# How fast do four Tire-Propellers have to rotate to lift 2000 lbs? [closed]

I just saw this come out, Goodyear's Tire-Propeller Concept – a combination tire that can also act as propeller (well, more like a ducted fan I guess).

There's a neat video that goes with it, describing how the fan blades need to bend while being used as a tire. (Otherwise, one winter in New England would take them out!)

Any ideas how fast something with say four 18 inch diameter wheel/propellers, and 3-4 inch fan blades, would have to spin to generate enough downforce to actually get 2000 pounds off the ground?

The weight is the vehicle with 4 motors, 300 pounds of batteries, 600 pounds of passengers & baggage; comparable to my 2500 lb max gross weight airplane, but with weight savings of no wings or tail structure, and batteries vs fuel. 4 motor/fans, and allowing (requiring?) the lift off process to be from a stop, so that all 4 motor/fans could be rotated from road use to propeller use (vs rolling on 2 like the video shows). I imagine 14,000 RPM or some high speed would make quite a lot of noise, compared to the takeoff noise of my 180 horse power engine and 86 inch propeller.

Edit: This rim-driven propeller wheel doesn't look like it would have much total blade are for moving air. That would take even more speed, wouldn't it?

The Moller Skycar seemed quite noisy for a comparison.

• Your going to get supersonic tip somewhere around 13500RPM which would be limiting for a rubber propeller
– Dave
Mar 7, 2019 at 16:36
• Got to imagine that would be pretty irritating to listen to if nearby. Mar 7, 2019 at 16:39

Let's do a back-of-the-envelope estimate. As a simplifying assumption, assume that a propeller generates lift by accelerating air through an opening of a particular size, at one particular speed. The formula for the amount of lift generated in this case is:

$$\text{Lift} = \text{Opening area} \cdot \text{Air density} \cdot \text{Exhaust speed}^2$$

For each propeller, the needed lift is 500 pounds force, and the area of each opening is $$\pi \cdot (\text{9 inches})^2$$.

So we can calculate the needed exhaust speed using Wolfram Alpha: square root of (500 pounds force / (density of air * pi * (9 inches)^2))

The answer comes out to 230 mph. That's the speed that the air has to move downward through the rotors. I don't know how fast the propeller will have to move in order to make this happen, but it will be a comparable speed.

The amount of power required can be found by multiplying this number again by the amount of lift. According to Wolfram Alpha, that comes out to 307 horsepower (229 kilowatts) per rotor. So about 1,200 horsepower for the entire aircraft, if the rotors were 100% efficient.

Good luck with that.

• 1,200 HP ~ 1 PT6A turbine engine, so it seems in the neighborhood of reasonable. The AW609 tiltrotor's got 3,880 HP total. Mar 7, 2019 at 18:46
• Hmm, that doesn't seem reasonable to achieve with the simulations shown in the video, especially as one gets the impression the tires are being rotated from the rim using a magnetic repulsion kind of idea. Or maybe that was just the bearings, to allow high speed revolutions. Gotta get braking in there too somehow for road use. Mar 7, 2019 at 20:50
• The Williams FJ33 en.wikipedia.org/wiki/Williams_FJ33 used on the Cirrus jet FJ50 "produces between 1,000 lbf (4,400 N) and 1,800 lbf (8,000 N) static thrust", so is it correct to say one would need 2 of those to make 2000 pounds force to count the 2000 pound weight of the vehicle including 600 pounds for the motors themselves? The separate quad copter/road wheels sounds more feasible. Mar 7, 2019 at 21:00