When to shift

We’ll start out with something easy. I have heard plenty of times from people that you should shift when the power starts to fall off on the car to be “faster”. But the fact of the matter is that due to gear ratios and the mechanical advantage of the gears, more often than not redline is your best shift point, but to be sure you need to do the calculations

To begin with, lets talk about what makes you “faster.” Faster in this case I will hold to mean acceleration. In a drag race or most other types of races you are rarely holding a constant speed, so it is about maximum acceleration. Want to run a ¼ first or make it down the straightaway fastest? Have the most acceleration. So understanding that we are talking acceleration lets get some of the other variables out of the way.

In simple terms acceleration for a car is described by the equation F=m*a where F is force, m is mass, and a is acceleration. Since we’ll be comparing “within” a car, m remains constant, so the acceleration that we are interested is directly related to the force. The force in this case, is the force exerted at the ground by the driving wheels. The force at the ground is translated from the engine out to the wheels by the transmission. Since, again, we are staying within one car, the radius of the wheel, and the final drive, remain constant so only the gear ratio matters.

For our example, we will look at a 2^nd gear to 3^rd gear shift. Below is an actual dyno from a Mazda Protégé and the gear ratios for its transmission. From that information you can see that 2^nd gear is 1.842 and 3^rd gear is 1.310. What this means is that for a given point on the dyno we will multiply the torque value by the gear ratio. Looking at roughly 6,000 rpm’s the torque value is approximately 100 ft-lbs. So in 2^nd gear this is translating to the final drive at a value of roughly 1.842*100 or 184.2 ft-lbs (this is not the value applied at the ground as the final drive and the wheel radius are not compensated for, but remember that multiplier, whatever it is, remains constant so only this input value to that system matters for our discussion). But for 3^rd gear the input value calculates to 1.310*100 or 131 ft-lbs. That means that 2^nd gear exerts 40% more force to the ground at that point than 3^rd gear is capable of.

1st gear 3.307:1

2nd gear 1.842:1
3rd gear 1.310:1
4th gear 0.970:1
5th gear 0.755:1
Final Drive 4.105:1

This difference then becomes the critical point of our discussion. Looking again at the graph you can see that at redline the torque has fallen to ~82 ft-lbs. At the peak of torque at about 4,000 rpm’s it is 125 ft-lbs. So second gear at redline creates 151 ft-lbs of torque while 3^rd gear at peak torque creates 163 ft-lbs. While this case looks like shifting a little earlier would help you it ignores two important issues:

1)Actual engine speed and where the gears will “pick up” on a shift

2)Other gear ratios

We’ll start with #2. Here are the same values under best/worst case as given above for the other shifts:

1-2: 1^st @ redline = 271 ft-lb versus 2^nd @ peak = 230.2 ft-lbs

3-4: 3^rd @ redline = 107 ft-lb versus 4^th @ peak = 121.25 ft-lbs

4-5: 4^th @ redline = 79.5 ft-lb versus 5^th @ peak = 94.4 ft-lbs

So as we can see the mechanical advantage in 1^st gear is so significant that the huge torque fall off still doesn’t matter, but in the closer, higher gears it still looks like shifting early would be better. But now when we go back to point #1 you’ll understand why this is not the case.

To address number one we need to figure out the change in rpm’s. Since at the time of the shift we can roughly assume the speed to be constant, then the rpm’s will change by the gear ratio amount. So in our 2-3 example you calculate it as 6800/1.842=x/1.310 where x yields are new shift rpm which is 4836 in this case.

So now redoing our math, we know we shifted at redline in 2nd at 151 ft-lbs of torque to the final drive, but now when we pick back up in 3^rd gear we do so at 4836 which yield a converted torque value of 118*1.310 or 154 ft-lb’s. In other words nearly identical. Now if you shifted earlier to try to obtain peak torque more, so lets say you want to start in 3^rd gear at 3800 so you start in and go through the max torque curve portion for this engine, that would mean you’d have to shift from 2^nd gear at 5343. This results in starting 3^rd gear at 3800 rpm’s with 163 ft-lbs going into the final drive which is nice because you are starting out with more than the 154 we got with the redline shift so we are accelerating harder at the start of 3rd gear, BUT you’ve now cut out of 2^nd gear when you were at 206 ft-lbs. So while you've started 3rd gear with more oomph, you gave up a LOT more oomph at the end of 2nd gear.

You can do the math on the other gears and you’ll find that in general the same trend will hold but fades as you get close to the top gears. The 3^rd gear to 4^th gear shift is essentially a wash with a case for shifting about 100-200 rpm’s early that can be made, but 4^th to 5^th changes things enough to start to matter due to the overdrive situation. In that case optimum shifting would appear to be at about 6400, or 400 short of redline. This effect is due to the mechanical advantage of the gears relative to one another:

1st has 80% advantage over 2nd

2nd has 41% advantage over 3rd

3rd has 35% advantage over 4th

And 4th has 28.5% advantage over 5th.

So in essentially all cases with this dyno graph you want to shift at or extremely close to redline. This becomes even more apparent when you take a vehicle like the turbocharged Mazdaspeed Protégé that has a flatter torque curve and less fall off toward redline. This also gets into why keeping the torque curve up throughout the power band matters (and that will obviously result in higher horsepower numbers as well). We’ll save torque and horsepower and area under the curves during a gear for another time as that is an even longer discussion.