pic: The Cow that Jumped over the moon!

[John]

Using the method from that website:

Area = #rotors * diameter^2 * pi/4
Area = 4 * 14in^2 * pi/4 = 616 in^2 = 4.28 ft^2

Power will be limited by the battery and main breaker, with a maximum continuous power of:

Power = voltage* current = 12 V * 120 A = 1440 W = 1.93 HP

PL = Power/Area = .452 HP/ft^2

TL = 8.6859 * PL^(-.3107) = 11.12 lb/HP

Thrust = Power * TL = 11.12 lb/HP * 1.93 HP = 21.5 lb

The battery alone is 12 lb. Each CIM is 2.8 lb. If I did the math correctly, I don't think you can fly.

[JamesCH95]

Originally Posted by Jim Wilks

6 CIM's all working near full power? You'd trip the main 120A breaker in a few seconds.

Bingo.

Originally Posted by nighterfighter

Each CIM wouldn't be experiencing very heavy loads though. The stall current is 133 Amps, and the free current is 2.7 Amps.

While they wouldn't be at free current, they wouldn't be drawing massive amounts of current if they are powering a light-weight rotor, would they?

If the CIM is operating at full power (337W) with, let's assume 0.65 efficiency (max efficiency, anti-conservative, but I'm only making a point) draws about 43A, which can pop each CIM's breaker, and with just 3 (let alone 6) you're running the risk of blowing the main breaker too.

Originally Posted by John

Using the method from that website:

Area = #rotors * diameter^2 * pi/4
Area = 4 * 14in^2 * pi/4 = 616 in^2 = 4.28 ft^2

Power will be limited by the battery and main breaker, with a maximum continuous power of:

Power = voltage* current = 12 V * 120 A = 1440 W = 1.93 HP

PL = Power/Area = .452 HP/ft^2

TL = 8.6859 * PL^(-.3107) = 11.12 lb/HP

Thrust = Power * TL = 11.12 lb/HP * 1.93 HP = 21.5 lb

The battery alone is 12 lb. Each CIM is 2.8 lb. If I did the math correctly, I don't think you can fly.

This too.

[artdutra04]

Originally Posted by JesseK

There's a fundamental design flaw with this, if you continue to use the sheet.

http://dsc.discovery.com/tv-shows/my...p-lift-off.htm

They demonstrated this with miniature helicopters -- the helos could lift themselves in unison when all connected to the same wooden lattice. Yet once they mounted a sheet underneath the lattice, the helos couldn't get off the ground.

Technically, if the surface underneath the bees/mini-helicopters was curvedl, they could use the Coanda Effect for lift.

Originally Posted by nighterfighter

Each CIM wouldn't be experiencing very heavy loads though. The stall current is 133 Amps, and the free current is 2.7 Amps.

While they wouldn't be at free current, they wouldn't be drawing massive amounts of current if they are powering a light-weight rotor, would they?

Nope. The mechanical power output is found by multiplying the torque times the rotational speed. For a DC brushed motor, torque and speed are inversely related, thus the output power forms a parabola with a maximum value at the mid-point. A CIM Motor at peak output power will be drawing 67.9 amps.

[Camren]

I am fairly sure this is illegal because you blocking the sight of a drivers station and it is not safe at all. plus with disks flying and hitting the wall you better hope that is a stable system or everyone would get to see a cliche ladder tipping scene.