# Working out RPM from car engine's lb ft?

Using an old copy of an automobile magazine, these are the cars with hp and lb ft I am trying to work out the RPM for:

2007 Honda Accord 2.4 sedan, 2.4-liter 187hp 4-cylinder, 164 lb ft pulling power

The math I did to work out RPM:

187 x 5252/164 = 5,988.56 (rounded up to 6000rpm, how far should i round up in significant figures or decimal places)?

2007 Peugeot 207 1.6 HDi 110 GT, 1.6-liter 110hp 4-cylinder, 177 lb ft pulling power

The math:

110 x 5252/177 = 3,263.95 (rounded up to 3,264 rpm, or should it be 3000 rpm?)

I used these formulas:

Torque and horsepower relations: T = HP * 5252 / RPM

HP = T * RPM / 5252

RPM = HP * 5252 / T

In my two examples, have I got the math correct for car engines? With RPM, when do I need to round up correctly and what significant figure level do I need?

• As far as I know, such data about cars states the max horsepower and max torque, which are usually not at the same RPM, so what RPM are you calculating? Jun 20, 2018 at 18:00
• i'm trying to work out the RPM for the engines, which was ''not'' stated in the car price guide at the time, just the max horsepower and max torque were listed. Jun 20, 2018 at 18:02
• right, what I mean is that max horsepower is usually delivered at different rpm than max torque, to get for example the RPM at which max torque is delivered, you would need the corresponding horsepower, not the max horsepower Jun 20, 2018 at 18:08
• You can’t always do this - i fitted a camshaft that was designed to give a “flat” torque response from 1500 rpm to 4500 rpm in my v8 which avoided a peaky power delivery - so where would you start with that.. Jun 21, 2018 at 4:32
• Whilst I can work out what is meant here, I find it confusing that you are using RPM as both the quantity and as the units (and also at some points as a variable name). The quantity here is rotational speed, the units are revolutions per minute (rpm).
– welf
Jul 22, 2018 at 11:44

As I already mentioned in my comment, the problem in your calculations is that maximum torque and maximum power are at different RPM. The picture below is an example for a torque-power-diagram (excuse the sloppy quality, I just took a photo of old lecture notes with badly edited translations) of a combustion engine. The solid line is the torque, the dashed line is the power.

You can clearly see that the maximum torque and maximum power are at least 1000 RPM apart. To calculate the RPM at which maximum torque is provided, you cannot use the maximum power from the data sheet. You need to use the power that corresponds to the RPM you are searching for, so in the diagram, the power where the dashed line intersects with the red line.

Because of this, usually car data sheets tell at which RPM the maximum power and torque are delivered. I hope I could make it a bit clearer why your calculations give some value, but not a meaningful RPM value.

Max HP will always lag max torque

HP = torque x rpm

Unless torque drops off massively fast HP peak will lag (higher rpm).

HP peak cannot be a lower rpm.

This can be proved mathematically but this site does not support Latex.

• engineering.SE does support Latex, use single dollar signs for inline equations and double dollar signs for blocks Jun 22, 2018 at 9:52
• @OpticalResonator I know I proved it in college but cannot remember. The words are a logical proof in my opinion. Jun 22, 2018 at 21:44
• I wasn't referring to the proof but to your claim that this site does not support Latex Jun 22, 2018 at 22:02
• @OpticalResonator I am just telling you why I did not try Latex here. Jun 22, 2018 at 22:03
• First sentence doesn't make sense. Max HP can't lag itself. I suspect you wanted one of those to say torque. Jul 24, 2018 at 10:25