# Calculating the force of an average car

• karabiner98k

#### karabiner98k

Hi everyone,

I'm interested to know how much kilograms force an average car can produce. I made the following calculations to answer my question but I'm not sure if I'm right or wrong!

Imagine we have a 1.8L engine that can produce a max torque of 155 nm @ 4250 RPM.

1st gear ratio = 3.455 Final drive ratio = 4.529 Tire size = 185/65R14 (Rolling Tire Radius = 0.289 meters)

It is a FWD car so we have a drivetrain loss of around 10%.

I calculated the acceleration force as follows:

Wheel torque = 155 x 3.455 x 4.529 = 2425 nm Assuming 10% drivetrain loss, we will end up with 2183 nm of torque at the wheels.

Now, to calculate acceleration force, I divided 2183 by tire radius (0.289):
2183 / 0.289 = 7553 Newtons of acceleration force.

Finally, to convert Newton to kilograms force:
7553 / 9.8066 = 770 kgf

Does it mean that this car can lift 770 kg of weight if its engine and transmission (both as a single unit) are used as an industrial elevator?

Imagine we have a 1.8L engine that can produce a max torque of 155 nm @ 4250 RPM.
Why didn't you stop there?

A Newton is a measure of force. If that torque of 155Nm acts through a 1 meter lever arm, how much force is produced?

And are you trying to design an elevator, or was that just an example that you are using...?

• karabiner98k
Why didn't you stop there?
Because I wanted to add the multiplication effect of transmission gears.

A Newton is a measure of force. If that torque of 155Nm acts through a 1 meter lever arm, how much force is produced?
Force = Torque / Distance

And are you trying to design an elevator, or was that just an example that you are using...?
Yes, I used an elevator as an example. I just want to know the amount of kilogram force the car wheels can produce.

Does it mean that this car can lift 770 kg of weight if its engine and transmission (both as a single unit) are used as an industrial elevator?
I would say the car can hold 770 kg of weight if used as an industrial elevator. (and that would imply some slipping somewhere, leading to more losses.) To move the weight, you need to accelerate it and that would require more force.

• karabiner98k
I would say the car can hold 770 kg of weight if used as an industrial elevator. (and that would imply some slipping somewhere, leading to more losses.) To move the weight, you need to accelerate it and that would require more force.
Can the same car apply almost 770 kg of force to a wall if it moves horizontally? (Assuming tire friction is zero)
Or we can put it this way:
If a human being is between the car and the wall, he will experience a crushing force of around 770 kg.

Can the same car apply almost 770 kg of force to a wall if it moves horizontally? (Assuming tire friction is zero)
Or we can put it this way:
If a human being is between the car and the wall, he will experience a crushing force of around 770 kg.
Tire friction is static, so it needs to be high. An easy assumption is 70% of the weight of the car is the max driving force.

• karabiner98k
What is confusing for me is that if we ignore the multiplication effect of the transmission, the engine itself seems to be as weak as or even weaker than a human being ! The engine produces 155 nm and with a tire radius of 0.289 meters it can produce 536 Newtons of force which is about 54 kilograms of force.

Does it mean the engine alone can lift no more weight than an average adult male?!
How is it possible that the explosion of petrol in 4 large cylinders can't be stronger than the arms of a human being?!

Last edited:
If an average car is ##1000 kg## and has an acceleration of ##3\ m/s^2##, then that represents a force of ##3\ kN##. That would be enough to accelerate a ##75\ kg## adult human at ##4g## or ##3g## against gravity.

You'd need mighty strong arms for that!

• karabiner98k
PS power and energy are really what matters in any case. The energy can only come from the engine. A small force at high speed represents significant power, potentially.

• karabiner98k and russ_watters
What is confusing for me is that if we ignore the multiplication effect of the transmission, the engine itself seems to be as weak as or even weaker than a human being ! The engine produces 155 nm and with a tire radius of 0.289 meters it can produce 536 Newtons of force which is about 54 kilograms of force.

Does it mean the engine alone can lift no more weight than an average adult male?!
How is it possible that the explosion of petrol in 4 large cylinders can't be stronger than the arms of a human being?!
No, you're ignoring the fact that that happens at a certain rpm. Maybe 3,000. Vs a bike at maybe 100 rpm. Then you gear it to match the needed wheel speed, which changes the torque.

For applications where you need a low rpm they do make very low rpm, very high torque engines (for ships, heavy equipment).

Combine rpm and torque and you get power, as @PeroK pointed out.

• karabiner98k and PeroK
Power is the rate at which a certain amount of work is done. It is not a force.
In this topic, I'm only interested to know about sheer force an internal combustion engine can produce and I have come into big problems!
According to my research, an average IC engine produces about 1 ton of force (1000 kgf) when each explosion pushes the piston down.
The crank radius is around 8 centimeters on average.
When crank angle is 90 degrees (Sin 90 = 1), an engine can produce lots of torque:

1000 kg x 0.08 meter = 80 kgm or 784 nm!
This is only for a single piston. Therefore, a 4 cylinder engine should be able to produce 3136 nm of torque (4 x 784).

However, the confusing part here is that most naturally aspirated 4 cylinder engines produce no more than 150 - 200 nm which is extremely low.

It just doesn't make sense to me!

Also, as an aside, what ultimately prevents a car or bicycle from continuing to accelerate is the speed relative to the road. At constant power output, the force reduces in inverse proportion to speed. Resistance forces increase, of course, but if you cycle with a tailwind the acceleration stops before you feel much wind in your face.
$$P = Fv$$

• karabiner98k and russ_watters
Power is the rate at which a certain amount of work is done. It is not a force.
In this topic, I'm only interested to know about sheer force an internal combustion engine can produce and I have come into big problems!...

It just doesn't make sense to me!
You just can't ignore the speed if you want to find/understand the answer. The way you are approaching the question the answer could be literally anything.

• karabiner98k
Can the same car apply almost 770 kg of force to a wall if it moves horizontally? (Assuming tire friction is zero)
Or we can put it this way:
If a human being is between the car and the wall, he will experience a crushing force of around 770 kg.
Assuming the tire friction force can take this force, yes.

You can get any force you want from any power source you want. It is just a matter of using the proper transmission (lever arm) to convert rpm to torque (velocity to force).
The engine produces 155 nm and with a tire radius of 0.289 meters it can produce 536 Newtons of force which is about 54 kilograms of force.

Does it mean the engine alone can lift no more weight than an average adult male?!
How is it possible that the explosion of petrol in 4 large cylinders can't be stronger than the arms of a human being?!
Imagine that the same engine connected to a tire radius of 2.89 meters produces only 5.4 kilograms of force. Use a tire radius of 0.0289 meters instead and you get a whopping 540 kilograms of force. What do you think about that? Hint: Mechanical advantage.

1000 kg x 0.08 meter = 80 kgm or 784 nm!
This is only for a single piston. Therefore, a 4 cylinder engine should be able to produce 3136 nm of torque (4 x 784).

However, the confusing part here is that most naturally aspirated 4 cylinder engines produce no more than 150 - 200 nm which is extremely low.
And if the crank radius is set to 0.8 meters, you would get 7840 N.m! And if set to 0.08 meters, only 78.4 N.m. What do you think about that? Hint: Mechanical advantage.

Also, 784 N.m is a maximum value and 150 - 200 N.m is an average value. Don't forget that the same piston produces no torque (even negative value) for one and a half revolutions in a four-stroke engine. The maximum force is absorbed and stored as potential energy within a flywheel that releases it more evenly throughout the cycle.

The best way to see this is by looking at a hit-and-miss engine (my favorite one!). The engine produces a single explosion only when the engine rpm drops below a pre-determined value. The more resistance you put on the engine crankshaft, the more explosions it makes. You can hear (and see!) each explosion distinctively. You can appreciate more the job done by the flywheel to spread out the energy over every revolution:

• karabiner98k, Tom.G, Lnewqban and 3 others
All you need to know is the dynamic coefficient of friction between the tires and the surface it sits on and the weight of the car to get the maximum possible horizonal force the car can produce.

Last edited:
• karabiner98k
...
How is it possible that the explosion of petrol in 4 large cylinders can't be stronger than the arms of a human being?!
It is one cylinder at a time, every two turns.
Each cylinder can’t deliver more work than the gas expansion delta pressure times the top area of the piston during one of the four strokes.
Its force goes down into the crankshaft at several angles, fighting internal friction and the valves train, oil pump, etc.

It is one cylinder at a time, every two turns.
Each cylinder can’t deliver more work than the gas expansion delta pressure times the top area of the piston during one of the four strokes.
Its force goes down into the crankshaft at several angles, fighting internal friction and the valves train, oil pump, etc.
...and somewhere on the order of a hundredth of a milliliter of fuel.

• Lnewqban
1000 kg x 0.08 meter = 80 kgm or 784 nm!
This is only for a single piston. Therefore, a 4 cylinder engine should be able to produce 3136 nm of torque (4 x 784).

However, the confusing part here is that most naturally aspirated 4 cylinder engines produce no more than 150 - 200 nm which is extremely low.
I'm not an expert in this area, but here are a few things I think may help:
• The cylinders aren't firing simultaneously. The timing is spread out so that there isn't one large jerk followed by a prolonged period of nothing while the pistons cycle back to their power stroke.
• The flywheel and various other parts of the engine and powertrain are accelerated with each power stroke, sapping torque, energy, and power.
• The maximum force on the piston occurs quite early in the power stroke, often between 10-20 degrees of crank angle. As the piston moves the pressure in the chamber falls quite rapidly, and by the time the crank angle reaches 45 degrees the chamber pressure has dropped to about 65% of max. At 90 degrees the chamber pressure has dropped to about 20% of max. Peak torque should occur between 30-40 degrees of crank angle (not an expert here, just eyeballed some graphs).
• The peak torque is NOT what is given on engine specs. That is something called brake torque. Brake torque is the time-averaged torque after you account for not only the average power stroke torque, but the negative torque associated with the other strokes of the piston(s). After all, you have to expend a good deal of energy, power, and torque to compress all that gas in the first place along with moving it in/out of the cylinder.
Let's say a force of 1,000 kgf occurs on the piston at a 35 degree crank angle, and that the crank radius is 0.08 m. That's only 450 Nm of torque. If we assume that the 1000 Kgf is the peak force, which occurs at an earlier crank angle (5-10 degrees) then our peak torque at 35 degree crank angle is even less at around 350 Nm.

But this is just the peak torque. A huge torque applied for a microsecond does next to nothing, so it's not the peak torque that really matters but the average torque. And the average torque is MUCH less. Over half the power stroke is operating at less than 20% of peak chamber pressure after all, and any force applied while the crank angle is near 0 or 180 contributes next to nothing to the torque thanks to sine losses (that's sine as in the math function).

The engine produces 155 nm and with a tire radius of 0.289 meters it can produce 536 Newtons of force which is about 54 kilograms of force.

Does it mean the engine alone can lift no more weight than an average adult male?!
How is it possible that the explosion of petrol in 4 large cylinders can't be stronger than the arms of a human being?!
You don't have the engine alone. You have attached it to a wheel. Besides, the two aren't directly comparable.

• Lnewqban
It is one cylinder at a time, every two turns.
Each cylinder can’t deliver more work than the gas expansion delta pressure times the top area of the piston during one of the four strokes.

• The cylinders aren't firing simultaneously. The timing is spread out so that there isn't one large jerk followed by a prolonged period of nothing while the pistons cycle back to their power stroke.
• The flywheel and various other parts of the engine and powertrain are accelerated with each power
You are right but the peak torque is reached in 4250 RPM and at this RPM the engine is turning so fast that the difference in time between the explosion of each cylinder becomes so negligible (less than hundredth of milliseconds) that we can almost think of it as a linear-force motor like an electric motor. I mean at high RPMs, it is like all 4 cylinders are almost firing simultaneously.

But this is just the peak torque. A huge torque applied for a microsecond does next to nothing, so it's not the peak torque that really matters but the average torque. And the average torque is MUCH less. Over half the power stroke is operating at less than 20% of peak chamber pressure after all, and any force applied while the crank angle is near 0 or 180 contributes next to nothing to the torque thanks to sine losses (that's sine as in the math function).
I also believe that sine function of crank angle is the main reason that kills peak torque in IC engines. Peak cylinder pressure is usually reached in 15 - 16 degrees ATDC and at this point, the leverage is low. Sin (16 degrees) = 0.275

• Drakkith