Juggling efficiency and power at partial throttle (cruise). - Page 3 - RC Groups
 Aug 03, 2006, 12:26 PM Dieselized User
Aug 03, 2006, 12:39 PM
Registered User
Quote:
 Originally Posted by gkamysz
Can you please clarify what the graph is showing? Where did you measure the motor voltage? The motor-side shunt was in 1 leg of the 3 ESC wires? And that went to an RMS meter?

Consider the case motor volts = 4V. Does that mean 8V at roughly 4A being pulled from the supply and that translated to 4V measured at a motor terminal using an RMS meter? And that there was 9A measured in one leg of the ESC?
Last edited by matttay; Aug 03, 2006 at 12:40 PM. Reason: Adding RMS clarification to meter
 Aug 03, 2006, 12:43 PM Registered User Patrick at Castle alluded to this... You can't simply "limit" a 10A controller (or motor) to a 50% throttle level at 10A, and assume that you have a handy "overdrive" you can use occasionally. The controller will see full-throttle current spikes all the time (say 20A, for argument's sake), and you will quickly destroy the controller, or motor. (controller will be the first to go though...) This is putting aside the natual inefficiency of part-throttle operation as it is!
 Aug 03, 2006, 12:45 PM Got more toys than my kid Heh...the obvious answer: both effective RMS voltage AND current change. Nonetheless, it looks like there is something to be said for "propping" for efficiency at 2/3 throttle, as long as the resulting prop is not too much at full-throttle. It's just we can't effectively simulate the best values yet. - Jim
 Aug 03, 2006, 12:57 PM Dieselized User Mattay, that graph is for a DC motor and controller at part throttle. Power supply was set at 8V and motor voltage, source current, and motor current were measured. All were measured with DC volt meters. Greg
Aug 03, 2006, 02:35 PM
Senile Member
Quote:
 Originally Posted by matttay Also, you comment in here (and Bruce also noted) that the currents aren't equal on both sides of the ESC. This is true for peak currents, but the average RMS currents absolutely are the same.
I'm sorry, but this isn't so. We're dealing with a series circuit here and it is a fundamental law that current in a series circuit is the same at any point in the circuit at any given time. At partial throttle the current will be pulsed on both sides of the ESC and be of equal amplitude. So it really doesn't matter where in the circuit you take your current reading, it is going to be the same. Of course if you take this reading with a meter it is going to show average current.

Voltage in a series circuit is a different matter. Each component in the circuit will drop a voltage equal to the instantaneous current times the resistance of that component. When you add up all of these individual voltage drops they will equal the applied voltage.

Because of this you will see different voltages on each side of the ESC. On the input side you will see the full battery voltage when the FETs are turned off. When the FETs turn on the voltage will be the battery voltage minus the current times the battery internal resistance. That is why the battery output decreases as current increases. So the amount of voltage fluctuation at the input to the ESC is dependent on the battery internal resistance and the throttle setting, i.e. the current. This is a DC voltage with a small ripple.

The voltage at the output of the ESC will be the input voltage minus the current times the resistance of the ESC when the FETs are turned on. The output voltage will be zero when the FETs are turned off.

Even when the ESC is being pulsed at a 50% duty cycle the the input voltage is still essentially a DC voltage equal to battery voltage while the output of the ESC will be a pulsed DC which will have an average voltage of half the battery voltage at the 50% duty cycle. This is the reason you cannot use the ESC input voltage to determine the actual voltage being applied to the motor.

Larry
 Aug 03, 2006, 02:58 PM Dieselized User I think I've been looking at this the wrong way in the past. I'll do some math and some tests to see if this works out. I'll post the results. Greg
 Aug 03, 2006, 03:30 PM Dieselized User No, I haven't been looking at it incorrectly. Though Larry's post almost changed my mind. I found this post by Bruce to prove it. I'm not an electronics guy. I know just enough not to burn up everything I experiment with. https://www.rcgroups.com/forums/show...9&postcount=33 Greg
Aug 03, 2006, 03:37 PM
Registered User
Quote:
 Originally Posted by matttay Also, you comment in here (and Bruce also noted) that the currents aren't equal on both sides of the ESC. This is true for peak currents, but the average RMS currents absolutely are the same.
They aren't.
That's one of the puzzles.

Let's say that you have a really simple setup.

A 1v battery, a perfect Brushed type speed controller, and a one ohm resistor.

At full power, there is one volt across it and its one watt. No one disagrees.

At half duty cycle, the current is an amp half the time, and nothing half the time.

The average current is half.

The average voltage across the resistor is half.

So half the current and half the voltage is one quarter the power right? But the battery is still providing full voltage at half the current... so that half the power right?

How can half the power be one quarter the power?

There is more going on than you suppose..and when the load is partly inductive, it gets even curiouser.
 Aug 03, 2006, 03:41 PM Got more toys than my kid Man, I need better tools. So, let's say I have a power station to replace the battery with variable voltage. As we know, voltage affects the RPM, and whatever current is required will be drawn. So, let's use matttay's data since he's the only one to do the leg-work so far (thanks!): BP21 motor with GWS 0843 on 3-cell pack A. 11.3V 9.3A 9990 73.4% (full throttle) B. 11.1V 7.2A 9180 75.0% C. 11.3V 6.5A 8880 73.9% D. 11.5V 4.1A 7380 66.2% E. 12.1V 2.3A 5520 47.1% (lowest throttle) Alright, this link gives full-throttle data. You can see his prop and the numbers are really close: http://www.peakeff.com/GraphKIR.aspx...+Hobbies+BP-21 Now, let's try matching throttle "D." I found a good match at about 8.0V: http://www.peakeff.com/GraphKIR.aspx...+Hobbies+BP-21 See his prop? 8.0V and about 6 amps draw. His battery shows 11.5V at 4.1Amps. The Watts ARE NEARLY THE SAME. Look at the efficiency! NEARLY THE SAME. I think we may have something here. HOWEVER!!! The battery sees 4.1amps, but the motor see 6 amps! What causes heat, voltage or current? That's right, current! That's why a motor will NOT run much cooler if you over-prop and just use partial throttle. There is more current than what the wattmeter shows. This supports what Bruce Abbott explained to us. The increased current vs voltage also explains why we usually see a drop in efficiency (as well as the increased switching losses from the ESC) at partial throttle versus propping to permit lower current at max voltage. Let's try extremely low and use throttle "E." I found a match to RPM at roughly 6V. http://www.peakeff.com/GraphKIR.aspx...+Hobbies+BP-21 So, I get about the same RPm with the same prop at 6V and 4 amps at near 60% efficiency. Matttay measured 12.1V, 2.3A, 43% efficiency. Hmmmm, he was burning more watts at lower efficiency. So, we could theorize that switching losses get worse as you go to lower throttles. It's hard for me to get exact values from the chart, for the watts difference is not as great as the efficiency difference (and they should be related if I matched the RPM properly). In any case, it looks like the motor is responding similarly to the equations. It's jsut a different ratio of voltage/current than what the battery is supplying. With this knowledge, I could size my motor for cruise. There will be different styles and simpler rules of thumb to follow based on the type of plane. Let's take a stab. I have an 8oz 3D plane (227g) that I want to power. Because I need crazy vertical, I'll say that "cruise" is when thrust=weight. I need 227g of thrust. That means spinning a GWS 8x4HD prop at 6800RPM. That requires 18.9W of power. Let's look at the Billet Bullet Singl motor again. I get 229g of thrust near 6800 RPM with that prop at 6.3V and 4.2amps: http://www.peakeff.com/GraphKIR.aspx...ngle+Cool+Wind Not too bad, but once again it looks like this motor's sweet spot is something between the 7x3.5 and 8x4. Also...This is almost half voltage! However, it does show that this would be close to the peak efficiency at cruise with the 8x4 prop, so this motor would be a good choice. As mentioned earlier, it can handle this prop at full throttle as well. For giggles, let's try the GWS 7x3.5 prop. For 227g of thrust, I need about 9500 RPM. I get 228g of thrust on the Billet Bullet at 8.1v and about 3.2 amps: http://www.peakeff.com/GraphKIR.aspx...ngle+Cool+Wind Whoa....the top of the efficiency curve! Fewer watts too. Also, the voltage drop isn't as low, so this might be closer to 2/3 throttle (more efficient at 2/3 than 1/2 for switching losses). Very interesting. Let's open it up to 11.1V with that prop: http://www.peakeff.com/GraphKIR.aspx...ngle+Cool+Wind 411g of thrust? That should be sufficient! Also, it is not that far from max efficiency. Yep, I think I found my prop. However, if my model was heavier, than the 8x4 might be a better choice. I think I might have a system here. We just need to confirm it with real data. Meanwhile, I need to get a new prop! A pity ryanl2006 is on vacation this week. I'd like to hear what he has to say to this. - Jim
 Aug 03, 2006, 03:47 PM Registered User Rember iron losses are proprtional to voltage/RPM. As you throttle back they go down too. Half throttle is generally about 1/4 power...half the volts and half the amps AT THE MOTOR. The motor inductance will reduce current PEAKS to less than you might expect. Its a really really complicated scenario, which is why I haven't made a clear statement.
 Aug 03, 2006, 03:57 PM Got more toys than my kid Yah, that's why I try to shoot for 2/3 throttle, since it feels like "half power" roughly. I'm trying to find a workable "rule-of-thumb" using existing tools. I think my method may be a reasonable starting point, but there is probably a lot of tweaking to do. I still think that shooting for 2/3 voltage may be a reasonable way to find a good efficient prop setting. I usually use this tool to figure out what prop at what RPM to give me the thrust and speed I want: http://www.badcock.net/cgi-bin/power...++20&Temp=23.0 From there, I could plug in 8V for the efficiency calculator and start fiddling with the kV rating until I get the proper RPM for that prop. From there, I could start looking at motors of that kV and start comparing what would be best for cruise. Ugh.....there is something to be said for the simplicity of prop and motor selection of gassers. Oh well! - Jim
Aug 03, 2006, 04:24 PM
Senile Member
Quote:
 Originally Posted by vintage1 They aren't. That's one of the puzzles. Let's say that you have a really simple setup. A 1v battery, a perfect Brushed type speed controller, and a one ohm resistor. At full power, there is one volt across it and its one watt. No one disagrees. At half duty cycle, the current is an amp half the time, and nothing half the time. The average current is half. The average voltage across the resistor is half. So half the current and half the voltage is one quarter the power right? But the battery is still providing full voltage at half the current... so that half the power right? How can half the power be one quarter the power? There is more going on than you suppose..and when the load is partly inductive, it gets even curiouser.
Vintage, you had me going there for a while

When you measure current and voltage at the input to the ESC you will see full battery voltage and an average current which is half of full current. Multiply them together and you get half power disipation. This part I agree with. However, for determining the power consumed by the resistor you cannot use average power and current, you must use peak power and current. Because for half the time the resistor will see the full voltage, pull 1 amp and disapate 1 watt and for the other half of the time the resistor will see zero volts, pull no current and disapate no power. This then averages out to half power disapation just like the input measurements show.
Aug 03, 2006, 04:29 PM
Dieselized User
Yes, jim this is the way I understand it.

I have in a spread sheet with a chart just for this. I've had the sheet for years and it has gone through many changes. Attached is a sample. The S400 6V motor is at 7V and 10A. The BP12 is at 10V and 7A. This was done to get roughly the same input power. The constants I have for the BP12 were swiped off the net somewhere so I have no idea how valid they are.

The question is how much additional(other than motor calculations would indicate) loss is there at part throttle and how do we predict it?

Greg