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  #16   Spotlight this post!  
Unread 10-01-2013, 10:36
dondanvm dondanvm is offline
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Re: pic: The Cow that Jumped over the moon!

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
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Unread 10-01-2013, 10:40
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Re: pic: The Cow that Jumped over the moon!

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Originally Posted by dondanvm View Post
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?
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Unread 10-01-2013, 11:10
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Re: pic: The Cow that Jumped over the moon!

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Originally Posted by JamesCH95 View Post
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
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Unread 10-01-2013, 11:17
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Re: pic: The Cow that Jumped over the moon!

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Originally Posted by dondanvm View Post
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
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Unread 10-01-2013, 11:27
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Re: pic: The Cow that Jumped over the moon!

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).
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Unread 10-01-2013, 11:29
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Re: pic: The Cow that Jumped over the moon!

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Originally Posted by Toa Circuit View Post
Okay, so here's the rundown of things:

4. Sit on the pyramid at the end of the game for maximum bonus points.
Just wanted to address this real quick; you can't sit on /top/ of the pyramid, where the chain hangs so that Frisbees can be thrown and caught in the top basket, because that would be interfering with game elements. You would have to hover and fly really close to the Level 3 bar, then latch on.

Also, what happens if you're not hanging onto the tower at the end of the match and your power gets cut? I'm guessing it wouldn't be well-liked to watch a 100 lb robot falling 10 ft to the ground
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Unread 10-01-2013, 11:30
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Re: pic: The Cow that Jumped over the moon!

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Originally Posted by MARS_James View Post
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.
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Unread 10-01-2013, 11:37
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Re: pic: The Cow that Jumped over the moon!

Quote:
Originally Posted by dondanvm View Post
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
Yes it does look technically feasible. My calculations show 6 CIMs running at peak power on a 4ft diameter rotor (just fits within the 54" cylinder with a bit of room to spare) can generate 140 lbs of thrust. With a superlight robot, you probably could make that work. This is the most efficient rotor configuration for area, but would be nearly impossible to control well. A hex-rotor configuration is more controllable, with 1 CIM each, but you'd lose a lot of rotor area, which starts to eat into your thrust capabilities.

That being said, if you run 6 CIMs at full power continuous a) your battery will probably drop to at least half-voltage within a minute and b) the motors will likely overheat and start losing power.

All of this ignores the safety aspect, which is that FIRST probably wouldn't allow a flying robot onto their field on safety grounds. The nets are designed to contain frisbees, not runaway helicopter bots.
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Unread 10-01-2013, 11:44
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Re: pic: The Cow that Jumped over the moon!

6 CIM's all working near full power? You'd trip the main 120A breaker in a few seconds.
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Unread 10-01-2013, 11:49
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Re: pic: The Cow that Jumped over the moon!

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Originally Posted by Jim Wilks View Post
6 CIM's all working near full power? You'd trip the main 120A breaker in a few seconds.
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?
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Unread 10-01-2013, 11:51
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Re: pic: The Cow that Jumped over the moon!

The point of this conversation being... Someday it will be possible. Try it outside of the competition for fun!!!
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Unread 10-01-2013, 11:52
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Re: pic: The Cow that Jumped over the moon!

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.
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Unread 10-01-2013, 12:37
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Re: pic: The Cow that Jumped over the moon!

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Originally Posted by Jim Wilks View Post
6 CIM's all working near full power? You'd trip the main 120A breaker in a few seconds.
Bingo.

Quote:
Originally Posted by nighterfighter View Post
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.

Quote:
Originally Posted by John View Post
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.
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Unread 10-01-2013, 12:37
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Re: pic: The Cow that Jumped over the moon!

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Originally Posted by JesseK View Post
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.

Quote:
Originally Posted by nighterfighter View Post
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.
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Unread 12-01-2013, 09:38
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Re: pic: The Cow that Jumped over the moon!

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.
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