pic: The Cow that Jumped over the moon!

[JesseK]

Oh! I didn't realize the sheet was vertical. In that case, there isn't a design flaw so long as the plate is cut to allow air flow. There's a strategy flaw with point # 5 though:

Originally Posted by R08

4.1.2 Safety & Damage Prevention

R08 ROBOT parts shall not be made from hazardous materials, be unsafe, cause an unsafe condition, or interfere with the operation of other ROBOTS.

<blue box>
Examples of items that will violate R08 include (but are not limited to):
A. Shields, curtains, or any other devices or materials designed or used to obstruct or limit the vision of any drivers and/or coaches and/or interfere with their ability to safely control their ROBOT

[nighterfighter]

Originally Posted by JesseK

Oh! I didn't realize the sheet was vertical. In that case, there isn't a design flaw so long as the plate is cut to allow air flow. There's a strategy flaw with point # 5 though:

It would be safe as long as they put some sort of shroud over the blades.

[JesseK]

Originally Posted by nighterfighter

It would be safe as long as they put some sort of shroud over the blades.

The safety issue isn't in the spinning propellors. The safety issue is the vertical sheet (shown as cow print in their CAD picture) preventing the opposing drivers from controlling their robot -- as called out in the blue box beneath R08. I emboldened the appropriate wording above.

[nighterfighter]

Originally Posted by JesseK

The safety issue isn't in the spinning propellors. The safety issue is the vertical sheet (shown as cow print in their CAD picture) preventing the opposing drivers from controlling their robot -- as called out in the blue box beneath R08. I emboldened the appropriate wording above.

Ah. Yes, I wasn't quite sure what that was.

That vertical sheet could be made out of a mesh or something instead. It would still serve the purpose of blocking shots, but not interfere with vision.

On another note: Do you think that the FRC battery will be able to provide enough power to keep the robot flying at acceptable altitudes? If you've ever flown one of those remote control helicopters, you'll notice that as the battery gets drained the helicopter's maximum altitude dramatically falls, and as the battery gets even lower you can no longer gain altitude.

But I think it would be awesome to see a flying FRC robot!

[MARS_James]

I am more interested in which motors you will use for the propellers

[JamesCH95]

Do the power calculations required to fly and/or hover the minimum required equipment in a robot given the available motors and battery. It will soon become obvious why there has never been a flying FRC robot before.

[dondanvm]

This idea is legal. The lower hanging "shield" that is in a crazy cow print would not be needed. If you have flown or been on a helicopter, you need the open bottom or the robot to get air flow. With the open bottom, the prop wash would create a disturbance and affect the flying ability of all frisbee's that fly underneath it, providing great "D".

This being said, it ruins the spirit of the game.

Using the battery and CIM motor's, it is possible to lift that much weight with a helicopter style blade

[JamesCH95]

Originally Posted by dondanvm

This idea is legal. The lower hanging "shield" that is in a crazy cow print would not be needed. If you have flown or been on a helicopter, you need the open bottom or the robot to get air flow. With the open bottom, the prop wash would create a disturbance and affect the flying ability of all frisbee's that fly underneath it, providing great "D".

This being said, it ruins the spirit of the game.

Using the battery and CIM motor's, it is possible to lift that much weight with a helicopter style blade

Care to throw some numbers at this?

[dondanvm]

Originally Posted by JamesCH95

Care to throw some numbers at this?

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

[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.