A body's speed limit due to air resistance?

[pconstantino]

Hello everyone.

I've got a question that has been bugging me ...

Do bodies have speed limit here on earth?

Like cars, why do they have speed limits? is it because they are made this way for safety or is it because air resistance increases as speed increases?

Not only cars but all bodies in general like jets, can it accelerate non stop ? will air resistance increase and balance it so that it will have a constant speed ( as happens ) ?

And like... if a car moves with a resistance due to air + friction of 100N but is being pushed by a force of 150N, it should accelerate forever, if there was a car with physical capabilities, would it accelerate forever?

Thank you.

 

[mgb_phys]

Ignoring relativistic effects (so at speeds much less than the speed of light) the limit is when the air resistance is equal to the engine power.

The air resistance increases with the square of speed, so at 200km/h you need 4x as much power as at 100km/h - which is the main reason for a limit.

The fastest speed for real vehicles (ie not bullets) are probably for rocket powered sledges run on rails to test missiles, you have an enormous amount of power (a rocket engine).

 

[pconstantino]

Mbg, thank you for your reply.

You said the resistance(Newtons) equals the power (watts) of the engine? how is that? i didn't understand the mix of units, did you use a different connotation?

 

[mgb_phys]

In lose terms - terminal velocity is when the motive force equals the drag force.

For a wheel driven vehicle the drag would equal the force of the tire on road, which multiplied by the distance the wheel rotates divide by time = force * distance /time = energy/time = power.

 

[russ_watters]

The resistance equals the force generated at the wheels. The required engine power is actually a cube function of speed, though. Wind resistance is a square function of speed, whereas engine power is a linear function of speed (ie, if you hold torque constant, engine power doubles with a doubling of speed). Combine the two and you get a cube function.

Also, you asked about speed limits in a way that implied you were asking about the law. The law is about safety.

 

[pconstantino]

so energy = work ?

 

[russ_watters]

so energy = work ?

Work is a mechanical form of energy.

 

[Lsos]

As has been said, it comes down to air resistance. To go slowly doesn't take much power. A car may need maybe a few horsepower to go 30mph. But the need for power goes up QUICKLY. Jets, for example, need literally tens of thousands of horsepower to be able to fly as fast as they do...and it's simply because of air resistance.

As you said, if a car moves with an air resistance and friction of 100N, but is being pushed with a force of 150N, it will indeed accelerate...except as it accelerates, the air resistance and friction rises. QUICKLY. And very soon indeed you will be at a point where the car is being pushed by 150N, but is also being slowed by the air and friction by 150N...and that's when you have reached your speed limit.

 

[jack action]

The drag force felt by a body due to air resistance:

[tex]F_d= \frac{1}{2} \rho C_d A v^2[/tex]

Where:

[tex]\rho[/tex] = air density
[tex]C_d[/tex] = drag coefficient (depends on the body' shape)
[tex]A[/tex] = frontal area of the body
[tex]v[/tex] = velocity of the body

When the maximum force of whatever propelled the body is equal to the drag force, then the speed is constant (acceleration = 0). That speed is known as the terminal velocity.

If you multiply the drag force by the velocity of that body, you get the power needed to maintain the body at that speed, i.e.:

[tex]P_d= \frac{1}{2} \rho C_d A v^3[/tex]

For example, someone who jumps from an airplane will feel the air resistance. Due to is body shape, the jumper will have a drag coefficient and a frontal area - both will be different depending on its position (flat or "diving") - and the force pushing him downward will be its own weight (mg). Hence the terminal velocity of the jumper will be:

[tex]v=\sqrt{ \frac{mg}{\frac{1}{2} \rho C_d A}}[/tex]

 

[Mazerakham]

Yep. Liked the last post, btw.

And yea, you know you've reached the "speed limit" when the sum of the forces equals zero. I just presented an example problem which encapsulates all of this, which is on the thread (also made my pConstantino) "What's the use of power" The only difference was, I based my problem in a universe where air resistance is proportional to v, not v-squared, so it's less realistic. If you watch the solution to the problem, you'll see that, if we were to make it realistic and have air resistance proportional to v^2, then we'd end up with a cubic equation to solve (and who wants that??)

RIP Cardano

--Jake

 

[rcgldr]

drag_force = 1/2 rho Cd A v^2 is an approximation for speeds below about mach .4. Above that, and the math gets complicated:

wiki_ballistics_models_compared.html

Rocket sled land speed record video (6400+mph in about 6 seconds). I recall some mention of a clear plastic tunnel near the end of the track filled with helium on these tests, not sure if it was used for this test.

 

[pconstantino]

The drag force felt by a body due to air resistance:

[tex]F_d= \frac{1}{2} \rho C_d A v^2[/tex]

Where:

[tex]\rho[/tex] = air density
[tex]C_d[/tex] = drag coefficient (depends on the body' shape)
[tex]A[/tex] = frontal area of the body
[tex]v[/tex] = velocity of the body

When the maximum force of whatever propelled the body is equal to the drag force, then the speed is constant (acceleration = 0). That speed is known as the terminal velocity.

If you multiply the drag force by the velocity of that body, you get the power needed to maintain the body at that speed, i.e.:

[tex]P_d= \frac{1}{2} \rho C_d A v^3[/tex]

For example, someone who jumps from an airplane will feel the air resistance. Due to is body shape, the jumper will have a drag coefficient and a frontal area - both will be different depending on its position (flat or "diving") - and the force pushing him downward will be its own weight (mg). Hence the terminal velocity of the jumper will be:

[tex]v=\sqrt{ \frac{mg}{\frac{1}{2} \rho C_d A}}[/tex]

Is Power ever constant? like for a body, will its power be constant and hence can we use this power we found when this body is moving under different friction forces?

 

[jack action]

Is Power ever constant? like for a body, will its power be constant and hence can we use this power we found when this body is moving under different friction forces?

I'm not sure I understand your question, but let me try an answer.

Imagine you have a car with a known maximum power. With your gas pedal, you can adjust the power output of your engine anywhere from 0 to P_max. The force that propels the vehicle will be the power divided by its speed. Your car will encounter all sort of resistances, but the most important one will be the drag force. So, equating all forces will give:

[tex]m_{car}a=\frac{P_{car}}{v_{car}}-F_{drag}[/tex]

No matter what speed you're driving at, if it is constant it means that the acceleration a = 0, so both terms on the right side will be equal. If at that point you decide to press on the gas pedal to get more power, then you will have more power than you need to fight the drag force and this will transform into an acceleration. And as the drag force increases, the acceleration will decrease until it reaches zero again, thus this will be your new constant speed. Obviously, Once you will reach maximum power, you will also reach a maximum speed.

 

[pconstantino]

I'm not sure I understand your question, but let me try an answer.

Imagine you have a car with a known maximum power. With your gas pedal, you can adjust the power output of your engine anywhere from 0 to P_max. The force that propels the vehicle will be the power divided by its speed. Your car will encounter all sort of resistances, but the most important one will be the drag force. So, equating all forces will give:

[tex]m_{car}a=\frac{P_{car}}{v_{car}}-F_{drag}[/tex]

No matter what speed you're driving at, if it is constant it means that the acceleration a = 0, so both terms on the right side will be equal. If at that point you decide to press on the gas pedal to get more power, then you will have more power than you need to fight the drag force and this will transform into an acceleration. And as the drag force increases, the acceleration will decrease until it reaches zero again, thus this will be your new constant speed. Obviously, Once you will reach maximum power, you will also reach a maximum speed.

Jack thank you very much for your reply.

How do we calculate the max power though?

What I don't understand is how can power dictate how fast a car will accelerate under different friction forces and air resistance, i understand power is just a rate of doing work and hence power changes depending on the speed and forces, i don't get how speed changes according to friction or air resistance, don't know how long it takes for air resistance to balance a body's force and if it depends on the original force of the body

Thank you

 

[rcgldr]

How do we calculate the max power though?

Assume it's included in the engine specification, and estimate about 85% to 87% efficiency of the drive train (15% to 13% losses).

power changes depending on the speed and forces

Power = force x speed.

1 watt = 1 Newton meter / second
1 horsepower = 550 pound feet / second

 

[jack action]

Jack thank you very much for your reply.

How do we calculate the max power though?

What I don't understand is how can power dictate how fast a car will accelerate under different friction forces and air resistance, i understand power is just a rate of doing work and hence power changes depending on the speed and forces, i don't get how speed changes according to friction or air resistance, don't know how long it takes for air resistance to balance a body's force and if it depends on the original force of the body

Thank you

I've seen your https://www.physicsforums.com/showthread.php?t=391600" and I can see you have a problem with power. So let's look at a real problem, step by step.

Find the power needed for the vehicle with the following:

[tex]\rho[/tex] = 1.22 kg/m³
[tex]C_d[/tex] = 0.35
[tex]A[/tex] = 2 m²

[tex]\left( \frac{1}{2}\rho C_d A = 0.427 \: kg/m \right)[/tex]

[tex]m[/tex] = 1500 kg
[tex]v[/tex] = 20 m/s

The power is:

[tex]P = \frac{1}{2}\rho C_d A v^{3} = 0.427 (20)^{3}= 3416 \: W[/tex]

So, to maintain a constant speed of 20 m/s, that vehicle needs to develop 3416 W.

Now, you decide to press the gas pedal such that your car develops 10000 W. What will be the new speed you will eventually achieve?

[tex]v = \sqrt[3]{\frac{P}{\frac{1}{2}\rho C_d A}}} = \sqrt[3]{\frac{10000}{0.427}} = 28.6 \: m/s[/tex]

This is the new constant speed that you will, eventually, achieve to balance all forces. What you don't seem to understand is what happen in between those 2 events. Let's look at it:

Find the acceleration just after you pressed the gas pedal:

[tex]a=\frac{P}{mv}-\frac{\frac{1}{2}\rho C_d Av^2}{m}}=\frac{10000}{1500 (20)}-\frac{0.427(20)^2}{1500}} = 0.22 \: m/s^2[/tex]

This means that for each second, your speed will increase 0.22 m/s. Hence, after 1 second, your new speed will be 20.22 m/s. Finding your new acceleration:

[tex]a=\frac{P}{mv}-\frac{\frac{1}{2}\rho C_d Av^2}{m}}=\frac{10000}{1500 (20.22)}-\frac{0.427(20.22)^2}{1500}} = 0.21 \: m/s^2[/tex]

Thus, after the 2^nd second, your speed will be 20.22 + 0.21 = 20.43 m/s

Repeating with the new value for speed:

[tex]a=\frac{P}{mv}-\frac{\frac{1}{2}\rho C_d Av^2}{m}}=\frac{10000}{1500 (20.43)}-\frac{0.427(20.43)^2}{1500}} = 0.208 \: m/s^2[/tex]

And so on. This will give the following graph:

You can see that it will take you around 200 s before you can achieve your new terminal velocity of 28.6 m/s.