The following treatment answers the question that Malacandra asked above regarding the power demands the propeller places on the wheels.

Notes:

1: The even 1.0hp number used below was chosen for ease of explanation. The actual HP numbers are dramatically higher on the Blackbird. The principles of course scale without issue.

2: For illustrative purposes, let’s take our DDWFTTW vehicle and remove the chain drive that links the drive axle with the prop shaft and replace it with a generator on the drive axle connected to an electric motor on the prop shaft.

3: Initially in the treatment we will ignore all losses in the system. This is just for ease of illustration. I will address the real world losses (which always exist) near the end of the post.

Alrighty – following is a simple energy/force analysis for a small propeller equipped DDWFTTW vehicle. In this analysis, we show that in a 27.5ftsec (~20mph) tailwind, and with the vehicle traveling DDW over the ground at 55ft/sec, the retardant force on the wheels needed to drive the propeller is less than the propeller needs to keep the vehicle at that speed.

A:

1.0 HP = 550 foot-pounds per second

(http://en.wikipedia.org/wiki/Horsepower)

This means that if we have 10lbs of force exerted on our chassis pulling it downwind @55ft/sec, we can harvest 1.0hp from our wheels with our lossless generator. (we’ll deal with losses below)

B:

1/2HP = 275 foot-pounds per second

This means that at 27.5ft/sec, 10lbs of force can be produced by a lossless propeller consuming 1/2hp. (we’ll deal with losses below)

C:

If the wind is blowing at 27.5ft/sec and our vehicle is traveling DDW at 2x the speed of the wind, the vehicle is traveling over the ground at 55ft/sec and through the air at 27.5ft/sec.

D:

Through the establishment of “A”, we know we can pull 1.0 HP from the wheels of the vehicle if it’s propelled by 10lbs of force, and through the establishement of “B” we can see that the propeller in the relative tailwind only needs 1/2HP input to produce that same 10lbs of force.

E:

We subtract the 1/2hp that the prop needs to produce its 10lbs of thrust from the 1.0hp that the wheels can produce from that same 10lbs and you have 0.5hp left over for the system losses.(told you we would get to losses).

In the real world we don’t have lossless components of course. If you consider an 85% efficient propeller (easy to achieve) and an 85% drive train (even easier to achieve with chain drive) we’ve still got nearly 1/4 HP left over for the Crr of the tires (very low for high pressure bike tires) and aero drag (which is also very low as our relative headwind is slight).

Do the same calcs on a no wind day and it’s easy to see that the wheels still produce 1.0HP at 55ft/sec, but now the propeller is forced to do work at 55ft/sec rather than 27.5ft/sec and it now takes a full 1.0HP at the prop to produce the 10lbs of force. This of course means that there is nothing left for losses and since there are *always* losses, the vehicle simply can’t motivate itself when there is no wind.

Malacandra, the short version of the above is this:

In a the above tailwind, the vehicle is moving over the ground twice as fast as it’s moving through the air. Using simple work=force/distance calculations it can be seen that this means the wheels can produce twice as much power as is needed by the prop to maintain that 2x speed.

This differential between the speed over the ground and speed through the air is of course the key to the entire brainteaser.

JB