I'm a bit curious about how exactly the torque and RPM could be determined from an engine. One website states that bigger displacement engines develop more torque while smaller ones spin faster, which means they can have a higher rpm and therefore make more horsepower.

For example, a 5.5-liter twin-turbo diesel V12 can make 725 hp. A 4.0-liter V8 can make 470 hp. Do the number of cylinders contribute to torque as well? I'm just unsure of this.


3 Answers 3


Generally large displacement engines tend to lower the rpms. The reason is that the inertial forces tend increase for larger displacements (assuming the stroke size remains the same) and eventually exceed the available material strengths.

Also, in general the power of an engine will be strongly correlated with the fuel consumed. This is the driving factor. However, because :

$$P = M\cdot \omega =M\cdot \frac{2\pi\cdot n}{60} $$


  • P is the power
  • M is the torque
  • $\omega= \frac{2\pi\cdot n}{60} $is the angular velocity in [rad/s], and $n$ is the rpm of the engine

Also, having more cylinders will increase the available torque.


The following data is from a widely available data set in R and python, called the mtcars or Motor Trend US magazine dataset (see reference for the variables here). Although this is very old (1974), it still shows the trends between the different variables.

The following data is from a publicly available data (auto-mpg). It has a lot more observations so the trends are much mover visible compared to the mtcars dataset, I posted originally.

Horsepower vs. Cylinders

enter image description here

Horsepower vs Displacement

enter image description here

horsepower vs. Fuel consumption - US

In the US the common measure for fuel consumption is miles per gallon. So higher values indicate better fuel consumption. As the displacement increases the fuel consumption becomes worse.

Horsepower vs Displacement

horsepower vs. Fuel consumption - European

The following graph is the similar to the above, however it presents the fuel consumption in an inverse manner. I.e. in Europe its more common to quote the fuel consumption in $\frac{liters}{100.km}$, and lower values show better fuel consumption.

enter image description here

Again the better fuel consumption is for smaller displacement.

(I probably prefer this graph because I am more used to this concept).

Fuel Consumption (EU) vs. Displacement

Again from this plot you can see, that for increasing displacement, the fuel consumption (in EU terms) increases.

enter image description here

The following is the code I used to produce the above diagrams in Python. You can play around and produce your own plots which can be very informative.

import statsmodels.api as sm
import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
import seaborn as sns

def get_mtcars():
    '''mtcars dataset from 1974 (Motor Trend Magazine)'''
    return sm.datasets.get_rdataset("mtcars", "datasets", cache=True).data

def auto_mpg_data():
    '''Slightly newer dataset (circa 1993)
    url = 'http://archive.ics.uci.edu/ml/machine-learning-databases/auto-mpg/auto-mpg.data'
    column_names = ['mpg', 'cyl', 'disp', 'hp', 'wt', 'qsec', 'Model Year', 'Origin']
    print("The data is being loaded")
    print("The column names have been defined")
    raw_dataset = pd.read_csv(url, names=column_names, na_values='?', comment='\t', sep=' ', skipinitialspace=True)
    return raw_dataset.copy()

# mtcars = get_mtcars() # uncomment this line for the 1974 original mtcars dataset 
mtcars = auto_mpg_data()

# %%
mtcars.plot.scatter(x='cyl', y='disp')

# %%
plt.plot(mtcars.cyl, mtcars.hp, lw=0, marker='o')
plt.xlabel("cylinders []", fontsize=14)
plt.ylabel("horsepower [hp]", fontsize=14)

# %%
plt.plot(mtcars.disp, mtcars.hp, lw=0, marker='o')
plt.xlabel("Displacement [cu.in]", fontsize=14)
plt.ylabel("horsepower [hp]", fontsize=14)

# %%
plt.plot(mtcars.mpg, mtcars.hp, lw=0, marker='o')
plt.xlabel("miles per US gallon [mpg]", fontsize=14)
plt.ylabel("horsepower [hp]", fontsize=14)

# %%
plt.plot(1/mtcars.mpg*235.2146, mtcars.hp, lw=0, marker='o')
# plt.xlabel("US gallon per mile [gallons/100 mile]", fontsize=14)
plt.xlabel("Fuel Consumption in European [L/100 km]", fontsize=14)
plt.ylabel("horsepower [hp]", fontsize=14)

# %%
plt.plot( mtcars.disp,1/mtcars.mpg*235.2146, lw=0, marker='o')
# plt.xlabel("US gallon per mile [gallons/100 mile]", fontsize=14)
plt.xlabel("Displacement [cu.in]", fontsize=14)
plt.ylabel("Fuel Consumption in European [L/100 km]", fontsize=14)

  • $\begingroup$ Oh, okay. So just to be clear, a larger displacement would mean lower RPMs? $\endgroup$ Commented Sep 2, 2021 at 18:32
  • $\begingroup$ And longer stroke means more torque and, of course, more displacement. $\endgroup$ Commented Sep 2, 2021 at 18:42
  • $\begingroup$ @ZytekB08 its not necessary that larger displacement means lower rpms. There are other considerations also. However, for all other things being equal larger displacement would lead to a restriction on the rpms (there are others though that could give a more complete and knowledgable answer, because this is really a very extensive field) . $\endgroup$
    – NMech
    Commented Sep 2, 2021 at 19:07
  • $\begingroup$ I am confused here and hope somebody can explain or point out my misconception. In the equation, if the torque remains constant, the higher the power will yield a higher rpm (n), doesn't it? $\endgroup$
    – r13
    Commented Sep 2, 2021 at 19:11
  • 1
    $\begingroup$ @r13 Well that is somewhat simplified. The units are not necessarily correct. So if RPM = power/torque, and the value of torque can't change because you are keeping it constant, if power goes up, what does RPM have to do to keep the equal sign true? If I have 20 = 10/2, and I decide 2 must remain the same, if I increase 10, what does the 20 have to do? $\endgroup$
    – DKNguyen
    Commented Sep 2, 2021 at 20:23

Of course engine size affects torque, but RPM (& power) is more complicated.

First of all, the thing that matters in an engine is power. Power is what gets things done and is the best measure of an engine's performance. Power is torque times RPM, so half the torque at twice the speed is the same. We can get whatever torque we want at the wheels through gearing. We aren't going to dig into the shape of torque curves and engine tuning for driveability versus power. Suffice it to say that you wouldn't enjoy driving a top fuel dragster around town. In terms of torque vs. size, if we doubled all the dimensions of an engine we would double the torque (roughly) but the size would mean its top speed (RPM) would be reduced. It is figuring out just how to balance all engine characteristics that keeps automotive engineers working.

Let's limit our discussion to spark-ignition (gasoline) engines. The power of a spark-ignition engine is generally related to the amount of air you can get into it. The higher an engine revs, the more air it can ingest, so the better its power will be. The rev limit isn't so much related to engine size as it is related to both piston size and the engine's stroke. The reciprocating nature of piston engines creates large forces as the pistons and valves go up and down. So a lighter piston traveling shorter distances is better. Displacement is bore times stroke, so a shorter stroke & larger bore is better for high rpm. So what does this mean for engine displacement? Again, speaking in generalities, an engine with more cylinders for a given displacement will have lighter cylinders and thus be able to rev higher and thus produce more power.

If you are comparing the installed base of engines this gets harder, because more cylinders means both a more expensive engine (more parts) & vehicle (larger engine bay to fit the different form factor), and expensive cars will tend to have improved features across the board including making the engine more powerful in general.

All of this is driven by the overall engine design. A 2021 Formula 1 V-6 engine is only 1.6 liters displacement yet makes somewhere above 850 horsepower. This is a displacement less than all but the smallest subcompact cars. So it isn't just size that makes an engine perform. It's all the components together. There is no single (or even dual set) measure of an engine (certainly not size) that can tell us an engine's performance level.


Compare a 5" bore piston undergoing 20+ atm of explosion pressure in a diesel engine with a 2.5" bore of a gas engine with 7.1 atm, of course, the larger engine produces more torque and power.

However, because of the large parts moving up and down in reciprocal movement of the crankshaft the larger engines can not have high RPMs, usually about 2000.

The rule of thumb is the more torque and more RPM means more power.

Generally, the trucks and heavy engines have better torque at lower RPMs and car engines, or more so race car engines deliver more power at low torque bu high RPMs. I had a Husquvarna 400 dirtbike that could deliver 40HP at 7800RPM because of its bee-sounding RPM.

  • $\begingroup$ Larger engines are definitely not limited to low RPMs. GM has crate engines (672 for example) with 4.5" bore pistons that spin to 7,000 RPMs, and those are pretty tame by race engine standards. Pro Mod engines can have > 5" bores and see 10,000 RPM. $\endgroup$
    – Bert
    Commented Aug 4, 2022 at 14:33

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