large rigger

[pda75]

Hi

I am thinking about a large rigger. The hull will be home made, with the following set up

10S 2P 5000 mA cells

ESC : YGE 200

weight of the hull : 6 kg

Length : 1,20 m (eg 45")

The objective is to run 5 minutes at 130 km/h (eg 80 mph). My assumption is that I need 4000 W for this weight and this speed.

Given the price, I will not buy a motor for "a test" and I need to choose the right KV with limited risk of mistake. So I would like to get your advice on the best KV .

kv = 750 (lehner 3040/14). Speed would be 80 mph with an 1667 octura prop assuming 15% slipery and 3,4 V per cells

Kv = 808 (Lehner 3040/13). Speed would be 80 mph with an 1467 octura prop assuming 15% slipery and 3,4 V per cells

Kv = 875 (Lehner 3040/12). Speed would be 80 mph with an 1462 octura prop assuming 15% slipery and 3,4 V per cells

The only true data point I have comes from the nitro guys. They get 80 mph with 25 000 rpm and a 1667 octura with an equivalent hull.

off course, I can target more rpm and a smaller prop but I am not sure that it will lead to sufficient propeller traction and that slipery will not dramatically increase.

Thanks for your help.

cheers

Pierre

PS: a lehner 3060/9 has a 778 kv. It could be an option but it adds some more weight.

 

[Dan Hydro]

Pierre,

there may be some guy's that might be able to help you here but I beleive you would get a lot more answers on the FE Forum.

Dan.

 

[shawdaddy]

Get with a guy on this board named Steve Reesor. He has a set up that runs in the mid 80s and will run a full nitro heat race. He also has a video on youtube of the boat running. Also try him on a FE board. Hope this helps, Ryon Shaw

 

[MitjaJ]

You should probably account for more slip - around 20%.

25000 rpm/60 x 67/1000x1.6 = 160 km/h

80miles = 128 km/h

slip = 1-128/160 = 0.2 or 20 %

A CMB .91 has 6.8 HP (5kW) at 26000 rpm - that is 5000W/(2*pi*26000/60)=1.8 Nm of torque.

You are limited to 10Ah/(5min/60)=120 amps of average current to get 5 minutes ride time (maybe less since 5000mAh batteries are more like 4500 mAh batteries in reality).

You can get the Kt (torque constant of an electric motor) from KV like this:

Kt = 1 / (KV/60 x 2 x 3.14 )

Torque M is then Kt x current : M = Kt * I

From the above equations you get KV = 60 * I/ (M * 2 * 3.14)

If you enter your data you can see that you need a motor with KV ~= 636 rpm/V

So the motors you choosed are about right - the only thing is to decide you want to use more rpm and a smaller prop

or less rpm and a bigger prop.

Also the boat will probably weigth less than 6 kg. A lehner is about 1 kg, 10S2P about 2.6 kg, and the boat with all hardware should be less than 2 kg.

 

[pda75]

thanks all of you

Mitjaj : Where did you get the formula Kt = 1 / (KV/60 x 2 x 3.14 ) ? does it derives from Power = Torque x 2 x 3.14 x rotation speed = Voltage x Amps ?

P = 2πN x C

where C= torque & N = rpm

P= 2π Kv x U x C = 2π Kv x U x Kt x I

But also P = U x I

That’s why 2π Kv x Kt = 1

I want to check it because C = Kt x I is a linear proxy.

Cheers

Pierre

 

[MitjaJ]

For DC motors Kt (torque constant) is 1/Kv. (You should read a book about DC motors to get more info on this.)

Since "our" KV is in rpm/V you need to convert it in proper units [(radians/second)/V]

To get radians/sec from rpm you:

(RPM/60) x 2 x Pi - since one turn is 2 x Pi radians and one minute has 60 seconds

So a 2000 rpm/V motor has a Kt of:

Kt = 1/((2000/60) x 2 x 3.14) = 0.00478 Nm/A - for every amp of current flowing trough the motor you get 0.00478 Nm of torque.