


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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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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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 


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. 


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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I try to understand anything of interest to me down to the microdetails. 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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Ill write that down also, cause it is very usefull.



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Any dyno engineers or designers out there? 


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



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



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.



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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