I really like the simplified explanation of how much energy to takes to move the air out of the way (v)^3 - I've heard this stated for vehicles before which is why roof racks are so bad for mileage.
Does anyone know of a similarly simple explanation of how much energy a vehicle must use relative to it's weight?
The best way to think of simple back-of-the-envelope calculations is this:
To go from point a to point b at a cruising speed of v, you need to speed up to v, maintain v, and brake to a complete stop. Analyze each separately:
1. Speeding up. Since this takes so little distance (compared to the total length), let's ignore resistance. This speedup takes KE = ½·m·v² of energy to do.
2. Maintain speed. This means spending energy to resist air friction (drag) and ground friction. Drag is ½·ρ·v²·Cd·A [0]; friction is (at most) μ·m·g. Energy is force × distance, and since the distance at top speed is effectively constant, it's going to be O(m + v²) in this case. [1]
3. Braking. You can theoretically recover -½·m·v² here, but most of the energy is going to be lost to heat.
So, when you sum these up, the energy it takes to drive a vehicle a certain distance is linear in mass and quadratic in speed.
[0] The drag coefficient, Cd, is really weird. At low speeds, it's actually proportional to 1/v instead of a constant (i.e., drag is linear in velocity, not quadratic).
[1] Yeah, the article says that energy is proportional to v³. It's actually wrong--the power is proportional to v³. Whether or not the total energy is proportional to v³ or v² depends on what you hold constant: if you hold total time constant, it's v³, but if you hold distance, it's v².
That's actually a bit oversimplified for highly aerodynamic objects like aircraft. It is known that such shapes can approach (v)^2. A (v)^3 characteristic is more representative of shapes that actually do have to accelerate all the air up to the speed of the vehicle such as the proverbial flat plate.
This doesn't change the conclusion; you still end up with a low spot on the graph. It is just that the graph looks more like a parabola.
Comments
I really like the simplified explanation of how much energy to takes to move the air out of the way (v)^3 - I've heard this stated for vehicles before which is why roof racks are so bad for mileage.
Does anyone know of a similarly simple explanation of how much energy a vehicle must use relative to it's weight?
The best way to think of simple back-of-the-envelope calculations is this:
To go from point a to point b at a cruising speed of v, you need to speed up to v, maintain v, and brake to a complete stop. Analyze each separately:
1. Speeding up. Since this takes so little distance (compared to the total length), let's ignore resistance. This speedup takes KE = ½·m·v² of energy to do.
2. Maintain speed. This means spending energy to resist air friction (drag) and ground friction. Drag is ½·ρ·v²·Cd·A [0]; friction is (at most) μ·m·g. Energy is force × distance, and since the distance at top speed is effectively constant, it's going to be O(m + v²) in this case. [1]
3. Braking. You can theoretically recover -½·m·v² here, but most of the energy is going to be lost to heat.
So, when you sum these up, the energy it takes to drive a vehicle a certain distance is linear in mass and quadratic in speed.
[0] The drag coefficient, Cd, is really weird. At low speeds, it's actually proportional to 1/v instead of a constant (i.e., drag is linear in velocity, not quadratic).
[1] Yeah, the article says that energy is proportional to v³. It's actually wrong--the power is proportional to v³. Whether or not the total energy is proportional to v³ or v² depends on what you hold constant: if you hold total time constant, it's v³, but if you hold distance, it's v².
Airplanes can just fly a bit higher so the atmosphere is less dense and the drag is less of a problem. Non-flying vehicles do not have this luxury.
That's actually a bit oversimplified for highly aerodynamic objects like aircraft. It is known that such shapes can approach (v)^2. A (v)^3 characteristic is more representative of shapes that actually do have to accelerate all the air up to the speed of the vehicle such as the proverbial flat plate.
This doesn't change the conclusion; you still end up with a low spot on the graph. It is just that the graph looks more like a parabola.