definition just by looking at the numbers). So, from some point in the mid RPM range on up Car B will start to pull on Car A. When Car A has to shift to 2nd (giving up torque multiplication for speed) Car B still has about 1000 RPM left in 1st and begins to pull away as the speed increases. As long as the RPM's are high, Car B, by definition, will have the advantage.

Another example would be the BMW E36 vs the Honda S2000 example I gave earlier. The BMW pulls harder than the S2000 although its torque advantage is diminished somewhat because of its extra weight. The real advantage however is the S2000 has another 800 RPM left at the point the BMW has to shift. Remember, a car accelerates fastest at its torque peak. In the case of the M3 this occurs at 3800 RPM. On the S2000 peak acceleration doesn’t occur until 7500 RPM and peak h.p. is achieved at 8300 RPM. Where the BMW is forced to shift at 6000 RPM (its h.p. peak) the S2000 is still pulling towards its torque peak and eventually its h.p. peak. So, while the S2000 doesn’t pull harder than the BMW its high torque peak allows it to pull longer.

There are a lot of examples of this. The Acura Integra Type R for example is faster than the regular Integra. Not because it pulls harder (it doesn't) but because it pulls longer. It doesn't necessarily feel faster, but it is.

Now, let's look at top speed. Horsepower wins again. Why? because making more torque at high RPM means you can use a taller gear for any given speed and thus have more effective torque at the wheels.

Finally, (assuming you are still with me) an engine that's operating at its horsepower peak means its doing the absolute best it can at any given speed, measuring torque at the drive wheels. I know I said acceleration follows the torque curve in any given gear but if you factor in gearing vs the cars speed the horsepower peak is 'it'. One more example of Car B will illustrate this. If you take it up to its torque peak of 3600 RPM in any gear it will generate X amount of torque (340 ft. lbs. of torque X whatever the overall gearing is) at the wheels which is the most it will do in that gear. Meaning that's where it's pulling the hardest in that gear.

Now, if you re-gear the car so it's operating at its horsepower peak of 5000 RPM (at the same speed) it will have more torque at the wheels because you'll need to gear it up by almost 39% (or 5000/3600) while the engines torque has only dropped by a little over 7% (or 315/340). In other words you'll have a net 29% gain in drive wheel torque at the horsepower peak vs the torque peak at a given speed.

Any RPM (other than the horsepower peak) at a given speed will net you lower torque at the driven wheels. This is true of any car so its theoretical top speed will always occur when a car is operating at its horsepower peak.

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