Two exact same cars are running at a constant speed. Car A: 2200 rpm, 2nd gear, 20mph. Car B: 2200 rpm, 3rd gear, 40mph.

1. Same power. Car A has low speed, high torque (due to lower gear). Car B has high speed, low torque. Why does Car A need more torque to run at a constant lower speed and why do you need less torque at higher speeds?
2. Suppose, we provide more gas/power to Car A and crank it to 3500 rpm. The engine provides more force to the gear and the gear/car accelerates. After some time, Car A reaches 40mph at 3500rpm and maintains a constant speed. On the other hand, Car B is running at 40mph as well but at 2200 rpm. So, both cars are now running at same constant speed but Car A is at higher power/higher torque and car B is at low power/low torque. The torque "T" provided by Car B was enough to overcome the resistive forces at 40mph and keep the car running at a constant speed. Since, Car A has more torque than T now at same speed, what is that "extra" torque being utilized for? Car A is not accelerating as well.
3. The above situation is similar to downshifting we do while overtaking. See this for example. We're running at 3rd gear 40mph but we crank our engine keeping the speed same (holding the clutch), lower the gear to 2nd, and release the clutch. Now, at 2nd gear and same speed, we have more power and higher torque. BUT the car doesn't accelerate until we press the gas pedal again. Why? The torque we had at 3rd gear was enough to keep the car at constant speed but as soon as we increase the torque by lowering the gear, shouldn't the car accelerate without needing to press the accelerator again?

EDIT: I think I caught my mistake after some due diligence. I was thinking about power in terms of just rpm. Rpm dictates the speed of the engine which gets transmitted to the gears according to the gear ratio. Torque is something that engine produces separately. Both multiplied give me the engine's power. Engine's torque varies a lot depending on the resistance or the gas pedal. The engine's torque then gets converted to wheel's torque according to gear ratios. Under constant speed, engine produces torque equivalent to external resistance. On giving more gas, torque produced by engine makes the wheel's torque greater than the resistance and the car accelerates.

• Power relates to top speed and torque relates to acceleration. Jan 19 at 8:10
• I guess. When you change to the third gear, the rpm of the engine with the same power becomes lower. If the rpm of the engine is lower than the rpm of the wheels, the engine will not be able to speed up the car. Jan 19 at 8:39
• Please be specific when you mention torque, as to where torque is measured is very important. For example in your into paragraph you mean "torque at the wheels" and not "torque at the engine", although that depends if the cars are coasting or accelerating at full throttle. Jan 19 at 13:38

I will offer a simplified mathematical model of the car system, in hopes to provide some clarity.

1. A car of mass $$m$$ (in kilograms) has speed $$v$$ (in meters per second)
2. The wheel diameter $$d_W$$ (in meters) results in the wheel rotation (in radians per second) of $$\Omega_W = \frac{v}{d_W/2}$$
3. The driveline gearing (final drive + gear selected) with ratio $$\gamma : 1$$ results in the engine rotation $$\Omega_E = \gamma\, \Omega_W = \gamma \frac{v}{d_W/2}$$
4. Based on throttle input, the engine produces torque $$T_E$$ (in Newton-meter), up to a maximum value described by the engine torque curve $$T_{\rm max}(\Omega_E)$$ which is a function of engine speed.
5. The power (in Watts) produced by the engine is $$P_E = T_E \Omega_E$$
6. The torque at the wheels $$T_W$$ (in Newton-meter) is $$T_W = \gamma\, T_E = \gamma\, \frac{P_E}{\Omega_E} = \gamma\, \frac{P_E}{\gamma\, \Omega_W} = \frac{P_E}{\Omega_W}$$
7. The tractive force (in Newtons) at the wheels is $$F_W = \frac{T_W}{d_W/2} = \frac{\frac{P_E}{\Omega_W}}{d_W/2} = \frac{\frac{P_E}{\frac{v}{d_W/2}}}{d_W/2} = \frac{P_E}{v}$$
8. The force of resistance $$F_D$$ (in Newtons) due to air resistance is $$F_D = \beta \, v^2$$ where the coefficient $$\beta$$ is a factor of frontal area, shape, and air density. It can be estimated if the top speed of the car is known.
9. The acceleration $$a$$ (in meters per second squared) at this instant is $$a = \frac{P_E}{m\,v} - \frac{F_D}{m} = \frac{P_E - \beta v^3}{m v}$$
10. If you know the top speed is $$v_{\rm top}$$ then use $$P_E - \beta v_{\rm top}^3 = 0$$ to find $$\beta$$ (in kilograms per meter) $$\beta = \frac{P_E}{v_{\rm top}^3}$$

So when you convert things to power, the gearing does not matter anymore, other than max torque is a function of engine speed.

In the above analysis driveline losses are ignored, as well as may other simplifications just to get a very basic level of understanding of how things are related.

To ignore a driveline loss of 15% for example, you would multiply $$T_W$$ with $$(1-0.15) = 0.85$$ to reduce its value. The same with engine power where applicable.

The above acceleration expression can be integrated over constant power $$P_E$$ to give a distance $$x$$ vs. speed $$v$$ relationship, using $$x = \int \frac{v}{a}\,{\rm d}v$$, and this is $$v(x) = e^{-\beta x/m} \left( \frac{P_E}{\beta} \left(e^{-3 \beta x/m} -1 \right) + v_1^3 \right)^{(1/3)}$$ where $$v_1$$ would be starting speed of where constant power is applied.

Read also this related post here.

• All is good but the drag force at any moment is indicative of engine power, because through it only power is drawn if neglect road friction. Jan 20 at 10:44
• @NeilLibertine - drag force is independent of engine power in general. Only when you have a steady speed you can equate $P_E \approx \beta v^3$. Jan 20 at 14:07
• Drag force is given by square of speed and that speed comes from power of engine. So after a moment, in any gear or speed if pedal remains static then power and drag are same with extra rate of energy delivery, speed. Jan 20 at 14:30
• @NeilLibertine - At any moment in time position and speed is given. Speed is a function of acceleration over a large time period. But for purposes of analysis of systems, you look at a single time frame and find out how acceleration relates to time, position and speed. $$a = a(t,x,v)$$ is the general definition of a mechanical simulation. Jan 20 at 14:37
• Acceleration is change and exists momentarily. So whenever pedal is pressed, more gas for combustion giving more rpm and power. That transmitted to wheel as more speed. As change in speed is not instantaneously so you have to wait for moment, then fuel burning in engine gives constant speed which is equal to drag force. Actually drag force means limit of speed achieve by force causing motion. Jan 20 at 14:44

Two exact same cars are running at a constant speed. Car A: 2200 rpm, 2nd gear, 20mph. Car B: 2200 rpm, 3rd gear, 40mph.

Same power.

Assuming that you intend there to be no resistive forces then coasting at the same speed means no acceleration which means no force so $$P=\vec F \cdot \vec v=0$$ for both. Same power (0) is implied by coasting assuming no resistance.

The alternative assumption is that the resistive forces are less on the faster car. That doesn’t make sense, but it would have been helpful if you had been explicit about the No-resistance forces assumption.

Why does Car A need more torque to run at a constant lower speed and why do you need less torque at higher speeds?

It doesn’t. They both require 0 torque. The only way you would need more torque at a lower speed is if there were more resistive forces on the lower speed car. For example if the lower speed car were driving with sticky tires or into a strong headwind or uphill.

Car A reaches 40mph at 3500rpm and maintains a constant speed. … Car A has more torque than T now at same speed, what is that "extra" torque being utilized for?

Assuming no resistance forces, Car A does not have more torque. If they are maintaining constant speed then they both have 0 torque. They are both coasting. You would just be spinning the engine at high RPM and not delivering any power to the wheels.

Now, at 2nd gear and same speed, we have more power and higher torque. BUT the car doesn't accelerate until we press the gas pedal again. Why?

You don’t have the higher torque until you press the pedal again. You are just spinning your engine faster without delivering additional torque.

Consider two engines, no transmission, just engines directly running a shaft which can be loaded with different loads. What is the difference between a high load at a given RPM and a low load at the same RPM? The engine RPM is the same in both cases but the torque is higher in the high load case. Delivering the additional torque at the same RPM means that the force delivered by the pistons is higher. A higher force in the piston is produced by burning more fuel in each stroke. So a given engine at a given RPM can produce different torques by burning more or less fuel. The fuel consumption and the torque depends on the load, not just the RPM. You are appearing to think that torque and RPM are in some fixed relationship.

• Thanks for replying! I think I caught my mistake after some due diligence. I was thinking about power in terms of just rpm. Rpm dictates the speed of the engine which gets transmitted to the gears according to the gear ratio. Torque is something that engine produces separately. Both multiplied give me the engine's power. Engine's torque varies a lot depending on the resistance or the gas pedal. Under constant speed, engine produces torque equivalent to external resistance. On giving more gas, torque produced by engine becomes greater than resistance and the car accelerates. Jan 19 at 13:09
• So if vehicle is moving at constant speed then there is no force or torque, then try to off the engine why waste fuel. Jan 20 at 10:45
• Yes. If there were no resistance forces and the ground were level then that would indeed work fine
– Dale
Jan 20 at 12:14
• @Dale If there is no friction then there is no measure of force. So you mean once you heated an object, then it gives light for eternity if heat is removed because here there is no friction. Jan 20 at 14:47
• Your “so you mean …” comment is completely unrelated to anything that I said or implied as well as being unrelated to the topic of the thread. You are welcome to have the last comment if you want, I will not respond further to you here
– Dale
Jan 21 at 14:34

First address point 1 and 2. Why do we need torque. We need torque in given situation if we have to carry more load, going uphill or moving over muddy surface or where traction is needed. Now to point 3, where you assume that while in third gear when shift down to second, why there is need to press gas. Because lower gears are bigger and they need more rpm to run at same speed of higher gears.

Torque is intrinsic quality of ICE, that depends upon kind of fuel used. Diesel has more torque and thus more efficient. Reason is that diesel is more heavy or dense so less entropy. If diesel cars are made for racing then they can run faster than petrol. Only drawback is that they have small range plateau of torque, they need hogher rpm. That can be solved by increasing number of cylinders.

• Do dense materials have less entropy? How is entropy related to engine efficiency? Jan 19 at 13:41
• @AccidentalTaylorExpansion Dense materials have lesser numbers of molecules, that is why high octane petrol is more efficient because of long chain of hydrocarbon molecule. Entropy is related to unused energy, for an engine it is due to rise in temperature, for within system due to expansion of volume. Jan 19 at 14:23