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Old 04-20-2011, 01:58 AM
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rear end differental questions, need knowledge!

ok final drive ive almost got the jist of, like the last gear in the transmission connect to the rear is very important, cause this determines how quick your car can be (acceleration) or top speed. I dont know gear ratios though and what sizes do, so i know the statement, size does matter. Can anyone help me understand the gear ratios and what they can benifit a car in performance?

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Old 04-20-2011, 05:21 AM
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tci has a page of calculators to figure out what you need

http://www.tciauto.com/Content/Stati...kieSupport=1#5
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Old 04-20-2011, 11:30 PM
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Quote:
Originally Posted by chevy302builder18
Can anyone help me understand the gear ratios and what they can benifit a car in performance?
I'll take a stab at it. Nobody has ever taught me what I'm about to say, it all comes from my mind and I might be full of mush.

In a V8 motor, each one turn of the crank (360 degrees) there are 4 cylinders that fire. One cylinder fires for each 90 degrees of crank rotation, so there are 4 firings for each full turn (360 degrees). This tells us that in order for all 8 cylinders to fire, we must turn the crankshaft two full revolutions (720 degrees). I'm just trying to get you wrapped around the basics first.

Any internal combustion engine has a basic operating range where it is effective. This range is mostly determined by the camshaft timing. The motor might operate from idle to 4000, making peak torque at 3000 or it might operate from 5000 to 8500, making peak torque at 6500. Since torque is what moves the vehicle, we try to use a torque converter, gear ratio and tire size that will maximize the torque peak to our benefit.

If we have a 3.73:1 rear gear, that means that the crankshaft will turn 3.73 times for each time the tire turns once. Now, since we know that there are 4 power pulses for each revolution of the crank, we can determine that with a 3.73 gear, the tire will receive 14.92 power pulses for each 1 revolution that it makes (4 X 3.73 = 14.92).

Now, let's pull that gear out and install a 4.56:1 gear. If we follow the same thinking and multiply our players, we find that the tire gets 18.24 power pulses for each one turn of the tire.

Now, let's say that this motor is producing 400 ft/lbs of torque at its torque peak. That would mean that each cylinder is producing 50 ft/lbs of torque per power pulse.

With the 3.73 gear, we can see that we have applied 746 ft/lbs of torque to the tire during one turn of the tire. With the 4.56 gear, we can see that we have applied 912 ft/lbs of torque to the tire during one turn of the tire. Will the car accelerate slower or quicker with more torque applied to the tire in the same amount of time?

At the same time, a shorter gear (numerically larger) will get the motor up to its torque peak sooner than a taller gear (numerically smaller), so that more power is applied to the tire sooner.

That's how I see it. If someone else has a better explanation, I would be very interested in reading it. I learn something new every day and today should be no exception.
2.80, 3.00, 3.23 = tall gear
4.10, 4.44, 4.56 = short gear
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Old 04-21-2011, 03:04 PM
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Quote:
Originally Posted by techinspector1
If we have a 3.73:1 rear gear, that means that the crankshaft will turn 3.73 times for each time the tire turns once.
The crankshaft motion would first need to go through the transmission, which would also apply a scaling to it, e.g. 6th gear OD of a T56 is 0.5, so each crankshaft revolution is only a half revolution at the transmission output shaft, the driveshaft, and differential input shaft, too because there is no gear drive in between.
From thereon, the 3.73:1 of the diff would make 1.865 revolutions at the axle shaft / at the tire.

So as far as for the rotational speed, it will be reduced by the differential gearing. Speaking of rotational force (=torque), the rear end gear ratio is utilized because of it's torque multiplication capability. Numerical example again, look over yonder. Got any Qs just post up.

Never heard of the concept of the power pulses, but then again that doesn't mean it's incorrect

Greetz
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Old 04-21-2011, 04:28 PM
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My explanation was aimed at explaining the relative advantages/disadvantages of different rear differential gear ratios and assumed a 1:1 relationship at the transmission for simplicity of explanation.

When the spark plug fires, a pulse of power is applied to the crankshaft. I don't know of another way to say it when trying to explain it to newbies. If you have a better term, post it and I'll use it.
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Old 04-21-2011, 04:36 PM
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well thats wicked. What if i had 3.73.1 gears, that means 3 to 1, so that would have to displace more torque to both wheels instead of one. Hmm seems good thank you!
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Old 04-21-2011, 06:08 PM
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Quote:
Originally Posted by techinspector1
Now, let's say that this motor is producing 400 ft/lbs of torque at its torque peak. That would mean that each cylinder is producing 50 ft/lbs of torque per power pulse.
TechInspector, that was a lucid explanation - if you don't mind I'd like to copy that into my notebook of such pieces of clarity and use it when someone comes along to the shop and asks about such things - saves me the time of explaining it when they can read it instead and in this case your explanation as written would be better than mine.

I try to understand anything of interest to me down to the micro-details. Concerning the above statement about torque, I am now not sure I understand about each cylinder's role in producing what I understand to be torque, or at least how it is measured.

I always thought torque was a force applied at an instant in time. If I understand your statement about each cylinder producing 1/8 of the rated torque output of the V8, then in order to achieve the full rated torque of the engine, one would have wait until all 8 cylinders have fired (two crank revolutions). It would follow that all that does not happen in one instant of time - it would seem instead to wander into the realm of work done or having moved something over those two revs, which is related to horsepower as I understand it.

The question that I respectfully ask is, if we assume that torque is a force applied at an instant of time, in this example does one cylinder pulse produce a force of 50 lb/ft, or 400? My understanding is that when we put an engine on the dyno, the dyno is measuring torque forces, and mathematically calculating horsepower based on RPM at the instant torque is sampled.

Okay, I'll stop rambling now. Tons of stuff have been written on the relationship between torque and horsepower in automobile engines, not always with consistency hence why I ask at the risk of hijacking the thread.

In any event, I appreciate you taking the time to answer the OP's question at that level of detail.
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Old 04-21-2011, 07:36 PM
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Quote:
Originally Posted by cucumber1949


I always thought torque was a force applied at an instant in time. If I understand your statement about each cylinder producing 1/8 of the rated torque output of the V8, then in order to achieve the full rated torque of the engine, one would have wait until all 8 cylinders have fired (two crank revolutions). It would follow that all that does not happen in one instant of time - it would seem instead to wander into the realm of work done or having moved something over those two revs, which is related to horsepower as I understand it.

The question that I respectfully ask is, if we assume that torque is a force applied at an instant of time, in this example does one cylinder pulse produce a force of 50 lb/ft, or 400? My understanding is that when we put an engine on the dyno, the dyno is measuring torque forces, and mathematically calculating horsepower based on RPM at the instant torque is sampled.
You know, some of this stuff tends to get away from me too. I didn't say I was the sharpest tool in the shed, there are others that I look up to on this site who are considerably sharper than I. But here's the way I see this 50 ft/lbs thing. Let's say the motor was on the dyno and showing 400 ft/lbs. Now, I walk in with my trusty Sawzall and hack away 7 of the cylinders off the motor, so that we have only one cylinder left. Making a stretch and assuming that the motor will still run, will it still make 400 ft/lbs or will it make 50 ft/lbs?
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Old 04-21-2011, 08:28 PM
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Ill write that down also, cause it is very usefull.
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Old 04-21-2011, 08:34 PM
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Quote:
Originally Posted by techinspector1
You know, some of this stuff tends to get away from me too. I didn't say I was the sharpest tool in the shed, there are others that I look up to on this site who are considerably sharper than I. But here's the way I see this 50 ft/lbs thing. Let's say the motor was on the dyno and showing 400 ft/lbs. Now, I walk in with my trusty Sawzall and hack away 7 of the cylinders off the motor, so that we have only one cylinder left. Making a stretch and assuming that the motor will still run, will it still make 400 ft/lbs or will it make 50 ft/lbs?
Yeah, I guess it all comes down to how the dyno samples torque, or perhaps defines what torque is for measurement in the automotive application. Can it distinguish individual cylinder pulses or does it take a composite sample comprising multiple pulses? As I attempt to hammer this through the old Mark 0.0 Brain, it would seem to argue that the dyno samples or measures torque based on a series of cylinder pulses. The illustration of taking 7 cylinders out of the equation (good visual by the way :-) ) would seem to effect the ability to get work done (less work) although at a given instant of time (again, a faulty assumption on my part?) the force applied ('torque') by the individual cylinder is the same. Hmmmmm...

Any dyno engineers or designers out there?
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Old 04-21-2011, 08:37 PM
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this would apply to any engine. v10 would have to have 40 ft pounds every cylinder to make 400 ft pounds on one revolution, v12 would have to have 33. ft. Pounds on every other cylinder to one revolution. Meaning it would take more time to let a V 12 take one true revolution.. This is really intresting this means america has built a better engine- V8
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Old 04-21-2011, 08:39 PM
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this would apply to any engine. v10 would have to have 40 ft pounds every cylinder to make 400 ft pounds on one revolution, v12 would have to have 33. ft. Pounds on every other cylinder to one revolution. Meaning it would take more time to let a V 12 take one true revolution.. This is really intresting this means america has built a better engine for torque- V8
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Old 04-22-2011, 05:34 AM
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I'm not sure how the nascar guys dyno but they tune every cylinder on an engine. I'm told that each cylinder has it's own compression ratio, spark advance and probably porting.
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Old 04-22-2011, 06:28 AM
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what was said about torque being a time instant is true. it has absolutely nothing to do with time. torque is just a rotational force. if it was calculated over time, you would get power instead.
no offense, but i also think that the concept of the power pulses is somewhat questionable. you can't just take an engine and divide torque by the number of cylinders and assume that's the torque each individual produces... since we're working with cross-plane 4-stroke V8 here, this means that for each piston at TDC there must be another one at TDC and two more at BDC.
Due to this fact, the rotational force induced by the power stroke of one cylinder (piston forced downwards to BDC after ignition at TDC) is also absorbed, to a certain extent, by the power needed for the suction (intake) stroke on another cylinder. Furthermore, during the same cycle, two pistons would also be doing an upwards motion, either for the exhaust (pumping) stroke, or the compression stroke, which both take off a certain amount of force from the crankshaft as well.

What i'm getting at is this: looking at one cylinder just doesn't make sense. At least when explaining gear ratios
Again, no offense. If anyone else wants to enlighten me why this would still be a good way to look at it, feel free to teach me.
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Old 04-24-2011, 05:44 PM
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that is true but what about rotating assembly? Wouldnt that turn into energy spinning in place? I mean fifty ft. Pounds of energy on top of each cylinder turning alot of wieght below, doesnt that make any energy?
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