How does vehicle weight effect a transmissions torque rating? To put it another way, why is GVW included in transmission specs?

 

Because 300tq in a 4000 lb car puts more stress on the transmission than 300tq in a 3000 lb car.

 

I appreciate your quick answer, but isn't 500ft/lbs just that, 500ft/lbs? or not? I really don't get it. I understand that a 4000lb car has more mass and takes more power to get to the same velocity as a 3000lb car, but if the engine puts out 500ft/lbs max, then shouldn't it really be up to the tires to decide whether to put the power down or just to spin?

 

Punch a pillow as hard as you can, and then do the same to a brick wall. Ensure you apply the same amount of force to each one. Discuss your results.

 

hahaha. Yeah, I'll try that tomorrow... When I'm cruising along, the torque put on the drivetrain is really small. If I were to dump the clutch (as your example implies) then the full force of the engine would be put onto the drivetrain within a very short time. But if the transmission is rated at 500ft/lbs and the engine puts out 500ft/lbs, what's the problem? Do you guys know any of the math involved? I've got a math brain so I can't let this go unless I have proof. Do you know how I would go about converting from a published spec to my own specs?

 

Ok, so here's my thought process. If I were to connect the output shaft of my transmission to an immovable object, then the forces would tear the transmission apart. But why? Does the immovable object send torque back through the tranny the opposite way? Actually, wouldn't that just stall out the engine?

The only reason I can come up with is that there is somehow a lot more torque getting to the transmission. But how? Where does it come from? If I were to slowly let the clutch off when the tranny was connected to the immovable object, then I would give it more power and the clutch would keep spinning. If I were to keep going, there would be a point where the clutch would engage and the engine would bog down and stop. Right? My instinct tells me that if I were to dump the clutch with my foot all the way down on the gas, it would break. I just don't get it. I keep thinking the mass of the engine spinning adds a lot more torque. I dunno. Any physics majors out there?

 

Newton's third law.

 

Ok. Whatever. If you guys don't know, I'll just figure it out by myself. If anyone else out there cares to know, I'll be glad to explain it once I figure it out.

 

Um..he just gave u the answer you were looking for. Your over thinking this. If the vehicle is heavier, there will be more stress on the transimision getting it to move. If the engine is putting out 500ft/lbs and the tranny is rated for 500ft/lbs, but the vehicle is grossly heavy, there may well be more than 500ft/lbs of stress on that tranny trying to make it go.

Another way to look at it. Get on a bicycle and put it in the lowest (read easiest to pedal gear), and try to accelerate. Easy right? Now put it in the highest (read hardest to pedal gear) and do the same. Suxs doesnt it? The high gear acceleration represents moving a heavier object. The stress on you represents the stress on the transmission of the car. Basically the heavier the car is, the more stress is put on the transmision, so basically if you have two cars with equal power outputs but one is 1000lbs heavier than the other, the heavier ones transmision will die sooner unless it is strenthened in some way. Hope all that helps.

 

Wow. looking back at my post, I was an ***. I didn't mean it that way. Sorry.

Yeah I understand the concept, but I need to figure it out mathematically to be content. Maybe I am over thinking it, but I can't live with a basic answer. Anyway, I'm thinkin' the second law (F=MA) has more to do with it than the third, though. For the third law: the engine puts out it's force which is equal to whatever the combustion creates plus the force of the rotating mass of the engine which is dependent on it's velocity. The car exerts the same amount of acceleration and force on the engine, but in the opposite direction. Which is why if you slowly accelerate, nothing breaks, but if you accelerate really quickly, stuff breaks. So there is double the force on the transmission. Half coming from each end.

In order to find out the actual force the engine puts out, you would need to know the force of the combustion, the mass of it's rotating and reciprocating parts, and it's velocity. Then you would need to know how fast the transmission is being brought up to the speed of the engine to find out the acceleration, and multiply it by the mass and add it to the force of the combustion. I think.

But then I found a F=MA calculator and plugged in the specs of a t56 to see what the maximum acceleration that it could take was and it was a dismal 13.34ft/s^2. Which is about 9mph/s. So that ain't right.

I dunno. I'll keep lernen and thinking and hopefully within a day or two, I'll be able to understand it thoroughly.

 

Yea idk about mathematical formulas. Math was never really my strong suit. Hands on and technical things where. Hope you find the answer your looking for.

 

Mr.Awesome, you dared ask a question I've always asked myself, and look at the replies you got.. You have my sympathy.
I've had a feeling that the GVW specification has something to do with continous duty power handling capability but I don't know. Accelleration is one thing, but hill climbing is another. Oil coolers will help but maybe there's more to it.
Please post your thoughts & findings, I'm very interested to know what this is really about.

 

Climbing a hill is the same as accelerating at a constant rate.

 

Certainly, except that accelleration neccessarily has to end within a specific (short) time span, whereas hill climbing has no specific end (depends on where you live) and the same goes for towing btw..

 

Actually, unless Apeiron and I are misunderstanding you, Apeiron is right. The formula for acceleration is change in velocity over time. So as long as you keep your velocity constant, the numerator will be 0, which will give a solution of 0, no matter the length of time. I suspect climbing a hill takes more power because of Earth's constant acceleration of 9.8m/s^2. To overcome that acceleration, you must constantly accelerate at 9.8m/s^2. So the normal formula (F=ma) turns into F=m(a+g). I'm pretty sure. It makes sense to me.

 

If you're trying to drive up the side of a vertical wall.

 

Haha. Yes. True. I forgot to add that part in. I just figured it would be easier to picture. To clarify, the gravitational acceleration (g) is the same no matter what, but the angle at which you are traveling will cause g to be distributed differently.

The following statements may be wrong! I am just using it as an example.
If you are traveling on flat pavement, all of g is pulling th car to the road (down) If you are traveling at an inclination of 45 degrees, half of g is pulling the car to the road while the other half is pulling it backwards. If you are traveling at a declination of 45 degrees, half of g is pulling the car to the road while the other half is pulling it forward (your direction of travel).

 

If you're on a 45 degree incline, 100% of the force of gravity is pulling directly down, none of it is pulling backwards.

 

Well, gravity will always push straight down, but on a 45 degree slope the car would effectively feel 6.9m/s^2 sticking it to the slope and opposing the climb up the hill rather than only the 9.8m/s^2 sticking it down on no slope.

I'm thinking of going by power dissipation with the transmission ratings. If you were to hold the tires still no matter what, the trans. would have to dissipate all of the engine power. If you let the tires freewheel, the trans. now dissipates very little engine power since it is mostly turned into motion.

 

Yeah. Thanks for that wonderful insight. Too bad I already said that. I chose my words carefully. Using "to the road" instead of down (except in the case of on a flat surface). I was clear.

I get the feeling that you are always right. No matter what anyone says, they are wrong. What better position than moderator for such a perfect person. I will cease posting on this thread as long as Apeiron keeps posting. He has sucked the fun out of a good question and has made the attempt to find the answer ridiculously arduous. Nor has he contributed anything. If anyone has any questions, or an answer to mine, pm me.

 

ok, look, mr awesome, its as simple as this.
the heavier the car, the more mass it has.
the more mass an object has, the more potential energy it takes to get it moving.
that energy can be expressed as power, or more specifically, torque.
a 3500 lb camaro accelerating from a dead stop at a moderate pace, is about the same as a 5000 lb H2 accelerating from that dead stop at a slow pace, which is the same as a 1900 lb lotus elise taking off at full speed...

what it comes down to is that the more energy you are asking the transmission to absorb, the stronger it has to be.
the torque ratings of transmissions are approximate.
obviously, if you are using a transmission rated for 350 ft-lbs behind an engine that puts out 300 ft-lbs in a 2000 lb car, you should be ok...
but if that transmission is behind a 300 ft-lb engine in a 7000 lb truck, you are going to have problems.





as for towing and climbing hills, yes, it is the same as accelerating in theory.
the normal force on the tires is: vehicle weight x cos(theta)
where theta is the angle of incline in degrees.

this is obvious by the fact that a car on flat pavement has a theta of zero, and cosine of zero is 1.

how does this matter?
the force pushing down the hill is: vehicle weight x sin(theta)

so we can see that if a car is on a 45* incline, and the car weighs 3000 lbs, there will be 2121 lbs pushing down on the ground and 2121 pushing the car down the hill.






go get yourself a physics book and read it...
it will all make sense in time
----------
how do you figure?
just wondering, because this makes no sense...

if you mean that gravity is pulling down, then duh, gravity always pulls down.
if you mean that the component of reaction force in the y direction from the tires on the ground is zero, then that is also not correct.

the reaction force is: weight x cos (theta)
as i said above