Diff Gear Change

[driggity]

Originally Posted by Riftur

Ok so will the 4.88 gear change make my 0-60 faster in an AT car?

You'll need an extra shift in order to get to 60 so probably not. 0-54 should be faster though.

[KoolBRZ]

Yes it will. But only if you let it shift automatically. A MT can actually get slower with a 4.88, since it needs to be shifted more often. If the driver can't make speedy shifts, 0-60 times would suffer. On an AT however, the transmission goes between gears so fast that there is no lag time between gears at all.

To say a MT with a 4.55 is about the same as an AT with a 4.88 isn't even close to reality. MT's have a humongous advantage over AT's with gearing. To match 3rd gear alone, one of the most important gears for acceleration, an AT would require a mythical FD of 5.72 to even come close.

Here's a comparison of just 6th gear alone;
Auto
I=3.538, II=2.060, III=1.104, IV=1.00, V=0.713, VI=0.582
0.582 @ 4.10 = 1954 RPM's @ 60MPH
0.582 @ 4.30 = 2049 RPM's @ 60MPH
0.582 @ 4.55 = 2168 RPM's @ 60MPH
0.582 @ 4.77 = 2273 RPM's @ 60MPH
0.582 @ 4.88 = 2326 RPM's @ 60MPH
(0.582 @ 5.404 = 2575 RPM's @ 60MPH) mythical
0.582 @ 5.71 = 2712 RPM's @ 60MPH

Manual
I=3.626, II=2.188, III=1.541, IV=1.213, V=1.000, VI=0.767
0.767 @ 4.10 = 2575 RPM's @ 60MPH
0.767 @ 4.30 = 2701 RPM's @ 60MPH
0.767 @ 4.55 = 2858 RPM's @ 60MPH
0.767 @ 4.77 = 2996 RPM's @ 60MPH
0.767 @ 4.88 = 3065 RPM's @ 60MPH

A 4.55 is the highest ratio I would recommend for a MT, unless you can shift really really fast. A 4.30 is what the new 2017's are going to come with, though a 5.30 is more like what an AT should come with. Weir gears has a 5.29 and a 5.71 available. I'm not sure they would fit into our rear carriers though.

[-Phil]

I haven't seen any mention of them but Mfactory makes some serious 4.44 and 4.67 Rear gears that are rated far past the Oem gears.

[Wayno]

Originally Posted by Riftur

Ok so will the 4.88 gear change make my 0-60 faster in an AT car?

Brochure AT 0-100ph (3rd gear) = 8.2 seconds
Brochure MT 0-100ph (3rd gear) = 7.6 seconds
Brochure AT 0-60mph (2rd gear) = 7.7 seconds
Brochure MT 0-60mph (2rd gear) = 6.7 seconds
AT 4.10 Stg2 98 0-100kph (3rd gear) = 7.3 seconds (auto)
AT 4.10 Stg2 98 0-60mph (3rd gear) = 7.0 seconds (auto)
AT 4.10 Stg2 98 0-100kph (2nd gear) = 7.1 seconds
AT 4.10 Stg2 98 0-60mph (2nd gear) = 6.8 seconds
AT 4.88 Stg2 98 0-100kph (3rd gear) = 6.9 seconds
AT 4.88 Stg2 98 0-60mph (3rd gear) = 6.7 seconds
AT 4.10 Stg2 E10 0-100kph (2nd gear) = 7.0 seconds
AT 4.10 Stg2 E10 0-60mph (2nd gear) = 6.7 seconds
AT 4.10 Stg2 E85 0-100kph (2nd gear) = 6.6 seconds
AT 4.10 Stg2 E85 0-60mph (2nd gear) = 6.3 seconds
AT 4.88 Stg2 E85 0-100kph (3rd gear) = 6.4 seconds
AT 4.88 Stg2 E85 0-60mph (3rd gear) = 6.2 seconds

Note: times are without brake torquing, thus do not rely on tyres, etc.

Short answer: It won't make you noticeably faster on a car that's tuned correctly.

And when comparing higher speed changes like 50-100, the difference is even more negligible, to the point that if you need an extra shift, it may add 0.1-0.2 sec to the measured time.

There's only one reasons to change FD, and that's to change the rpm you are in for certain corners.
- You might have tracks with long corners or daisy-chained corners where you are at 7000+ rpm in 2nd for a number of seconds and are shifting to 3rd mid corner exit, and want to be in 3rd.
- Or you might have super tight, uphill corners where you have to drop below 4000 rpm in 2nd.
- If you have short straights then you should try to gear so you don't need to use 5th, as it's super long.
- Be careful you're not just shifting those problems you have to another part of the track in another gear.

https://docs.google.com/spreadsheets...kT0/edit#gid=0

[Wayno]

Originally Posted by KoolBRZ

To say a MT with a 4.55 is about the same as an AT with a 4.88 isn't even close to reality. MT's have a humongous advantage over AT's with gearing. To match 3rd gear alone, one of the most important gears for acceleration, an AT would require a mythical FD of 5.72 to even come close.

Here's a comparison of just 6th gear alone;

I honestly don't know how you manage to talk so much crap at unsuspecting people all the time.
AT 4.88 almost exactly matches MT 4.56 in the first 3 gears
Similarly, AT 4.55 is roughly equivalent to MT 4.30 in 1-3.
Nobody give a shit about comparing in AT 6th to MT 6th.

[KoolBRZ]

Originally Posted by Wayno

I honestly don't know how you manage to talk so much crap at unsuspecting people all the time.
AT 4.88 almost exactly matches MT 4.56 in the first 3 gears
Similarly, AT 4.55 is roughly equivalent to MT 4.30 in 1-3.
Nobody give a shit about comparing in AT 6th to MT 6th.

What you completely ignored is the SCIENCE behind calculating how much horsepower there is per REVOLUTION of the tire. Force at the rear wheels, not horsepower, determines how fast our car accelerates at any moment.

In the quote I made above I showed the difference between gears in revolutions, not seconds. The comparison I was making was in rpm's.

MT 4.55 3rd gear @ 60 mph = 5723 rpm's
AT 4.88 3rd gear @ 60 mph = 4397 rpm's
MT 4.10 3rd gear @ 60 mph = 5157 rpm's
AT 5.71 3rd gear @ 60 mph = 5145 rpm's

The MT with a 4.55 FD is running 1326 rpm's faster than the AT with a 4.88 FD. That is a huge difference in horsepower per revolution of the tire. Even with an OEM FD of 4.10, the MT has more rpm's than an AT with a 5.71 FD!

You also completely ignored the fact that an AT in automatic mode can shift faster than most MT drivers can shift manually, and that's not even taking missed shifts into consideration.

[gramicci101]

I was bored, so here you go. The X Axis is MPH; the Y Axis is RPM. This is all in Excel, so I can do custom gear ratios and ring and pinion ratios if you would like.

Final drive of 4.10:1 on both manual and auto. Darker colors are the automatic gears. The reason you can't see fourth gear on the auto is because it's behind fifth gear on the manual. They're both 1:1 with a 4.10:1 ring and pinion.




Let's change the ring and pinion on the auto to 4.88:1, because this is a popular thing to do. Manual is still 4.10:1. Darker colors are still automatic.




Edit: Oh, all MPH was calculated using the OEM tire size of 215/45-17.

[KoolBRZ]

Where are you getting your gear ratios from? There's something wrong with the second chart. Here are the actual ratios.
Auto
I=3.538, II=2.060, III=1.104, IV=1.00, V=0.713, VI=0.582
Manual
I=3.626, II=2.188, III=1.541, IV=1.213, V=1.000, VI=0.767

MT with 4.10 in 3rd is 5157 rpm's @ 60 Mph
AT with 4.88 in 3rd is 4397 rpm's @ 60 Mph

[gramicci101]

Those are the ratios I have. Your MT at 4.1 in 3rd at 60 mph is right in line with the chart, but your AT in 3rd at 60 mph is the number for a 4.1 ring and pinion, not 4.88.

[scmil95eg]

Originally Posted by KoolBRZ

Force at the rear wheels, not horsepower, determines how fast our car accelerates at any moment.

Ever read Paul Yaw's comparison between an equally geared Diesel truck vs. a F1 car (torque v. hosrsepower)? Read. Understand. Enlighten.

(link has original pics and charts)
https://web.archive.org/web/20050207...om/tqvshp.html

Most Ridiculous Item of the Day

Have any of you ever watched Bill O’Rielly on the Fox news channel? You know, the guy with the huge ego, stating his opinion as fact to "save the country" from its own evils. Bill has a section of his daily show devoted ridiculous happenings in the world of politics.

If I were to do my own "Most Ridiculous" item in the world of racing it would be based on the following statement. "Horsepower sells motor cars, but torque wins motor races." This couldn’t be further from the truth.

Like it or not, everything that goes on around us is governed by the laws of physics, and these laws are non-negotiable. The good news is that we don’t have to be Einstein to apply the basic laws of physics to racing. The fact that too few do is the reason that such ridiculous statements are common in racing.

Let’s start with a few definitions. Webster’s dictionary describes torque as "A turning or twisting force." Note that the definition does not imply motion. As applied to an engine, it is simply a measure of the twisting force at the crank/eccentric shaft. Torque is normally rated in Lbs.-Ft. Since pounds feet doesn’t exactly roll off the tongue, most of us refer to it as foot pounds.

Notice that there are two terms. Force (In lbs.) and distance (In ft.). At first it may seem strange to describe a "Turning or twisting force." in terms of distance, but a more detailed description makes it clear. If I were to put a shaft in a bench vice, attach a 1-ft. long lever to the end, perpendicular to the shaft, and then hang a one pound weight off the end of the lever, I would be applying one ft.-lb. to that shaft. Notice that the shaft is not rotating even though a torque is applied to the shaft.

If I were to replace the one-foot lever with a 100-foot lever, I would now be applying 100 ft.-lbs. to the shaft with the same one lb. weight. As you can see, the amount of twisting force on the shaft will vary depending on the length of the arm, and that requires that we specify a measure of distance to properly describe the force seen at the center of the shaft.

Let’s say for instance that I pull the shaft from the vice, and ask you to hold it in your hand. If I do this with a one pound weight hanging from the end of the one foot lever, I will be applying a force of one ft.-lb. to your hand, and you will have no problem holding on to it. If I replace the lever with one that is 10 ft. long, with the same 1lb. weight on the end, (For all these scenarios, we assume that the lever itself is weightless.) You will now have a force of 10 ft.-lbs. applied to your hand, and it will be much harder to keep the shaft from rotating, even though you are still only resisting the one pound weight.

This would have exactly the same effect as setting a torque wrench to 10 ft.-lbs., attaching it to the end of the shaft, and applying force until the wrench clicks. Ten ft.-lbs. is ten ft.-lbs. whether it is applied with a one-foot lever and a ten-pound weight, or a ten-foot lever and a one-pound weight. Torque is equal to the weight, or force, times the length of the lever. It’s that simple.

If a particular engine has a peak torque rating of 200 ft.-lbs., that force is equivalent to attaching a one foot lever to the shaft, and hanging a 200 lb. Weight from the end of it. Or…any other combination of weight and lever length which has the product 200.

Notice that I can take you from easily holding on to the one pound weight, to not even having a chance of holding it just by changing the length of the lever. (Like a one-lb. weight, and a 100-ft. lever.) Of course you say, that’s just leverage! Well…you’re right! Keep that in mind, because it is that leverage that makes all the difference, and a gear is in fact just a clever way to apply leverage between two or more rotating devices.

Let’s say that the shaft used in our example is the input shaft of the transmission from a 1993 RX-7 with the following ratios.

1st 3.483 to 1

2nd 2.015 to 1

3rd 1.391 to 1

4th 1.0 to 1

5th .719 to 1

If the transmission is in 4th gear, one complete revolution of the input shaft will result in one complete revolution of the output shaft, just as if there were a solid shaft running all the way through. If we attach a 1-ft. lever to the input shaft with a 10-lb. weight on the end, the torque at the input shaft will equal 10 ft.-lbs. as we have already determined. Since we have a 1 to 1 ratio from input to output, we will also have 10 ft.-lbs. at the output shaft.

If we were to keep the same weight and lever on the input shaft, but switch the transmission to third gear, we would still have 10 ft.-lbs. at the input shaft, but we would now have 13.91 ft.-lbs. at the output shaft. This value is the product of the input torque and the gear ratio. (10-ft.-lbs. times 1.391 gear ratio equals 13.91 ft.-lbs.) If we were to switch the transmission into 1st gear, the result would be 34.83 ft.-lbs. at the output shaft.

As you can see, a gearbox gives us a simple way to vary the torque through leverage, and it is equivalent to changing the length of the lever. Thanks to gears, we can have any amount of torque that we want! In fact, a bone stock 12A making only 100 ft.-lbs. of torque could be geared to pull an 18-wheeler up a steep hill, as long as we are not in any big hurry to get the job done.

Let’s say that it takes 10,000 lbs of force to pull a heavy weight up a hill. No problem! We could even do it with our stock 1980-GS in 4th gear if we are willing to build a custom ring and pinion gear with a ratio of 100 to 1. (100-ft.-lbs. times transmission gear ratio of 1:1 times ring and pinion gear ratio of 100:1 equals 10,000 ft.-lbs.)

If we are using a tire with a diameter of 24", the distance from the axle center to the ground is exactly one foot, and so the force is equal to 10,000 lbs. Remember, the torque is equal to the lever length times the force. If we re-write that formula to solve for force, force is equal to torque divided by the lever length, and so that 10,000 ft.-lbs. at the rear axle results in 10,000 lbs. of force at the tire contact patch.

With this same information, we can also calculate the acceleration rate of the vehicle, but first we need to consider Newton’s second law of motion, which states that "Acceleration is proportional to force." and "Acceleration is inversely proportional to Mass." This law is normally stated more simply as "Force equals mass times acceleration." or F=MA. If we rewrite this to solve for acceleration, we get A=F/M. To find the rate of acceleration for a vehicle, we simply divide the force (In lbs. at the tire contact patch.) by the mass (Total weight of the vehicle in lbs.)

Let’s calculate the acceleration rate of a 1st. gen. RX-7. The engine has a torque peak of 100-ft.-lbs. In fourth gear, the ratio is 1 to 1, and so the torque at the output shaft is also 100-ft.-lbs. The ring and pinion ratio is 3.909 to 1, and so the torque at the rear axle will be (100 times 3.909) 390.9-ft.-lbs. The tire diameter is 24 inches, and so the lever length (Distance from the center of the axle to the ground.) is 12 inches, or one foot. The resulting force at the tire contact patch will be (390.9-ft.-lbs. of torque divided by lever length of one foot.) 390.9-lbs. of force. The total vehicle weight with a driver is 2600 lbs., and so the acceleration rate in G’s (The force of gravity.) will be force (390.9) divided by mass (2600) which equals .15 G’s.

If we do the same calculations for first gear acceleration, we find that the force at the contact patch is 1,436 lbs., and the acceleration rate is .55 G’s. It’s clear that we have used the gears for leverage, with the result being a greater rate of acceleration in 1st gear. Of course you knew that already, but now you know why.

By now it should be clear that the acceleration rate of a vehicle is determined by the weight, and the force at the contact patch, which is the result of the torque output of the engine, and all the levers/gears between it and the ground.

You’re probably thinking that we have just determined the acceleration rate of the vehicle, and even changed it with gearing, with no mention of horsepower. So torque really is the determining factor right? Wrong! We haven’t considered speed.

We can gain acceleration by changing the gear ratios, but we can’t go very fast in first gear, so what’s the point? We have effectively changed the amount of torque available to accelerate the vehicle, but our top speed is limited to about 25 mph.

Confused yet?

Read on.

OK, so we all know what torque is, now let’s get to horsepower.

Referring once again to Webster’s dictionary, horsepower is defined as "A foot/pound/second unit of power, equivalent to 550 foot/pounds per second."

Put more simply, horsepower is a measure of work done over time, or the rate at which work is done.

So now we have another term to confuse things. As if force and distance weren’t enough, we now have time involved, and the shaft must actually be spinning. Why… Well, if you are just standing there holding on to a shaft with a lever and a weight, you are doing no work. If you stand there long enough, you will feel like you are working, but in fact you are doing no such thing. Don’t believe me? Clamp the shaft back in the vice, and you can leave it there indefinitely without having to feed it, add gas, and any other means of supplying it with energy.

Torque, all by itself does nothing useful. In fact, the definition of torque does not even require that the shaft be moving. I am sure that all of you want your car to do something useful, like take you to the movies, or get you around the track before the other guy. In other words, you need your car to do some work, and you want it to do that work in the least amount of time possible.

If I give you a wagon full of cement blocks, and ask you to pull it one mile up a hill, you would agree that I am asking you to do work. If I ask you and your buddy to each pull a wagon full of cement blocks up the hill, you will both be doing the same amount of work. But…If it takes you an hour, and your buddy does it in 30 minutes we have a very different situation. You have both done the same amount of work, but your buddy, by completing the task in half the time, has proven that he can develop twice the horsepower that you can. You both traveled the same distance, and exerted the same amount of force, but the third term in the definition of horsepower, time, was different.

For those of you who are sticklers for details, the force required to pull the wagon up the hill at a steady speed is equal to the weight of the wagon, times the sine of the angle of the hill. If the wagon weighed 100-lbs., and the hill was at a 45-degree angle, the required force would be (Sine 45 Times 100 lbs.) equals 70.7 lbs.

If we were interested in moving the wagon by driving the wheels rather than pulling it by the handle, we could convert the force to torque by dividing the required force by the radius of the driven wheels. Let’s say that we have 6" diameter wheels. That would give a lever length (Distance from the center of the axle to the ground.) of 3 inches, or .25 feet. The required torque would then be (70.7 lbs. times .25 ft.) which equals 17.675 ft.-lbs.

James Watt, who spent the majority of his life perfecting the steam engine, created the term horsepower. He was looking for a way to measure the rate of work done by a horse so that he could make valid comparisons between horses which did most of the work in those days, and his steam engines which he hoped would do most of the work in the future.

Watt found that on average, a horse could lift 330-lbs of coal 100-ft in one minute. He then stated that the power available from one horse was equal to (330-lbs. times 100-ft.) or 33,000-lbs./ft./min. If you divide that by 60 to convert to lbs./ft./sec. you get 550-lbs./ft./sec. Watt called this one horsepower, which leaves most of us wondering why he didn’t call it one watt. I don’t have the answer to that, but I do know that 746.6 watts equals one horsepower. If you ever see an engine rated in watts, (This is still popular in some countries.) you can divide by 746.6 to determine the horsepower. Or, you can tell you pals that your bone stock 3rd. gen. RX-7 puts out One hundred ninety thousand, three hundred and eighty three watts.



So if one horsepower is equal to 33,000 lbs.-ft. per minute, we can rearrange that to say that horsepower equals torque times rpm, divided by 5252. How do we get there?

In the above formula, force and distance are stated in ft.-lbs., and time is stated in RPM, so we need to convert our terms. First we need to express that 33,000 lbs. of force as 33,000 ft-lbs. As you now know, that is equivalent to a 33,000-lb. weight hanging from a 1-ft. lever. Then we need to express the one-foot per minute as RPM.

The circumference of a circle is defined as the diameter of the circle times Pi, which is 3.14159. We have a one-foot lever, so if we were to spin the shaft, the outer edge of the lever would scribe a 2-foot diameter circle. The circumference of a 2 foot circle is (3.14159 times 2) 6.282 feet. If we divide 1 foot by the distance traveled in a complete revolution (1 divided by 6.282) we get .159 revolutions per minute, which is equal to one foot per minute.

So now we have: One horsepower equals 33,000 lbs.-ft. of torque per .159 RPM.

That’s still kind of ugly dealing with just a fraction of an rpm, so we divide both terms by .159 and we get: one horsepower equals 5252 lbs.-ft. of torque per 1 rpm.

This can be rewritten a few different ways that are valuable to us.

Horsepower equals torque times rpm divided by 5252. Horsepower = (Torque X RPM) / 5252

Torque equals horsepower times 5252 divided by rpm. Torque = (Horsepower X 5252) / RPM

RPM equals horsepower times 5252 divided by torque. RPM = (Horsepower X 5252) / Torque

If you know any two of the terms, you can calculate the third. You might also notice that torque and horsepower will always be equal at 5,252 rpm, horsepower will be greater than torque above 5252 RPM, and torque will be greater than horsepower below 5252 RPM. ALWAYS…NO EXCEPTION! Just look at any dyno sheet, and you will see what I mean. If you see a dyno sheet where this is not true, you can be sure that someone fudged the numbers to help sell a product.

Back to the issue at hand, I’m sure that the coal tugging horse and a wagon full of cement blocks probably doesn’t seem all that relevant to your racecar. So let’s look at things another way.

The definition of horsepower includes three terms. Force, distance, AND time, where torque is simply a force applied over a distance. In the case of Watt’s experiment, the force was exerted by the weight of the coal, which was being lifted from the mine. In a car, we are interested in acceleration, not the ability to lift an object. In our case, the force is exerted by the inertia of the vehicle, which resists acceleration.

So now I need to bore you with another definition. Back to Webster’s dictionary, inertia is defined as "A property of matter that causes it to resist changes in velocity." In more simple terms, your car would rather not be accelerated from 30 to 70 mph, and so an external force is required to make this happen. This force comes from your engine.

To accurately describe the acceleration capability of your vehicle, we must consider time. If we just considered force, and distance, we wouldn’t really be saying much about the car. If I tell you that my car can pull a 3,000-lb. weight 100-ft. up a hill, would you be impressed? Certainly not, because I haven’t really told you much. If I told you that I could do it in 10 seconds, while your car needed 15 seconds to do the same job, you might be impressed.

After all, what we are really interested in is the ability to cover distance in a period of time. The distance from the exit of one corner to the entry of the next, or the quarter mile, or maybe even the distance from one stoplight to the next.

If we consider the rate of acceleration, AND miles per hour, we have all three terms included in the definition of horsepower. Time, distance, and force. Force is the rate of acceleration, or the force of inertia. Time is in hours, and distance is in miles.

So now instead of just considering the rate of acceleration arbitrarily, let’s include miles per hour.

And while we’re at it, let’s consider the acceleration rate of two very different motors to illustrate the importance of horsepower, and the absolute irrelevance of torque.

At one extreme we have a Honda F1 motor which revs to 18,000 rpm, makes nearly 800 horsepower, but a measly 281 ft.-lbs. of torque. At the other end of the spectrum we have the Cummins turbo diesel available in the 2003 Dodge Ram which makes a whopping 555 ft.-lbs. of torque, but only 305 horsepower. So which one do you think will accelerate faster?

Everyone that you ask will answer that the Honda F1 engine will accelerate faster. Even your neighbor with the big block who claims that torque is the key to going fast. If torque were the determining factor, the Cummins diesel would win hands down. So what gives?



Let’s calculate the acceleration rate for both engines in a hypothetical 2500 lb. car using the transmission from the 1995 RX-7 with a two-foot diameter tire. Since we know that an F1 car will go 200 MPH, we will gear the car for that speed with both motors.

Starting with the F1 engine which redlines at 18,000 rpm, we need to calculate the required ring and pinion ratio to achieve 200 mph at redline in 5th gear.

First we convert miles per hour to miles per minute by dividing by 60.

200/60=3.333 miles per minute

The tire diameter is rated in feet, so we must convert this 3.333 miles per minute into feet per minute. There are 5280 feet in a mile, so:

3.333 X 5280 = 17,600 feet per minute

If our tire is 2 feet in diameter, the circumference is 2 feet times Pi

2 X 3.14159 = 6.28318 feet per revolution.

Now we divide the feet per minute, by the feet per revolution and we get:

17,600 / 6.28318 = 2801.13 tire revolutions per minute to achieve 200 MPH.

The engine redlines at 18,000 RPM, and in 5th gear the transmission ratio is .719 to 1. To determine the RPM of the output shaft at redline, we take the engine RPM divided by the gear ratio to get:

18,000 / .719 = 25,034 RPM

The output shaft is spinning 25,034 RPM, and we need the wheel to spin 2801.13 RPM to go 200 MPH. To find the correct ring and pinion ratio, we divide the output shaft RPM by the required tire RPM and we get:

25,034 / 2801.13 = 8.937 to 1

Going through all the same boring math for the Cummins diesel which is only spinning 3000 RPM at redline, we get a required ring and pinion gear of 1.489 to 1 to go 200 MPH at redline in 5th gear.

(Note that the F1 engine is spinning 6 times faster than the Cummins, and so the required ring and pinion ratio is exactly 6 times higher.)

With the transmission in first gear, both vehicles will be traveling at 13.76 miles per hour at the bottom of their powerband. (1,000 RPM for the Cummins, and 6,000 rpm for the Honda.) The following chart shows the acceleration rate of both engines in our hypothetical vehicle from that point to 200 MPH.



Note that at any point on the chart, the percent difference in the rate of acceleration is EXACTLY the difference in horsepower. For instance, at 200 mph, the Honda F1 engine is accelerating at a rate of .572 G’s, while the Cummins diesel is accelerating at a rate of .228 G’s. If we divide .228 into .572 we get 2.5, and so the acceleration rate of the Honda is 2.5 times greater than that of the Cummins.

The Cummins, at 3,000 rpm is making 305 horsepower, while the Honda is making 763 horsepower. The Honda is making 2.5 times the power of the Cummins, which is exactly the difference in the rate of acceleration. You can work this out at any point on the chart, and you will find that this direct relationship between horsepower and rate of acceleration always holds true.

I’m sure there is someone out there that still thinks I’m off my rocker, but as I stated earlier, the laws of physics are non-negotiable. After showing this article to a few people for proofreading, one person stated that the results aren’t valid because torque motors are for low rpm grunt, and aren’t meant to run at 200 MPH. Ignoring the fact that this is a ridiculous statement, let’s consider what would happen if we geared both combinations for a top speed of 100 MPH. It should occur to you that this would be a simple matter of doubling the ring and pinion ratio, and that would be correct.

The end result is that the acceleration figures would simply double across the board for both combinations. The difference in acceleration would still match the difference in horsepower, and ultimately the difference in performance would be the same.

So there it is. Horsepower is the determining factor in the rate of acceleration of any vehicle. The next article will go into more detail, and show you how these simple calculations can be used to choose appropriate gearing for any track.

Below are some useful definitions and formulas.

TORQUE IN LBS./FT. = (WEIGHT IN LBS. X LEVER ARM LENGTH IN FEET.)

1 HP = 550 LBS./FT./SEC.

1 HP = 33,000 LBS./FT./MIN.

HP = (TORQUE X RPM) / 5252

TORQUE = (HP X 5252) / RPM

TORQUE AT THE REAR WHEELS = (ENGINE TORQUE X TRANSMISSION GEAR RATIO X RING AND PINION RATIO)

ACCELERATIVE THRUST = (TORQUE AT THE REAR WHEELS / TIRE RADIUS IN FT.)

RATE OF ACCELERATION = (ACCELERATIVE THRUST / TOTAL VEHICLE WEIGHT.)

[KoolBRZ]

Originally Posted by gramicci101

Those are the ratios I have. Your MT at 4.1 in 3rd at 60 mph is right in line with the chart, but your AT in 3rd at 60 mph is the number for a 4.1 ring and pinion, not 4.88.

You might want to recheck that. AT with 4.10 in 3rd @60 mph = 3694, not 4397.

[gramicci101]

You're the one saying 4397 was 4.88:1.


So, automatic third gear is 1.404. Multiply that by 4.1 for the ring and pinion and you get a 5.7564 overall ratio. At 60 mph you go 1 mile in 1 minute, which makes math easy. The OEM tire rotates 819.26 times in one mile, also in one minute at 60 mph. Multiply that by your 5.7564 overall ratio, and you get an RPM of 4715.98.


For a 4.88:1, you have 1.404 x 4.88 x 819.26, which is an RPM of 5613.17. You're wrong on both counts.

[KoolBRZ]

Upon further research, I have been victim of a typo on my reference site. AT 3rd gear is actually 1.404, not 1.104 like I used in my calculations. I apologize. I used the wrong data to come to my conclusions. I also found 3 sites claiming the AT had the exact same 3rd gear ratio as the MT, so it's not hard to get the wrong data. See my signature below for what happens with the right calculations and the wrong data.

[gramicci101]

Originally Posted by KoolBRZ

Upon further research, I have been victim of a typo on my reference site. AT 3rd gear is actually 1.404, not 1.104 like I used in my calculations. I apologize. I used the wrong data to come to my conclusions. I also found 3 sites claiming the AT had the exact same 3rd gear ratio as the MT, so it's not hard to get the wrong data. See my signature below for what happens with the right calculations and the wrong data.

S'ok. The thread I found with gear ratios actually had MT and AT swapped. It took a bit of staring at the graph thinking "that doesn't look right..."