pic: The Cow that Jumped over the moon!

[MARS_James]

Originally Posted by dondanvm

I used the calculation method from this website assuming that each CIM is just under 1/2hp and weight roughly 3 lbs. You could use all 6 cims or just 1. The math is complicated, but it walks your through pretty well. You certainly would not have a lot of weight room left but it is possible from my calculations.

http://www.heli-chair.com/aerodynamics_101.html


Still, not in the spirit of the game

Did you factor in the weight of bumpers, metal, crio, motor controllers, etc.?

Cause I can't imagine this happening but I would love to be suprised

[dondanvm]

Yes. Light light light everything. How bout some carbon fiber!!! The battery is the biggest weight. The rest of the electronics are fairly light compared. Build the frame small and start with the props vertical so you do not need much in the way of metal/carbon fiber or bumpers. This is all conceptual, not anything you can or should try. It would get shut down for ruining the spirit of the game as well as safety (carbon fiber explodes).

[nighterfighter]

Originally Posted by MARS_James

Did you factor in the weight of bumpers, metal, crio, motor controllers, etc.?

Cause I can't imagine this happening but I would love to be suprised

Exactly. The bumpers, chassis, cRIO, wireless router, PD block, and all the wires will be weight prohibitive. Even if you got the robot flying, you wouldn't have enough weight left to make any sort of manipulator.

[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.