# Thread: Calculate the optimal shift points

1. ## Calculate the optimal shift points

Hey there,

is there some kind of tool that you can use in order to see the perfect shift points for each engine/gear ratio? I mean something where you can see the torque transfered to the wheels considering the current gear ratio and engine RPM's, so you can see whether and in which gears it is beneficial to shift a little earlier in order to have better acceleration instead of just redlining every time.

I noticed it on my old diesel car, the power falloff was quite large in between 4000 and 5000 RPM's, so on some cars it was beneficial to shift 500 rpm earlier to have more torque on the wheels...

EDIT:

Found a nice online tool here. You can enter some key points of your engine RPM's vs Torque and it will interpolate in between, so you can easily model the engine torque diagram. Of course it will tell you at which RPM to shift in order to have the best acceleration:

http://glennmessersmith.com/shiftpt.html  Reply With Quote

2. I use telemetry for a few laps to see when the power drops off. I try to make a connection between the engine sound at that exact rmp and go by ear after that (easier than keeping an eye on the revs during a race)

I am no expert at all but as far as I know, and the few I have asked, the drop off should be the same for all gears (?)  Reply With Quote

3. Thats a good approach, but I think the torque shown in the telemetry HUD is the torque the engine provides at a given RPM. Simplified you're having the best acceleration, when you shift, as soon as the next higher gear provides a higher torque on your wheels than the previous gear.

An example:
5000 RPM = 400 Nm
6000 RPM = 350 Nm

2nd gear ratio: 2.75
3rd gear ratio: 1.73

Now lets use a simplified calculation.
6000 RPM @ 2nd gear: 350 Nm * 2.75 = 962.5 Nm on your wheels
5000 RPM @ 3rd gear: 400 Nm * 1.73 = 692 Nm on your wheels

Means 6000 RPM/ 350 Nm @ second gear still accelerates faster than 5000 RPM/ 400 Nm @ third gear. So even whereas you engine is having less torque due to the high RPM's, you're still having a better acceleration due to the higher torque on your wheels.

Now the problem is that this depends on the torque diagram of your engine and of course on the gearing ratio and where the rev limiter kicks in.
If you can REV up to 7000 RPM, it might be smart to still stay in second gear until you hit the rev limiter. But again, depending on the torque diagram of your engine, you could loose even more torque. So you would have to calculate 7000 vs 6000 RPM as well to find out if it is STILL smart to stay in second gear. Because at some point/at a specific RPM range the engine looses too much torque, so you will actually have more torque in third gear than in second gear. But calculating all of this by hand is hell a lot of work, hence a calculator... because this strongly depends on the engines characteristics and gearing ratio...  Reply With Quote

4. When it comes to shifting points, don't look for the torque but for HP.
The theoretically perfect shifting point is when your current HP match the HP you'll get in the next gear.

Taking a fictive example:

5500 RPM = 680 HP
6000 RPM = 700 HP
6500 RPM = 715 HP (peak)
7000 RPM = 700 HP
7500 RPM = 650 HP (redline)

4th gear ratio: 1.20
5th gear ratio: 1.03
Gear ratio quotient: 1.03 / 1.20 = 0.86

Now let's find out the optimum gear shift point.

1) Shifting at peak HP: 715@6500 -> 6500 x 0.86 = 5590 RPM -> ~680 HP --> You're shifting too early because your HP drop too much
2) Shifting at 7000: 700@7000 -> 7000 x 0.86 = 6000 RPM -> ~700 HP --> Perfect shifting point because HP of 4th and 5th gear match up
3) Shifting at redline: 650@7500 -> 7500 x 0.86 = 6450 RPM -> ~715 HP --> You're shifting too late because your HP already dropped down too much  Reply With Quote

5. Originally Posted by Bealdor When it comes to shifting points, don't look for the torque but for HP.
The theoretically perfect shifting point is when your current HP match the HP you'll get in the next gear.

Taking a fictive example:

5500 RPM = 680 HP
6000 RPM = 700 HP
6500 RPM = 715 HP (peak)
7000 RPM = 700 HP
7500 RPM = 650 HP (redline)

4th gear ratio: 1.20
5th gear ratio: 1.03
Gear ratio quotient: 1.03 / 1.20 = 0.86

Now let's find out the optimum gear shift point.

1) Shifting at peak HP: 715@6500 -> 6500 x 0.86 = 5590 RPM -> ~680 HP --> You're shifting too early because your HP drop too much
2) Shifting at 7000: 700@7000 -> 7000 x 0.86 = 6000 RPM -> ~700 HP --> Perfect shifting point because HP of 4th and 5th gear match up
3) Shifting at redline: 650@7500 -> 7500 x 0.86 = 6450 RPM -> ~715 HP --> You're shifting too late because your HP already dropped down too much
Thats an interesting approach as well. When I look for various calculations, then some people do it by torque, some people use your HP method and there also seems to be a third method that also works using the HP values. I think after all this comes more or less to pretty simiar results because horsepower is calculated by torque and RPM's.

The formula is this: Horsepower = TORQUE x RPM ÷ 5252

And actually the torque is what moves your car forward, not the horsepower. Horsepower is just a value to describe the relationship in between torque and RPM's of an engine. This is also why a 150 HP diesel car is not faster than a gas car with 150 HP. The HP is constant, so the 150 HP diesel has less RPM but more Torque, and the gas car has less torque but more RPM than the diesel. So sure, the diesel car has more torque so it benefits from a faster acceleration. But due to the short RPM band you have to change the gearing ratio in order to reach same speeds as the gas car with a bigger RPM band. And after doing this, the torque both cars have on the wheels are more or less the same, and the acceleration is almost equal.

Hence people came up with the definiton of horsepower, because it makes engines of different types with different torque and RPM's comparable. without having to consider RPM and Torque alone. This is why I can't this bullshit talk people are having regarind 150 Hp diesel cars being faster than 150 HP gas cars...
I mean you could theoretically have an engine with 1000 Nm torque but maximum 1000 RPM. It will still not be faster than a car with 250 Nm torque and 4000 RPM.

Anyway. I think no matter whether you use HP or Torque as calculation, in the end both values are in a tight relationship to each other and people are calculating the same result by taking different variables and hence different calculations... But I somehow rather stick to torque because this is the physical force that actually drives you forward, whereas horsepower is just a value that describes the relationship between your engines RPM's and torque...

EDIT: Actually my formula was bullshit. Corrected it...  Reply With Quote

6. @Marlborofranz: Torque is only half the picture. It's the amount of work (torque) you're doing at a certain pace (rpm) that's important. That's why you should base your shifting point on HP; as soon as the HP for a certain gear drops below the HP of shifting to the next gear you'll be losing power, whatever the torque or RPM may be at that point. Bealdor is spot on.

Your reasoning above about the relation between torque and rpm is a bit flawed. It's not all about speed. When pulling a boat on a trailer I'd rather have a smaller diesel engine producing lots of torque at low rpm's than a bigger petrol engine that produces more HP through higher rpm's with lower torque.  Reply With Quote

7. This is true, but isn't it the result more or less the same? Serious question... Because by looking for the point where you have the most horsepower, you also have the point for the highest torque. Because horsepower is just a value calculated from Torque and RPM. It's just another formula to find where the torque curves of the lower and higher gear start to intersect each other, which is the optimum shift point.

About torque: Yes it is simplified, but assuming you have two cars with a 150 HP diesel and a 150 HP gas engine, both can pull exactly the same mass. Of course, if you use the same gearing ratio for both cars, the diesel engine will pull more mass and accelerate quicker, but has less top speed due to the limited RPM range. But if you adjust the gearing ratio so both cars have the same speed at red line, then they will accelerate the same and pull the same mass. At different RPM's due to engine design, but they can pull exact the same mass.

You can have the exact same effect as when using two identical cars/engines but change the gearing ratio towards top speed and towards acceleration. You will just adjust the amount of torque vs RPM arriving at your wheels. But the engine will do the same. It doesn't make a difference about the horsepower because the horsepower is always the same. Torque and RPM = horsepower.

So when calculating the optimum shift point based on horsepower and the gearing ratio, it is the same as doing this based on torque and the gearing ratio. Both calculations provide the RPM at which to shift, which is the RPM at which the torque line of gear one intersects with the torque line of gear two.  Reply With Quote

8. The general idea that I get with PC2, is that the game makes you shift up more quickly than what i've seen during real races,
just to preserve engine temps. Even with radiator openings to a max.

I always instinctively use rev sound + rev lights on the dash for upshifting.
But if you see this video onboard the audi for example, he pretty much always hits the very rev limit before shifting up:

https://youtu.be/RAbWVTKzRx0

Mostly when I shift like this, the engine will blow within a few laps.

Not to mention this kind of shifting ... :

https://youtu.be/L-6i8r2QceI  Reply With Quote

9. Originally Posted by Marlborofranz This is true, but isn't it the result more or less the same? Serious question... Because by looking for the point where you have the most horsepower, you also have the point for the highest torque. Because horsepower is just a value calculated from Torque and RPM. It's just another formula to find where the torque curves of the lower and higher gear start to intersect each other, which is the optimum shift point.

About torque: Yes it is simplified, but assuming you have two cars with a 150 HP diesel and a 150 HP gas engine, both can pull exactly the same mass. Of course, if you use the same gearing ratio for both cars, the diesel engine will pull more mass and accelerate quicker, but has less top speed due to the limited RPM range. But if you adjust the gearing ratio so both cars have the same speed at red line, then they will accelerate the same and pull the same mass. At different RPM's due to engine design, but they can pull exact the same mass.

You can have the exact same effect as when using two identical cars/engines but change the gearing ratio towards top speed and towards acceleration. You will just adjust the amount of torque vs RPM arriving at your wheels. But the engine will do the same. It doesn't make a difference about the horsepower because the horsepower is always the same. Torque and RPM = horsepower.

So when calculating the optimum shift point based on horsepower and the gearing ratio, it is the same as doing this based on torque and the gearing ratio. Both calculations provide the RPM at which to shift, which is the RPM at which the torque line of gear one intersects with the torque line of gear two.
The issue with torque is, that it doesn't take velocity (which is dictated by the gear ratio) into account.
Because gear ratio dictates the velocity, the acceleration will be lower the faster you go.

(Warning: Physics lecture) As you can see, the optimum a(cceleration) at a given velocity is dependant on P(ower), not T(orque).

Edit: Since I just saw that you're from Germany, check out this document.  Reply With Quote

10. Originally Posted by Marlborofranz ...Because by looking for the point where you have the most horsepower, you also have the point for the highest torque...
That's a false assumption and the main reason why your theory doesn't work.
Torque is only one part of the equation. The other one is RPM.

Check out this dyno plot for example: A combustion engine reaches max torque far earlier than max power and it also drops off earlier while the power still rises (because of rising RPM).  Reply With Quote

sms 