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Old Jul 24, 2006, 02:27 PM
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How to calculate motor Watts ?

Hello all,

Lets read the specs of a Speed 400, from an online shop:

[START]
Speed 400 SMC-M-3321
Nominal Voltage 6.0v - Voltage Range 2.4 - 7.2v - No load RPM 18,000 - No Load Current 0.7 A - Current @ Max Efficiency 4.0 A - Stalled Current 25A - Max Efficiency 70% - Case length mm 37.8 - Case dia. mm 27.7 - Free shaft length mm 13.8 - Shaft Dia. mm 2.30 - Weight (g)73
[END]

How can we calculate the max power that this motor can give ?
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Old Jul 24, 2006, 03:44 PM
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Dunno.. But I can tell you from experience that that motor will runn happily at 10 amps, and much less happily at 12 and not for v long at 14 amps.
So Voltage times amps give wattage insert your Voltages... remember tho that efficiency is approx 50% at 10 amps
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Old Jul 24, 2006, 05:42 PM
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East Anglia, UK
Joined Sep 2002
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Quote:
Originally Posted by dtrip
Hello all,

Lets read the specs of a Speed 400, from an online shop:

[START]
Speed 400 SMC-M-3321
Nominal Voltage 6.0v - Voltage Range 2.4 - 7.2v - No load RPM 18,000 - No Load Current 0.7 A - Current @ Max Efficiency 4.0 A - Stalled Current 25A - Max Efficiency 70% - Case length mm 37.8 - Case dia. mm 27.7 - Free shaft length mm 13.8 - Shaft Dia. mm 2.30 - Weight (g)73
[END]

How can we calculate the max power that this motor can give ?
Essentially you can't from those specs.

But I can tell you that a speed 400 will just about handle 100W input for about 60W output at 9A/3s (11v) LIPO sort of level.

Need to gear it though. Or run a 4" prop
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Old Jul 24, 2006, 06:11 PM
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Greece, South Aegean, Thira
Joined Nov 2004
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Hello people, thanks for replying,
since you have the experience, could you tell me directly:

I just want to calculate the Watts/kg ratio on my two airplanes.

1. Stock Easy Star (8.4V NiMH, Permax Speed 400 motor, direct-drive,
Gunther stock prop) TOW is 640g, what is the W (roughly) ?

2. Stock Nebula XL 2200 (8.4V NiMH, Speed 600 motor, direct-drive,
DONT-KNOW stock plastic folding prop with two blades). TOW is 1600g,
what is the W (roughly) ?

Tomorrow is the maiden flight of [2]. Im STRIVING to predict the power/weight
ratio, to have an indication of what to expect for the initial climb. Manual says
I should NOT run with it, or throw it hard. Just toss. I have not fired up the
motor yet, so I have not even an idea how strongly it pulls (have not charged the
HUGE battery yet, the EZ* ones are too small to try - but not too small to fry!
see I rimed )

Well, I hope you understand my case, if you could estimate anything, feel free
to post. And wish me luck for tomorrow, this airplane feels big and heavy!!!
(but so did the EZ*, 1.5 year ago )

In any case, a friend of mine with a video camera is coming along !
So I ll talk to you folks again in either "Electric gliders", or "Crash
discussion"

Take care,
Dimitris
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Old Jul 24, 2006, 08:20 PM
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East Anglia, UK
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Quote:
Originally Posted by dtrip
Hello people, thanks for replying,
since you have the experience, could you tell me directly:

I just want to calculate the Watts/kg ratio on my two airplanes.

1. Stock Easy Star (8.4V NiMH, Permax Speed 400 motor, direct-drive,
Gunther stock prop) TOW is 640g, what is the W (roughly) ?
I'd say probly 10A at 7v..roughly..70W?
Quote:
2. Stock Nebula XL 2200 (8.4V NiMH, Speed 600 motor, direct-drive,
DONT-KNOW stock plastic folding prop with two blades). TOW is 1600g,
what is the W (roughly) ?
About 22A...at 7v..150W give or take.
Quote:
Tomorrow is the maiden flight of [2]. Im STRIVING to predict the power/weight
ratio, to have an indication of what to expect for the initial climb. Manual says
I should NOT run with it, or throw it hard. Just toss. I have not fired up the
motor yet, so I have not even an idea how strongly it pulls (have not charged the
Firm steady LEVEL shove will get a plane like that in the air. Not a HUGE climb availabale....let it pick up speed before using any back stick.
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Old Jul 24, 2006, 10:24 PM
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P= I X E
P = power
I = amps
E = volts
Thus if Volts = 7.2 and Amps = 10 then power is 72 watts
This is quick and simple and pretty accurate.
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Old Jul 25, 2006, 07:23 AM
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nu49er wrote:
"Thus if Volts = 7.2 and Amps = 10 then power is 72 Watts.
This is quick and simple and pretty accurate."

Sorry but its not that simple.

1. Amps are not easy to calculate. Depend on the prop, which makes
the load on the motor.

2. The OP is about the MAXIMUM Watts (or Amps if you like) that the
motors can take (without burning, that is).

So either way, its not that simple (I wish it was ! )

Dimitris
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Old Jul 25, 2006, 07:26 AM
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Quote:
Originally Posted by vintage1
About 22A...at 7v..150W give or take.
Thanks, thats useful,
why do you say 7V, since the pack is 8.4V ?
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Old Jul 25, 2006, 08:05 AM
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Quote:
Originally Posted by dtrip
Thanks, thats useful,
why do you say 7V, since the pack is 8.4V ?
Not when you are sucking 20A out of it, it isn't
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Old Jul 25, 2006, 08:24 AM
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dtrip:

An Astroflight Whattmeter( aka wattmeter ) inserted between the flight battery and ESC gives actual voltage, current, wattage, and amphours. This is one of the basic tools of the trade. You should get one or one of the clones. It ends such speculations.

ciao -rjf
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Old Jul 25, 2006, 12:35 PM
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United States, MA, Webster
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Quote:
Originally Posted by ray foley
dtrip:

An Astroflight Whattmeter( aka wattmeter ) inserted between the flight battery and ESC gives actual voltage, current, wattage, and amphours. This is one of the basic tools of the trade. You should get one or one of the clones. It ends such speculations.

ciao -rjf
Truer words were hardly ever spoken, Ray ...

Russ
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Old Jul 25, 2006, 01:52 PM
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dtrip it is that simple, P = I x E the assumption is made that you know your amps and volts. It has nada to do with calculating a motors current draw as you
have to know that data to figure power. The question was how to figure power (watts) not the motors amp draw. Ohms law is pretty simple and accepted. So max power that a giver motor can give is volts times its max amps so lets say the motors max current is 25A and the voltage is 10V then power is 250 watts. I have to agree on getting a watt meter or other similar incarnation as it makes the actual measurement quick simple and accurate.
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Last edited by nu49er; Jul 25, 2006 at 02:11 PM.
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Old Jul 25, 2006, 02:05 PM
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accidental double post
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Old Jul 25, 2006, 02:43 PM
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Joined Apr 2004
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I believe you can calculate max. INPUT power in watts using the following motor traits. Amps is a linear function from no load to stall. Maximum power out occurs at 50% of stall current (or speed), and about 75% of no load speed is the max. efficiency point. In this case 12 amps. x 6 volts = about 72 watts input. This is max. power not max. efficiency. Our motors do not like to be run this hard. Bare's experience is dead on. Another way to tell is speed. 1/2 of no load speed is at max. power out. In this case 9000 RPM IF you prop the motor to run at this speed you will be right at max. power out. 70% is stated as max. efficiency. This occurs at about 75% of the no load speed. So, 75% x 18000 = 13,500 RPM. IF it were me, I would prop this motor to be no slower than 12,000 static and let it unload in the air towards the 13,500 figure.
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Last edited by KenSt; Jul 25, 2006 at 03:04 PM. Reason: incorrect info
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Old Jul 25, 2006, 04:11 PM
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Hello all,

vintage1: 7.0V --> 8.4V very interesting ! didnt know that

nu49er: No, the question was "how to know the MAX power a motor can give".
Power = Amps x Volts does not tell you when the motor will start smoking

ray: Yes you are right. But Im a beginner (and on a budget) so I cant afford all
the gadgets right now. BTW I was thinking to use an "Amp-meter" (thats what its
called in Greek at least), that professional electricians use (i.e its not for RC use only).
You clip it OUTSIDE the cable, and shows you the Amps running through it. This is
cool, because in what you describe I dont like the "inserted between" part
EDIT: Now that I think of it, I wouldnt even buy this. Just borrow it from a
friend once in a while to check things (I build 1 airplane per year more
or less).

KenSt: Thanks very useful!

Dimitris
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