I'm really confused about torque. I understand that it is the rotational force, but like I can't wrap my mind across why a Lamborghini goes faster than a truck when a truck has more torque. My friends tried to explain it in a simple way, but I literally don't understand it. I can't wrap my head around it.

Here's an example of what I'm talking about. A Ford F-250 has

  • 440 horsepower
  • 925lb of torque
  • max RPM 4200
  • 0-60 mph of 6.9 seconds
  • 7369 kg

A Lamborghini Aventador has

  • 740 horsepower
  • 507 lbs of torque
  • max RPM 5500
  • 0-60 mph of about 2.9 seconds
  • 1690 kg

The truck has way more torque than the Lamborghini, so I'm wondering how the Lambo can go faster then the truck with more torque?

  • $\begingroup$ By 'goes faster', do you mean faster acceleration of greater top speed? $\endgroup$ – innisfree Aug 18 '17 at 3:52
  • $\begingroup$ Can you add into your question some facts about Lambos versus trucks? E.g., power, torque, max RPM, weight, acceleration and max speed $\endgroup$ – innisfree Aug 18 '17 at 3:56
  • $\begingroup$ greater acceleration and top speed $\endgroup$ – Ethan Aug 18 '17 at 4:02
  • $\begingroup$ @innisfree yeah $\endgroup$ – Ethan Aug 18 '17 at 4:08
  • 1
    $\begingroup$ The pages to which you linked have faulty numbers. The 2017 Ford F250 has 925 lb-ft (1254 Nm) of torque at 1800 rpm rather than the cited 505 lb-ft (684 Nm). The question also has a faulty number. The Lambo redlines at 8500 rpm, not 5500. This is key. $\endgroup$ – David Hammen Aug 18 '17 at 12:33

The first thing to say is that the figures given the manufacturers as well as being informative are there to sell vehicles.
The next thing is that comparison can often be a nightmare because of the mix of units which are used.
For example it depends on your background as to whether you "appreciate" better what 100 horsepower (hp) or 75 kW is.

The specifications that you have given can certainly be used to give you an answer to your question but to make a better comparison more information needs to be provided.

The acceleration of a vehicle $a$ depends on its mass (m) and the force applied on it $F$ via tyre/ground contact $\Rightarrow F = ma$.

The power and torque specifications that you have given are for the engine of the vehicle at particular engine speeds which tend to be maxima.
For the engine power $P$ and torque $\tau$ are linked $P=\tau\, \omega\,$ where $\omega$ is the angular speed of the engine.
Please note that I have figures for the petrol version of the Ford and I have quoted figures for the diesel version later.

For the two vehicles in question the power and torque characteristics look something like this.
I have redrawn the graphs from my primary sources because of mixed units eg foot-pound and newton-metre for the torque and horsepower and kilowatt for the power.
Another problem is that these vehicles have different specifications for different engines and different years so what you see is a rough sketch.

enter image description here

You will note that the Lamborghini engine has a higher maximum engine speed.
As I stated before the mass of the vehicles is a contributory factor to their acceleration and as the Lamborghini (1690 kg) has a mass which is much less than than of the Ford (2745 kg - note the difference in mass from that given by the OP which was the gross value) for a given force the Lamborghini's acceleration will be much larger.
However the torque delivered by the engines changes as one progresses through the drive chain to the wheels.
One must first look at the gear ratios (which "decrease" the speed of revolution and "increase" the torque. then at the transmission ration.
To get the maximum torque delivered to the wheels you need the maximum possible gear ratio.

The Lamborghini has 7 forward gears with ratios

1 - 3.91:1 / 2 - 2.44:1 / 3 - 1.81:1 / 4 - 1.46:1 / 5 - 1.19:1 / 6 - 0.97:1 / 7 - 0.84:1

and a differential ratio of 3.54:1 so when the Lamborghini is in first gear all the torque shown in the graph have to be multiples by $3.91 \times 3.54 \approx 13.8$ assuming that there are losses in the gear trains.

The Ford has 6 forward gears with ratios

1 - 3.97:1 / 2 - 2.31:1 / 3 - 1.51:1 / 4 - 1.14:1 / 5 - 0.85:1 / 6 - 0.67:1

and a differential ratio of 3.73:1 so when the Ford is in first gear all the torque shown in the graph have to be multiples by $3.97 \times 3.73 \approx 14.8$ again assuming that there are losses in the gear trains.
This is slightly higher than that of the Lamborghini.
The turbo version of the Ford has a maximum torque of 925 ft-lb at 1800 rpm and has a gear ratio of 4.17:1 in first gear and can have a differential ratio of 4.10:1 which means that the torque delivered at the wheel is much larger.

There is a last component to consider and that is the radius of the wheels $R$ because

torque at wheels = accelerating force $\times$ radius of wheels

The Lamborghini is a four wheel drive vehicle and has front tyres 335/30YR20 and rear tyres 255/35YR19

Again we have mixed units as the first number is the width of the tyre in millimetres, the second number is the "height" of the tyre as a percentage of the width and the third number is the diameter of the tyre in inches.
So this works out as a radius of the wheels: front - 14 inches and rear - 13 inches.

For the Ford with tyre designation 265/75R16 the radius is 15.8 inches which represents a larger decrease in the force for a given torque.

Taking all this into account the Lamborghini wins hands down primarily because it is much lighter than the Ford and even though the diesel version delivers more torque.

Once the vehicles are moving drag forces play a part and the drag force is using assumed to be proportional to the square of the speed and is given by $F_{\rm drag} = \frac 12 c_D A \rho v^2$ where $\rho$ is the density of the air, $A$ is the frontal area of the vehicle and $v$ is the speed of the vehicle.

The drag coefficient $c_D$ is given by the manufacturers as 0.23 for the Lamborghini and 0.4 for the Ford.
As an aside note that a Formula One car can have a drag coefficient of over one because of the regulation regarding exposed tyres and the necessity to have a down force.
The Ford also loses out in terms of frontal area $8\,\rm m^2$ as opposed to the Lamborghini's $2.3\,\rm m^2$.

At the maximum speed for the Lamborghini or the Ford the power delivered by the engine is equal to the rate of working against the frictional forces.
So as it delivers more power, has a smaller drag coefficient and frontal area you would expect the Lamborghini (217 mph) to have a much greater top speed although one Ford was modified to get close.

The real difference between the two vehicles is their usage.
The Ford needs torque to move heavy loads and does this by adjusting the gear train to magnify the engine torque but as the engine has a relatively low maximum speed the gearing is spaced out so that a reasonable top speed can been achieved in top gear.
Because the Lamborghini is relatively light torque is not a major consideration but there is a need for speed so the gearing is arranged so that the engine speed does not change much and is always close to delivering maximum power.
If you look at the specification of the Lamborghini over the years you will note that the maximum torque does not increase by very much and but the maximum power is increased by making the maximum engine speed higher.

  • 2
    $\begingroup$ The radius of the tires isn't very critical. What is critical is the vehicle's moment of inertia about the axis defined by the line between the bottom of the rear wheels. The Lambo's mass is considerably smaller than that of the F250, and the Lambo's center of mass is considerably lower than that of the F250. The combination means that the Lambo accelerates like a bat out of **** while the F250 accelerates like ... Trying to get the right analogy. Elephants can accelerate quickly, so that's not quite right. $\endgroup$ – David Hammen Aug 19 '17 at 17:18

According to Newton's second law, $$m\mathbf{a} = \mathbf{F}$$ which means that the effect of applying a force on an object is making this object accelerate, and that the heavier an object is, the least it will accelerate.

Now, considering a car, the torque $T$ you're talking about is the one that makes the wheels rotate, which is translated into a force $F$ through friction on the road. You can show that $ T = RF$, where $R$ is the radius of the wheel, so finally using Newton's second law, we got

$$ a = \frac{T}{Rm}$$

Which explains why a truck accelerates less than a car. In fact, in your example the masses ratio is around 4, while the torques ratio is around 0.5, and wheels radii are almost similar in general.

The model I just presented is simple but false: the first reason is that I didn't take air friction into account. This force can be expressed as

$$ \lambda v^\alpha$$

where alpha is a positive number, usually 1 or 2, but it doesn't really matter here. Then, you got

$$ ma = \frac{T}{R} - \lambda v^\alpha$$

However, $a= \frac{\textrm{d}v}{\textrm{d}t} =\dot{v}$, and when maximum speed is reached, $\dot{v}=0$, so

$$ v_{max} = \sqrt[\alpha]{\frac{T}{R\lambda}}$$

Herr, you can understand the difference in maximum speed between a truck and a car. Indeed, a sport car shows much less wind resistance than a car!

There is a second road in my model: I assume that torque is constant, and equals maximum torque of the engine of each vehicle, and this is wrong. But taking this into account would not really help to understand what is going on...

  • Mass of the vehicles. All other things being equal, the higher mass car needs more power and more torque to accelerate at the same rate. I don't see the masses there.

  • Torque only matters at the point it's delivered (where the wheel turns on the road). To get from the figure on the engine, you'd need to multiply it by the gear ratio in the transmission (depending on the gear you're in), the final drive ratio, and divide by the radius of the drive wheels. I don't know those figures for these two vehicles. They may be similar, or they may be very different.

  • The torque figure you quote will be the maximum delivered, only at a particular engine speed. A high first gear might mean that one of the vehicles has to lug along for a bit longer than the other before reaching max torque. That would reduce acceleration initially.


The truck is twice heavier than the sports car. It takes a lot more power to get the truck to the same speed and a much higher torque to have the same acceleration. Also keep in mind that the power is not changed by the transmission, but the torque is proportional to the gear ratio. The torque numbers specified are for the engine, not at the wheels. So ultimately the torque is irrelevant, because conceptually a transmission can be designed to create any torque you need. Ultimately it is power vs. weight. And with a half the weight and more power the Lambo naturally gets there in a half the time.

  • 1
    $\begingroup$ I don't agree that torque numbers are irrelevant for acceleration or that power is more relevant than torque for acceleration. Power (which is ~ torque * RPM) is relevant for top speed (top speed is limited by the amount of work you can do against friction). Torque, which is proportional to force, is the relevant thing for acceleration. $\endgroup$ – innisfree Aug 18 '17 at 5:10
  • $\begingroup$ You can create any torque by changing the gear ratio in the transmission. This is exactly why the transmission is there in the first place, to increase the torque at lower speeds to get moving faster. According to the formula in your comment, you can double the torque by reducing the rpm twice in the transmillion (not in the engine). According to this formula you can replace torque with power/rpm and don't mention torque at all. I think the confusion may be in the fact that the torque at the engine is not at all the same as the torque after the transmission (except for the direct gear). $\endgroup$ – safesphere Aug 18 '17 at 5:33
  • $\begingroup$ Also, the weight of F-250 must be 7,369lbs, not kg. $\endgroup$ – safesphere Aug 18 '17 at 5:34
  • $\begingroup$ I think it's kg, check the link to be sure $\endgroup$ – innisfree Aug 18 '17 at 5:52
  • $\begingroup$ My equation is just power ~ torque * RPM or $P = \tau \cdot \omega$. This is a very well-known result in classical mechanics. $\endgroup$ – innisfree Aug 18 '17 at 6:14

You're comparing apples to oranges.

One, the turbocharged F250 diesel, is a somewhat expensive vehicle designed to tow ten+ tons of cargo at a wide but moderate range of velocities over a wide range of engine speeds (RPMs). This requires lots of torque; power is not as critical. The truck redlines at 4000 RPM, but it would be crazy to push the truck to the redline. The torque for this truck peaks at 1800 RPM, drops slowly until about 3000 RPM, and then falls rapidly. Power peaks at 2800 RPM and then falls rapidly. There is zero reason to redline a F250.

The other, the naturally aspirated Lambo, is a ridiculously expensive vehicle designed to carry less than 1/10 ton of cargo (the mass of the driver) at ridiculously high velocities and at ridiculously high engine speeds. This requires lots of power; torque is not as critical. The car redlines at 8500 RPM, with power peaking at 8400 RPM. That 8400 RPM is where the car is designed to operate. There is every reason to come very close to redlining the Lambo.

The truck engine would explode at half the Lambo's redline rate. Apples and oranges.


The analysis below ignores driveline losses & air resistance.

Suppose you have two cars, each weighing the same and each having the same peak torque value

  • Car A -

    • The peak torque of 300 ft-lbs occurs at 3100 rpm
    • At that engine speed the car makes 177 bhp
  • Car B -

    • The peak torque of 300 ft-lbs occurs at 6200 rpm
    • At that engine speed the car makes 354 bhp

Here is where you need to be careful. Do both cars have the same gearing? Let us examine both options

  1. Both cars have the same gearbox, and let's say with 4-th gear they will be moving with the following speeds

    • Car A - is at 30 mph accelerating with 0.734 gee.
    • Car B - is at 60 mph accelerating with 0.734 gee.
  2. Cars have different gearing such that peak torque in 4-th gear is at about 60 mph.

    • Car A - is at 60 mph accelerating with 0.367 gee.
    • Car B - is at 60 mph accelerating with 0.734 gee.


Power is the ability to accelerate at speed. The higher the power rating on an engine the higher speeds it can carry the acceleration (which is produced by torque).

In the examples above car A either has to move at half the speed to get the same acceleration as car B, or it has half the acceleration at the same speed.

Useful Formulas

$$ \mbox{(hp)} = \frac{ \mbox{(tq)} \mbox{(rpm)}}{5250} $$

$$ \mbox{(gee)} = 374.4 \frac{ \mbox{(hp)} }{\mbox{(weight)} \mbox{(mph)}} $$

  • $\begingroup$ weight is in lbs in the formula above. The following units are used 2.206 lbs per kg, 2.232 mph per (m/s), 745.7 W per hp and 9.80665 (m/s^2) per gee. $\endgroup$ – John Alexiou Aug 18 '17 at 20:34

In this answer I willfully ignore talking about the transmission in order to keep everything very straightforward. We can ignore the transmission because in practice any automatic or manual transmission can be approximated to a perfect torque converter which always use the engine at the maximum power to obtain the maximum torque at the wheel and thus the maximum acceleration in the vehicle speed range. The key concept here is that engine torque does not directly translate into wheel torque, the faster you go the longer the gear ratio the lower the torque at the wheel.

Consider the formula for kinetic energy $[E_k]$ : $$E_k = {1 \over 2}mv^2$$ Which can be solved for $[v]$ to obtain: $$ v = \sqrt{ {2 E_k} \over m }$$

This give us the speed as a function of kinetic energy and mass. To increase kinetic energy we need to be able to do work, engines produce work over time at a rate given by its power. The work produced over time is transferred to the car as kinetic energy by our ideal transmission and drive-train. By deriving over time we can get acceleration as function of power, velocity and mass:

$$ a = { P \over v m }$$

From this equation is straightforward to see that torque of the engine does not matter at all when an ideal transmission is used. Acceleration depends only on power/weight ratio.

If we want to calculate the maximum speed of the car we can use the same equation and add the (negative) acceleration provided by drag, let's ignore drive-train and ground-friction tire losses since those are low at high speed compared to drag. Usually air resistance force is approximated to $f = kv^2$ which can be integrated in the equation:

$$ a = { P \over v m } - {kv^2 \over m}$$ $$ k = \text drag coeff.$$

Maximum speed is reached when engine acceleration equals the drag deceleration, or simply we put $a = 0$ which give us the equation:

$$ { P \over v_m m } = {kv_m^2 \over m}$$

solved for max velocity $v_m$:

$$v_m = \sqrt[3] {P \over k}$$

This let us draw some conclusions:

  1. Maximum speed is a function of power and drag coefficient.
  2. Mass does not affect maximum speed.
  3. Torque of the engine does not affect maximum speed.
  4. Torque of the engine does not affect acceleration.
  5. The Lamborghini would be faster than the truck even at the same weight and drag coefficient.

But both are lower so it is a lot faster :)

Addendum, The ideal transmission approximation:

The ideal transmission can be thought as a lossless CVT (Continuous Variable Transmission) with no gear ratio limit. In practice a gearbox have a limited number of discrete ratios and this means that the engine will have to adapt the RPM to the current car speed and the active gear. The power loss here happens mostly in first gear when the car is so slow that it cannot be at peak power RPM, but at this stage, more than by power, cars are limited by tires grip, especially in the case of high powered cars; so it is a non-problem.

The ideal transmission perfectly convert power of the engine into torque at the wheel so that at 0 speed torque is infinite. As speed increases torque at the wheel decreases following the ideal equation: $$ τ_w = {P \over v} $$

In practice power varies as function of rpm and rpm varies as function of the gear and v, this makes everything unnecessarily complex and car-specific. This approximation, let us use very simple math to draw the above conclusions.

Addendum: Drive-train, tire and other ignored losses As speed increases drag, unlike other losses, increases with the square of the speed. Other losses are usually very small compared to drag at high speed. At low speed power is in excess compared to the grip tires offer so other losses do not really matter.


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