RPM's and max HP output

The graphs I've seen so far here aren't quite useful for figuring out the shift points. You really have to plot the rear wheel torque (RWTQ) vs speed, in each gear. Here is a quick and dirty example I made using a dyno sheet from MM&FF. The origonal dyno numbers I used are from the red headed step child series, the baseline dyno for a 1996 GT mustang with plenty of miles. I threw the numbers in excel and did the calculations for each gear. I chose this example because it should be very similar to my 97 GT. ;)

Where the lines cross indicate your ideal shift point. You can look at the mph of the shift on the graph, but you have to look at the data set to find the exact engine rpm.

From this example, here are the ideal shift points:
1st to 2nd: 5450 rpm ~ 36 mph
2nd to 3rd: 5150 rpm ~ 58 mph
3rd to 4th: 4900 rpm ~ 82 mph
4th to 5th: 5150 rpm ~ 115mph

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Same idea different torque curve. I do not have the best of torque curves, not sure where I got it. This is for '00 PI engine and shows shift points at a higher MPH. This should be similar to my '00 GT.

Edit: This is the same torque curve wheel torque as in my graph above but I scaled it so the peak torque numbers would be similar to yours.

Stan

ab-whtrq.gif
 
i know this is an old thread but i just saw a bunch of short shifting info being passed out here so i thought i would give my angle.

i cant believe that the guy above with VT stage II's is shifting by 6000rpms

i have about every proven mod but still have stock heads. i am running VT stage I's with 4.10's and i dont shift til 6500rpms.
have done that for the last 6yrs. now it is not my daily driver but i put a little less than 5000miles a year on it..............but almost everytime i drive the car is driven hard and to 6500rpms.
and yes i am all stock valvetrain.
i have not been to the track but i beat anything that runs higher than 12.3
i live in mo so we are at about 1000ft above sea level. (i hate smartazz responses from people at sea level about they run better).

and just so you know i ran across this searching for the highest somebody is running reliably since i will be getting twisted wedges soon and might want to shift a little higher as the power band moves up a little
 
i know this is an old thread but i just saw a bunch of short shifting info being passed out here so i thought i would give my angle.

i cant believe that the guy above with VT stage II's is shifting by 6000rpms

i have about every proven mod but still have stock heads. i am running VT stage I's with 4.10's and i dont shift til 6500rpms.
have done that for the last 6yrs. now it is not my daily driver but i put a little less than 5000miles a year on it..............but almost everytime i drive the car is driven hard and to 6500rpms.
and yes i am all stock valvetrain.
i have not been to the track but i beat anything that runs higher than 12.3
i live in mo so we are at about 1000ft above sea level. (i hate smartazz responses from people at sea level about they run better).

and just so you know i ran across this searching for the highest somebody is running reliably since i will be getting twisted wedges soon and might want to shift a little higher as the power band moves up a little

Shift at 6500 many times and you're gonna float a valve. I guarantee it. Seen it happen a many a times.

Also, it's funny to see my replies in this thread from 2 years ago and some of the "bad" advice that I was getting, and how much I've learned over the years. You need to shift based on your horsepower curves, NOT your torque curves.

And to the above poster ^^, I can almost guarantee you that shifting at 6500 rpm puts you out of your optimal horsepower band. A friend of mine's 01 Bullitt has EVERY possible bolt-on and stage II Hitechs, and his optimal shift point is something around 6250 rpm (and that's with the Bullitt intake manifold). I can't say for sure about your car as I haven't seen your dyno sheet, but from seeing other cammed 2 valve dynos I'm pretty sure of it. Even if it doesn't, the marginal horsepower gain from doing that is nowhere near enough to justify the chance at floating a valve.

To answer your last question, I don't know what the limits of the stock rotating assembly are, but your first obstruction is the stock valve springs. Upgrade to some Comp. valve springs and your top end will be good to 7k+.
 
I guess im the exception, i **** at 6800 and take to 7200 on dyno, but a built motor making 18lbs of boost is the exception.. sorry to carry away from the topic but i couldnt help it :nice:
 
It is possible to graph the rear wheel torque in each gear vs speed. You have to calculate the speed in each gear at each rpm, and then multiply the torque number on the dyno by the gear ratio for each rpm. After each gear is plotted out, you should see where the graphs of each gear cross. The graph of 1st gear should cross the graph of 2nd at some point, and that rpm is where your ideal shift is for that gear is.

Does all that make sense?

You guys are doing this the hard way. Yes, multiplying the torque curve by the tranny gear ratio, and seeing where it crosses the next gear's torque curve will give you the optimal shift point. However, you can save yourself a lot of work by looking at the HP curve. When the hp curve drops off, you simply shift where the next gear will be making the same power. The next gear will be on its way up towards peak power, while the gear you're in will continue to drop off.

It's 2 ways of getting at the same thing. Both methods will give you the exact same answer. Mine will take you about significantly less time to calculate, and you don't have to worry about torque multiplication; looking at power already factors that in for you.
 
Shift your car 500rpm pass peak HP. Put together a spreadsheet of the DA, 60ft, 330ft, 1/8 mile ET and trap speeds, 1/4 time and trap speeds of all your runs. Once you dial in the 60ft and have a consistant feel of the car, play with the different shift points by lowering/raising your shifts based on peek power, but make sure you factor in the DA each time you visit the track. I have a little safe box right next to my computer with probably 40 track passes from 200-775hp cars that I have owned in the last 4 years. I always look back and track the progress.

On my PI cars, I always shifted around 5800-6000rpms. Good luck.
 
You guys are doing this the hard way. Yes, multiplying the torque curve by the tranny gear ratio, and seeing where it crosses the next gear's torque curve will give you the optimal shift point. However, you can save yourself a lot of work by looking at the HP curve. When the hp curve drops off, you simply shift where the next gear will be making the same power. The next gear will be on its way up towards peak power, while the gear you're in will continue to drop off.

It's 2 ways of getting at the same thing. Both methods will give you the exact same answer. Mine will take you about significantly less time to calculate, and you don't have to worry about torque multiplication; looking at power already factors that in for you.

You are ignoring the fact that you accelerate harder in 2nd than 3rd because 2nd is a lower gear. It doesn't matter if the horsepower number is falling or not, the amount of torque from the engine * transmission gear ratio is the relevant number. I'm not saying you are wrong, I'm saying your method isn't necessarily correct.

Horsepower is simply torque times rpm. 200 lb-ft is 200 lb-ft, but 200 lb-ft at 3000 rpm is only 114 hp, while 200 lb-ft. at 6000 rpm is 228 hp. Despite what I stated in an older post on this thread, you will accelerate the same at 3000 rpm and 6000 rpm if the torque is the same at both rpms. Regardless of the horsepower number. And because of the multiplication of torque by your transmission, you will accelerate harder at 6000 rpms in 2nd than you will at 3000 rpms in 3rd.

IOW, stay in gear as long as possible as a rule of thumb. Use a dyno chart and your transmission gear ratios to verify this.
 
You are ignoring the fact that you accelerate harder in 2nd than 3rd because 2nd is a lower gear.
No I'm not. You will not be accelerating harder in 2nd gear if you're making less power than you could be making by shifting to 3rd. Work out the calculations your way and you'll see what I mean.

It doesn't matter if the horsepower number is falling or not, the amount of torque from the engine * transmission gear ratio is the relevant number.

While relevent, you're wasting your time with extra math that isn't necessary to arrive at the exact same conclusion. HP is equally as relevent, but it's also a more efficient approach.

I'm not saying you are wrong, I'm saying your method isn't necessarily correct.

It is exactly correct and also identical to your method... try it out for your self if you need justification. All you want to do is maximize average hp over the entire run. So you shift when hp at the shift point in one gear is identical to hp in the next gear, or when you reach your RPM limit, whichever comes first. If you're unclear about the method, I'll try to explain more articulately.

Horsepower is simply torque times rpm. 200 lb-ft is 200 lb-ft, but 200 lb-ft at 3000 rpm is only 114 hp, while 200 lb-ft. at 6000 rpm is 228 hp.

If your shift was actually going to drop you from 6k to 3k, which is in itself unrealistic, then by my method you wouldn't shift because you're still making more power at 6000 than you are at 3000. I don't need the details of the amount of torque you're making or what the multiple of that by your tranny gear is.

Despite what I stated in an older post on this thread, you will accelerate the same at 3000 rpm and 6000 rpm if the torque is the same at both rpms. Regardless of the horsepower number.
I know, in the same gear, of course.

More interestingly and to my point, if you're making 220hp at 6000 RPM in 1st gear (say, 3:1), and by shifting to second (say, 2:1) you'd be at 220hp at 4000 RPM, then you'd have equivalent acceleration in either gear, thus it would be your optimum shift point, assuming as in any normal engine going beyond 6000 would hurt your power output while going above 4000 would increase power output. Those gear ratios are not just arbitrary, they are pretty close to a T5's actual ratios, and they would result in that exact drop in RPM.

And because of the multiplication of torque by your transmission, you will accelerate harder at 6000 rpms in 2nd than you will at 3000 rpms in 3rd.

In your specific example, yes.

IOW, stay in gear as long as possible as a rule of thumb. Use a dyno chart and your transmission gear ratios to verify this.

Bad rule of thumb. It doesn't apply to the anemic stock 5.0 HO pushrod motors, for example, which ran out of breath and had an optimum shift point of about 5500 in 1st, 5250-5300 in 2nd, around 5150 in 3rd, and around 5300 again in 4th, even though redline was upwards of 6000 rpm and the limiter was at 6250. I could spin my billet internaled 331 to 7k+ RPM all day, but my street cam causes a rapid drop off after 6300 RPM. Spinning it over 7k would be senseless.

Chris
 
^ for starters.

Ideally, you want to use dyno numbers to calculate shift points to include a curve with the largest area under it.

I thought this was an interesting and correct point you made. Did you take calculus? What really is the area under the torque curve on a dyno chart? Since your X-axis is RPM and your Y-Axis is Torque, then the area under the curve is torque*RPM. What does that sound like? If you said "power" then you hit it right on the head :nice: The point here is that maximizing power is the same thing as maximizing the area under the torque curve on a dyno graph.

Chris
 

It's funny, about a week ago I read through some threads on another forum about this very thing, and you are wrong, at least on a couple of points.

You are ignoring the fact that you accelerate harder in 2nd than 3rd because 2nd is a lower gear.

You are sort of right here, but not exactly, and I don't know exactly how to explain it. Yes, our cars accelerate harder in 2nd than 3rd (at least to a certain rpm), but if a lower gear ratio (i.e. more torque multiplication) accelerates harder, then why don't cars come with much steeper transmission and rear gear ratios?

Coming from an agricultural background, I can tell you that on a 300 horsepower/2500 ft-lb torque tractor, a first gear with a top speed of 1.1 mph has almost NO acceleration. However, a higher gear has much greater acceleration, even though there's less torque multiplication from the gearing. There's some reason for this, although I don't know exactly.

Despite what I stated in an older post on this thread, you will accelerate the same at 3000 rpm and 6000 rpm if the torque is the same at both rpms. Regardless of the horsepower number.

Contrary to popular belief, this is absolutely not true, ever. Don't believe me? Go ride in a nitrous car (which makes peak torque at whatever rpm you begin spraying it at) and tell me if it accelerates the hardest at 2500 rpm or at 5500 rpm. I can tell ya, my car makes peak torque at about 3000 rpm, and peak horsepower at 5500. It accelerates MUCH harder at 5500 than 3000, and that's not an opinion. It's just the way it is. You accelerate the hardest at peak horsepower, and the horsepower is directly related to the acceleration.

And because of the multiplication of torque by your transmission, you will accelerate harder at 6000 rpms in 2nd than you will at 3000 rpms in 3rd.

Once again, I don't know exactly how this works. You accelerate according to your horsepower, but the gear multiplication obviously has something to do with it as well. BUT, too much gear multiplication isn't a good thing either, as demonstrated with my tractor example above. So I'm not exactly sure how you are supposed to use the horsepower graph (in combination with gear ratios) to determine shift points.
 
I didn't read all the details posted above, but there is a rediculous amount of mis-information. I am not here to get into a pissing contest, but rather try to clear up the confusion and actually have some technical content in this thread.

Acceleration is based on the amount of torque you are applying to the rear wheels. Take engine torque, then multiply by the transmission gear ratio and differential gear ratio. For example, the average 4.6 makes around 300ft*lbs of torque. In first gear (at peak torque, around 3500rpm), you get 300 * 3.35 first gear * 3.73 rear end = 3750 ft*lbs of torque (ignoring inertial and frictional losses). In 4th gear, you get 300 * 1.0 trans gear * 3.73 = 1120 ft*lbs at the wheels. This is why accelerating at 3500rpm in 1st feels so much stronger than accelerating 3500 in 4th. It is torque that provides acceleration.

Power is simply a function of rpm and torque. power = torque * rpm / 5250. The higher the rpm you can make good torque, the higher power is.

In a given gear ratio, the maximum acceleration is AT the peak torque. This is 100% true. However, if you are at a certain speed and you want to maximize your acceleration, you want to choose the gear that keeps you at peak power. In a continuously variable transmission, this is easy... it keeps you at peak power always.

Here is a tractive force graph I made for my 97 GT. It is the engine torque multipled by gear ratios, then subtracting out the losses. Where the lines cross represent the ideal shift point. You can make a good guess by taking peak power and adding 500rpm, but this is exact. The ideal shift point depends on the jump between gears. For example, the jump from 1st to 2nd is large, so you want to rev up higher on the 1-2 shift than the 3-4 shift.
97GTTractiveForce.webp
 
I didn't read all the details posted above, but there is a rediculous amount of mis-information.

I can't find fault with the information as you've laid it out, except that going through all of that is not the easiest way to arrive at the answer. Neither will you be able to find fault with anything I've stated in this thread. As arrogant as that sounds, this subject is so old it's no longer interesting.

Acceleration is based on the amount of torque you are applying to the rear wheels.
Hate to be a stickler, but it pays to be in physics. It's actually based on the linear force applied to the ground by the tires.

Take engine torque, then multiply by the transmission gear ratio and differential gear ratio. For example, the average 4.6 makes around 300ft*lbs of torque. In first gear (at peak torque, around 3500rpm), you get 300 * 3.35 first gear * 3.73 rear end = 3750 ft*lbs of torque (ignoring inertial and frictional losses). In 4th gear, you get 300 * 1.0 trans gear * 3.73 = 1120 ft*lbs at the wheels. This is why accelerating at 3500rpm in 1st feels so much stronger than accelerating 3500 in 4th. It is torque that provides acceleration.
Now, please do your calculations on the example I laid out before and verify my point. 1st gear of 3:1 at 6000 rpm with an engine that produces 220 HP there would be making identical torque at the rear wheels as the same engine in a 2nd gear of 2:1 at 4000rpm where the engine also produces 220hp. I haven't even worked out the math myself, but I know it's true... Like I said, old knowledge. When I was a student in college, I had never seen nor heard a good answer on how to determine optimum shift points. My first thought to determining optimum shift points was as you have done. I made the graphs. I even created an excel spreadsheet that does a damned good job at estimating quarter mile times based on physics formulas using the idea. I realized as I was doing this work how much more quickly you arrive at optimum shift points when you simply compare the hp at a high rpm in a gear and compare it with the hp at the same vehicle speed in the next gear. You need not multiply torque curves and overlay them. Trial and error using the hp method gives you an answer in about a minute...

After completing a project on this, I verified the results with guys who had made hundreds of trial runs to determine their shift points and the results were always nearly spot on.

I would add one thing of note. If you want a super accurate model to compute using either of our methods, you should consider a couple more things. First, the efficiency changes from one gear ratio to the next. This will slightly offset every gear other than the one that the dyno was actually performed with. This should have an impact, albeit small, on the optimum shift point. Shifting makes a difference too. Powershifting, and shifting done in automatic transmissions is going to result in a different shift-point from that of a granny-shifted car. The granny shifted car's optimum shift point will be slightly higher than when powershifting the same car. This difference is actually significant enough to differ, I'd guess, by a couple hundred RPM. I'm not going to type out the explanations for those unless it's really necessary.
 
bhuff, I hate disagreeing with you because honestly I've never known you to be wrong. BUT, I do disagree with you.

In a given gear ratio, the maximum acceleration is AT the peak torque. This is 100% true.

If this is true, then why does my car accelerate harder at 5500 rpm than at 3000? Why does my buddy's nitrous Bullitt pull WAY harder at redline than at 2500 rpm (even though it's making twice the torque at 2500?) Why does your average pickup truck accelerate much harder around 5000 rpm instead of at its peak torque at 2000 rpm where it's peak torque is?

If the torque is what causes acceleration, then why don't we put these in all of our cars?

ccc53-20---Steam-Engine.webp


I mean, it makes like 10,000 ft-lbs of torque.

Also, why do you say this:

In a given gear ratio, the maximum acceleration is AT the peak torque. This is 100% true.

and then say this?

However, if you are at a certain speed and you want to maximize your acceleration, you want to choose the gear that keeps you at peak power.

Am I missing something?

From the little bit of re-reading I did this afternoon, a car, at any given speed, will accelerate the hardest at the gear that places you at the highest horsepower. That's why the car accelerates the hardest at 20 mph in 1st gear; any other gear would lower the rpm and thus the applied horsepower.

As far as the shift points, using this ideology, you should shift when the horsepower you are making in a given gear drops below the horsepower you would be making if you shift to the next gear.

I haven't done the calculations (yet), but I think that the values we would obtain for a given car would be the exact same. We are just doing it two different ways.


If I am wrong somehow, I hope that someone points it out. I'm not at all too conceited to admit that I am wrong. I need to go back and re-read that thread from another forum, because there was one poster on there that really laid it down. He would be much better at explaining my point than I am, lol.
 
You are missing something sneaky. In a given gear you will accelerate hardest at peak torque, but if you could change gears at that same vehicle speed to any point that's making more hp, you will accelerate harder in that gear. When you multiply the torque made at the engine, by the new tranny gear ratio, you'll end up with more torque going to the rear-end.

Compare an insane 1050 ft-lbs torque tractor diesel at 1500 RPM - that's 300hp
Now consider an engine similar to an F1 engine that makes 350hp at 20,000 RPM - that's only ~92 ft-lbs of torque

At the same vehicle speed, which engine if optimally geared will make more torque at the rear end?

At a given speed, say they were rotating the driveshaft at 500 RPM (vehicle should be moving the same speed with identical rear gears, wheels, and tires). To get that driveshaft speed down to 500 RPM with the tractor motor, the gear ratio needed is only 3:1 --> 1500rpm/3 = 500rpm. At the same time, the torque to the rear will be multiplied by 3 making an astounding 3150 ft-lbs.

To get the F1 motor down to 500 rpm, the gear ratio needed is 40:1. 20,000/40 = 500 RPM. At the same time, the torque sent to the rear is multiplied by 40: 92*40= 3680 ft-lbs. The tiny F1 motor wins, exactly as predicted by the fact that the F1 motor is making more hp. With this knowledge we can just as surely predict that a new mustang 5.0 at 412hp @ 6200 (best guess) geared optimally will outperform the F1 motor. The certainty comes from the fact that the 5.0 makes more power. I need not actually do the torque multiplication to verify this result.