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crshbndct
 
from http://www.miata.net/sport/Physics/02-Tires-Stuck.html
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Finally, let's examine how engine torque becomes force at the drive wheels and finally acceleration. For this examination, we will need to know the mass of the car. Any equation in physics that involves mass is called dynamic, as opposed to kinematic. Let's say we have a Corvette that weighs 3200 pounds and produces 330 foot-pounds of torque at the crankshaft. The Corvette's automatic transmission has a first gear ratio of 3.06 (the auto is the trick set up for vettes-just ask Roger Johnson or Mark Thornton). A transmission is nothing but a set of circular, rotating levers, and the gear ratio is the leverage, multiplying the torque of the engine. So, at the output of the transmission, we have of torque. The differential is a further lever-multiplier, in the case of the Corvette by a factor of 3.07, yielding 3100 foot pounds at the center of the rear wheels (this is a lot of torque!). The distance from the center of the wheel to the ground is about 13 inches, or 1.08 feet, so the maximum force that the engine can put to the ground in a rearward direction (causing the ground to push back forward-remember part 1 of this series!) in first gear is Now, at rest, the car has about 50/50 weight distribution, so there is about 1600 pounds of load on the rear tires. You will remember from last month's article on tire adhesion that the tires cannot respond with a forward force much greater than the weight that is on them, so they simply will spin if you stomp on the throttle, asking them to give you 2870 pounds of force.
that is in 1st gear, which has 3.06 ratio, rear gear of 3.07, and wheel radius of 1.08 feet. on an engine producing 330 ft-lbs at the crank, it gives you 2870 ft-lbs of force. so if it were to be put on a dyno where the rollers weigh exactly 2870 lbs it would accelerate the roller @1ft/sec^2

for a dyno to give an accurate representation of how much power is being generated at the engine, it needs to have all these factors entered in. but by being able to measure the revs of the engine, it will know the exact reduction ratio of the drive train, and so will know at what revs the motor is turning, and hence can calculate the power and torque at the wheels.

this is why software such as homedyno is just as accurate as a chassis dyno when given all the correct inputs correctly.

usually the dyno machine and operator calculate all this for you.

also actual results on the race track/autox track/strip > dyno results
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Old 08-30-2004, 03:59 PM crshbndct is offline  
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demosh
 
Quote:
Originally Posted by crshbndct
from http://www.miata.net/sport/Physics/02-Tires-Stuck.html


that is in 1st gear, which has 3.06 ratio, rear gear of 3.07, and wheel radius of 1.08 feet. on an engine producing 330 ft-lbs at the crank, it gives you 2870 ft-lbs of force. so if it were to be put on a dyno where the rollers weigh exactly 2870 lbs it would accelerate the roller @1ft/sec^2

- 2870 lbs-ft of torque. lbs-ft are a measure of torque, not force.
- ft/sec^2 is a measure of linear acceleration, not rotational acceleration. Rotational acceleration is measured in radians per second squared.

You are jumping between rotational and linear motion.
However, that article seems pretty solidly written.
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Old 08-30-2004, 04:07 PM demosh is offline  
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#167  

ballorfall
 
wow... so much off-topic lol but anyways... lol no i dont have a fart can but i will maybe put a catback on it next summer... dunno yet.... and i swear my rotor isn't crack... i know it looks like it but it isn't...
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Old 08-30-2004, 04:14 PM ballorfall is offline  
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#168  

SFC
 
Quote:
Originally Posted by demosh
NOT AT THE FRONT WHEELS!

It theoretically makes 300 lbs-ft at the engine.


dude, it's AT THE FRONT WHEELS... there is no "Drivetrain" at the engine... wtf@yuo, how do you not get this? Dyno's are taking into account DRIVETRAIN LOSS! There is none at the engine, there is "at the wheels" that's why it's WHEEL HORSEPOWER and WHEEL TORQUE.


and furthermore a dyno could never get 425lb/ft of torque out of tex's setup because you aren't entering in your gear ratio...

"the HP in any gear would be the same, because while the torque is increased proportionally, the speed is decreased proportionally, so the power would otherwise stay the same."

"But the torque at the dyno and torque at the engine are 2 different things, and I think this may be where some confusion is. The dyno measures the applied torque at the drum, which is higher in first gear. But the speed at the drum is lower, so (again ignoring the losses) the HP *should be* the same. But then the dyno is also measuring the engine speed (RPM), so it takes the dyno measured power, then goes back and calculates the "engine torque" from the engine speed and dyno HP."


getting it yet? Engine tq with driveline loss==wtq.

Last edited by SFC; 08-30-2004 at 04:48 PM..
Old 08-30-2004, 04:30 PM SFC is offline  
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#169  

demosh
 
Quote:
Originally Posted by SFC
dude, it's AT THE FRONT WHEELS... there is no "Drivetrain" at the engine... wtf@yuo, how do you not get this? Dyno's are taking into account DRIVETRAIN LOSS! There is none at the engine, there is "at the wheels" that's why it's WHEEL HORSEPOWER and WHEEL TORQUE.

No it isnt at the front wheels. It is a theoretical value of torque at the engine, calculated with measurements of torque at the wheels.

Do you even know what torque is? Do you understand the principle of leverage? Gear reduction?

Lets leave the whole car out of this. Assume there are two shafts, connected with gears. Say the gears are set up such that the ratio is 10:1, so when you spin shaft A 10 times, shaft B makes one complete revolution. So shaft A spins ten times faster than shaft B.

Now let us say there is absolutely no friction. The shafts ride on perfect bearings, whatever. So if you apply a certain amount of power to shaft A, you get the same amount of power out of shaft B, since there is nowhere for it to go - there is no friction to overcome, right?

Now say the power you are putting in at shaft A is 1 hp, and it is spinning at 5250rpm. Now we agreed that for every 10 revolutions A makes, B makes one - so shaft B is going to be spinning at 525rpm. Simple mathematics, right?

Now, since there is one hp being put into shaft A, and we agreed that there is nowhere for it to go but shaft B, there must be 1hp coming out there. Shaft B spins at 525rpm because of the reduction. Lets plug this into the equation you posted, that describes the relationship between horsepower, torque, and rpm.

hp = torque * rpm / 5250

1 = T * 525/5250
1 = T * 1/10
T = 10

So torque on shaft B is 10 lbs-ft. This is something that should have been obvious without a lengthy example.

Now going back to cars, shaft A would be the engine crank, and shaft B would be the wheels. As we all know, power is lost in the drivetrain, so more power is being put into shaft A than comes out at shaft B.

Take some care with a 4th gear that is 1:1, and a final drive ratio of 3:1. So in this configuration, one revolution of the wheels means three revolutions of the engine crankshaft. Say we measure that the torque on the wheels is 300lbs-ft. In an ideal world, in one without driveline losses, with these particular gear ratios, there would be 100lbs-ft on the engine crankshaft.

But this isnt an ideal world. It takes power to overcome friction. Some torque is needed from the engine to fight against friction torque from bearings, gears, all that stuff. So based on our measurement of 300lbs-ft of torque on the wheels, we can deduce that there is 100lbs-ft on the engine crankshaft, plus whatever is needed to overcome the friction in the driveline.

Thus in conclusion, there is 300lbs-ft being applied to the wheels, and 100lbs-ft + x lbs-ft at the engine.
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Old 08-30-2004, 04:54 PM demosh is offline  
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#170  

Fear Night
Speed Limit?!?
 
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This argument is funny.
Old 08-30-2004, 04:56 PM Fear Night is offline  
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#171  

prak
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I thought this thread would have died after I left the office.

Old 08-30-2004, 04:56 PM prak is offline  
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#172  

LagPenguin
 
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Why can't my saab take pro stocks!?
Old 08-30-2004, 04:56 PM LagPenguin is offline  
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#173  

Why_Ask_Why
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I make Gen[M]ay notable
 
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my tiburon makes 10000lb ft
honestly, I swear it does and this proves it: eleventy^3 x 42 - canada's gnp/ mean sum of superman's flight speed= engine tq
Old 08-30-2004, 05:14 PM Why_Ask_Why is offline  
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#174  

demosh
 
Quote:
Originally Posted by LagPenguin
Why can't my saab take pro stocks!?

Because it only manages to put out a few hundred hp.
5250 lbs-ft of torque at 1rpm is just 1hp.
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Old 08-30-2004, 05:17 PM demosh is offline  
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#175  

SemperFly
 
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Quote:
Originally Posted by demosh
NOT AT THE FRONT WHEELS!

It theoretically makes 300 lbs-ft at the engine.
No
Old 08-30-2004, 05:53 PM SemperFly is offline  
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#176  

Why_Ask_Why
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Quote:
Originally Posted by demosh
Because it only manages to put out a few hundred hp.
5250 lbs-ft of torque at 1rpm is just 1hp.

how does one hit 1rpm in a car?
Old 08-30-2004, 05:58 PM Why_Ask_Why is offline  
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#177  

SFC
 
all I can wonder is after all this demosh.. how you can fail to understand a dyno measures WHEEL numbers... this isn't an ideal world, there is driveline loss, the dyno does NOT take that into consideration, it spits out what WHEEL TORQUE is, not engine torque. Fucking a, seriously, call dynojet. You obviously didn't read the page I posted, and if you did and are still arguing then you're a lost cause. Call them up, argue it with their engineers, I have a feeling even after THEY prove you wrong you'll still say "well their engineers don't know common middle school physics."
Old 08-30-2004, 06:08 PM SFC is offline  
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#178  

LagPenguin
 
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I'll go with the bandwagon on this one, #teen magazines, race engineers, and dyno techs can't be wrong. But hell, for the sake of argument I've got a dyno instructor at my school who has done nothing but dyno cars for the past 40 years, perhaps I'll bring this up to him.
Old 08-30-2004, 06:12 PM LagPenguin is offline  
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#179  

nonhuman
 
Quote:
Originally Posted by FlyNavy
No
Yes.

Gearing changes the amount of torque applied to the wheels. Think about it. If torque was always the same, then why wouldn't you always start from 5th gear?

rwtq isn't a measure of torque applied to the wheels; as demosh demonstrated, simple math shows that gearing changes torque. Instead, rwtq is a measure of torque the engine produces after drivetrain loss is factored in. If a dyno shows that an engine produces 200 ft/lb at 3000 rpm, where would you find that torque? Definitely not the wheels, since those are most likely not spinning at 3000rpm. The only thing spinning at 3000rpm is the crank and associated bits. Thus, it's engine torque that's being calculated.

Of course, demosh is a for nitpicking about this in the first place. But his argument still stands.
Old 08-30-2004, 06:14 PM nonhuman is offline  
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