Item 1584

BUY: Motor - RC Model - Plettenberg - Predator 37

 There is no page for Motor or RC Model, nor need for one ~ at present.

For Drawing see; 1520.dc for motor (in metric)

Predator 37 Notes:

Information on the Predator 37: http://www.plettenberg-motoren.com/UK/Motoren/aussen/Predator37/Motor.htm

The tall brother of the Predator 30/08 bribes with his incredible Power. It is currently the first and single engine on the market, that brings with only 1900 g (incl. Propmounting for 3 for Threebladprop) a Power of 15 KW (20.1 hp) - completely unrivalled. With the Predator 37/06 in his Raven, Bernd Beschorner won the German Acromasters and moved the professional world into astonishment. 1,900 g 4.2 lbs

The engine has 3 ballbearings 20 poles and clearly more Power than a Gasengine of 150 size does.

Weight: 1900g incl. propcouplesystem shaftdiameter: 10 mm cells: 30 - 40 cells / 10-14 LiPo

Relevant Notes on Predator 30:

For Predator 30 see 1520.html

From a forum:

Engine: Plettenberg Predator 37/6
Batteries: 14s3p Lithium-Polymer with 4900mAh
ESC: modified Future40-160H
Prop: Rasa 28,5/12, 3-Blade CFK

_________________________

Motor ~ Predator 37/6

Motorhersteller (Engine Manufacturer) ~ Plettenberg

Regler (Regulators) ~ Schulze Future 40-160

Reglerhersteller (Regulator Manufacturer) ~ Schulze

Luftschraube (Air screw) ~ 28,5 x 12 Zoll

Vollgasstrom (Full Power) ~ 300 A

Impeller (Impeller) ~

Flugakku (Air Battery) ~ FlightPower 4.900 mAh 14s3p

_________________________

they say in an email: "The biggest motor we produce is the Predator 37/6. The motor has with about 50V a power of 15 KW. It is developed to use it in big modell-planes. Because of the open housing and the badly needed cooling the motor can not be used in a bicycle or motorcycles."

_________________________

Relavent threads on R/Cforum"

http://www.rcgroups.com/forums/showthread.php?t=491653&page=2

http://www.rcgroups.com/forums/showthread.php?t=523329

Subject: Use of Predator37 motor

Question:
In order to calculate the gear box reduction ratio we need to know the exact motor RPM for a given power; for example we need 6.7 KW on shaft per motor; (that is 6.7/0.85=7.88 KW DC). What RPM do we need to have 6.7 KW on shaft? We have the same question for several operating points.

Would you have a curve that would give RPM versus power?

What would be the maximum power that could be achieved on shaft of a Preadtor 37, assuming that we use a Shulze Driver Ref: fut-xxl-40.300WK supplied at 39.2 V?

Calculation example:

From the two points available on your web page:

Predator 37

RPM with no load / volt: 170 1/min

current with no load about 11 V: 6,0 A

prop:

voltage

current

RPM 1/min

efficiency%

trust in N

CFK 28,5x12 RASA Kl. 3Bl

47,0

308,0

6300

    

CFK 29x12" Kl 3Bl

44,5

282,0

5900

    

We did some figures:

(47x308)x0.85= 12.306 KW on shaft, @ 6300 RPM, that is 18.65 N.m

(44.4x282)x0.85= 10.66 KW on shaft, @ 5900 RPM, that is 17.26 N.m

If we take an average of 18 N.m of torque, that means that to obtain 6.7 KW on shaft we need to turn at (6.7x9550)/18= 3554 RPM. Do you agree with this calculation? And in this case what would be the input voltage?

 

Reply:

The torque is dependent to the current not to the rpm, so you need approx 308A to get a torque of 18Nm

If you take the load point with 47V, 308A and 6.300 rpm you have a load rpm of 134 1/min/V.

If we calculate with 26V (3484 rpm), 308A and 85% efficiency you will get a torque of 18,6 Nm with a shaft power of 6,7Kw

Note: The motor is developed for higher voltages. This values are only theoretical, we don’t know the motor efficiency at this point and we can’t guarantee that the motor will run at this operating point. The motor is not suitable for a continuous run at this power. You have to pay special attention to the motor cooling.

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Latest revision; September 6, 2011