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Loosing weight..., more effective on our cars than others?

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Old May 4, 2003 | 05:18 PM
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Default Loosing weight..., more effective on our cars than others?

I was thinking something i've thought before. Torque helps to get weight moving, so theoretically the less torque you have, the more the weight of your car plays in the whole accelleration equation. Right??? In conclusion, reducing the weight on our car has more of an effect due to our low torque???

Brandon
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Old May 4, 2003 | 05:27 PM
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Yes but its not just torque weight relates to HP too. Picture if you have 2 of the exact same cars, same everything. One car has one person and the other car has two people. The car with one person is obviously going to go faster because there is less weight.
Replace stock parts with aftermarket parts that weight less and youll be able to go a little faster. Rims are the first thing that actually reduce alot of weight.
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Old May 4, 2003 | 06:23 PM
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Yes, i didn't mention the HP relation simply becuase thats the obvious one. Its all anyone ever talks about. Look at it like this, you have two cars that are equiped as followed: (Note: all #'s are theoretical and simply used for the purpose of making my point more evident.)

Car 1: Car 2:

Engine: 280 HP 300ft/lbs Engine: 280HP 180ft/lbs
Weight: 3100lbs Weight: 2800lbs

0-60: 5.2 0-60: 5.8

1/4: 13.3 1/4: 13.8

Ok, the times arn't the same, but the HP is, even though car two has less weight, it has WAY less torque. Now, lets subtract some weight from both the cars...... 300lbs worth.... thats a serious jenny craig plan for both cars

Weight of car 1: 2800lbs
Weight of car 2: 2500lbs

Still, car 1 may be faster, but i imagine the margains would be MUCH closer. More like 0-60 times of 4.8 and 5.1 with 1/4 times of 13 flat, and 13.2-13.3 Anyone get my point???? '

Brandon
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Old May 4, 2003 | 07:20 PM
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Umm... Actually, in your example, Car 2 is faster than Car 1 since it has a greater power to weight ratio. The fact that it has less torque at the crank doesn't matter, since it has the same torque at the wheels.

Here's a better example. Take two cars, each with 10lbs per hp. Car 1 is 4000lbs and 400hp, Car 2 is 2000lbs and 200hp. In a straight line, they're equal. Now, take 100lbs off. That's only 2.5% of Car 1's weight, but fully 5% of Car 2. Therefore, it'll have a significantly greater impact on Car 2.

Basically, any weight loss will have a greater impact on a car that's already lighter. It has nothing to do with how much crank torque a car has.
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Old May 4, 2003 | 07:35 PM
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?? I disagree, it does matter how much crank torque the car has. Forget the example above, hows this for simplicity. At 1lb(i'd use 0lbs but u can't have 0 as the starting point) torque is has basically no influence on accelleration, other than braking the wind(still a big deal though), correct? Now, lets go to the opposite end of the spectrum. The weight at which a car with 180ft/lbs of torque will NOT accellerate at ALL. Lets say 15000lbs. This is not linear though. This graph would be exponential. The car at 15000lbs wouldn't move a whole hell of a lot faster than it would at 19500lbs. Thats a 4500lbs difference. A car that goes from 6000lbs to 1500lbs would fly by comparison. Now, here is where the argument comes in, and why crank HP DOES MATTER. If the two cars at 6000lbs has the same weight, and HP, but one had twice as much torque, there would be a GREAT difference in accelleration. Now, subtract 4500lbs. The two cars still have the same HP, and car two still has half the torque, but the race would be MUCH MUCH MUCH closer. Of course you can't subtract 4500lbs from any 6000lbs vehicle and still have it running. I simply used large numbers to make the point easier to see. Is anybody following this??

Brandon
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Old May 4, 2003 | 07:42 PM
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Oh, BTW, as for your comparison. Yes, your correct about the weight subtraction, but again, thats obvious. Whats not obvious is how torque plays a role. If i had to place money on a race between a 400HP/400ft/lbs that weighed 4000lbs against a 200hp/200ft/lbs car that weighed 2000lbs my money would be on the heavier car that has double the torque. This car has more torque for braking the wind, thus it will begin to pull as the force required to brake the wind increases exponentially, which it does. I imagine around 60 it would be a car length ahead, and at the 1/4, several ahead. Hey, i could be wrong, somebody else jump in on this!

Brandon
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Old May 4, 2003 | 08:16 PM
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Assuming that two cars of equal ability to accelerate both cut their weight down by an equal percentage, please explain why you think the one with less torque sees a greater benefit from this reduction in weight.

I think they'll both increase their acceleration by an equal amount after the weight reduction.
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Old May 4, 2003 | 08:28 PM
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I figured i'd explained it already, but, in simpler terms.... the weight of the car and torque are directly related, especially for low RPM launches where you don't have much torque. A car with enough torque to get a car moving to a point where the HP generated is able to be used wouldn't bennefit as much from an equal reduction in weight as would a car thats weight is to much for the torque generated. In this case, its the low amount of torque that keeps the RPMS climbing(speed climbs when RPMS climb). This low torque is the limiting factor in accelleration, at least down low. The limiting factor in accelleration of the car with more torque(double say) would be the HP. Here's a good example. A BIG OL FORD truck with a V-10 and 450ft/lbs of torque and 315HP. Put this against a 315HP/200ft/lbs S2000 that weighs close to half a ton less. Up to 30 i can bet the truck would be ahead even though its immensely heavier. As soon as the HP "catches up" with the torque though, its game over and the S2000 will walk all over the Ford. This is getting kinda crazy, and i'm loosing sight of my origonal idea, so i'm going to bed.... night all.

Brandon

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Old May 4, 2003 | 10:06 PM
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Think physics. F=ma. Move things around a little bit and you have a=F/m. Therefore, to determine an object's acceleration you take the force acting on it and divide that by the object's mass. Now consider the slightly more specific case of a car. Its mass is pretty obvious. But ask yourself where does the force acting on it come from? It's in the contact between the tires and the road. Now how can we determine that force? We can take the engine's crank torque and multiply it by the gearing it goes through to get to the rear wheels and ultimately the ground. So whether that crank torque starts off as 50 lb-ft or 20,000 lb-ft, as long as we have 500 lb-ft at the rear wheels, the car will accelerate the same no matter what.

[QUOTE]Originally posted by BrandonS
[B]??
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Old May 5, 2003 | 04:01 AM
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Interesting speculations, but another very important consideration is the torque and HP curve. All that have been mentioned are peak numbers.

I would put my money on a car with lower torque if it could sustain that level over a wider RPM range than a different car with higher peak torque, but that could only sustain it briefly.
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