Am i right to think that if you use a low kv motor or higher gearing and a higer cellcount with lower mah you are running more effiecent then a higer amp setup like is common now?
And would it not also increase the amount of cycles a pack will do?

Cmulder:

Greetings from an Irish lad in Toledo, Ohio.

All cell chemistries have internal resistance, some more than others. Nicd generally the lowest, nimh a little higher, and lipo more than the rest. There are other chemistries but these are the contenders in Er/c today.

The battery test vault shows that as current from a battery increases the voltage decreases generally speaking. So as the current draw goes up and the voltage goes down, there comes a point where the power transferred from the pack to the motor reaches a maximum power level. This is called, in the electrical parlance, the matching load for the power source. That is, the effective motor load where the Re=Vin/Iin. That is, where the Re value best approximates the internal resistance of the battery pack.

This is complicated by the fact that our load is really a 3 phase motor and inductance enters into the mess in an AC way. That not withstanding there will be a best load point or range more like it for really good operation.

Even if you prop for high current at static load conditions, the motor load drops as a function of air speed, and most people prop up a motor for high current but fly at much reduced power settings(closer to Re). So your general contention is basically true, however, finding the Re value is quite a bit more difficult in the real application when cells, motors, props and the rest must be purchased.

There are many luminaries in the Er/c hobby who agree with your assertion. Pack size, weight, capacity, configuration and performance level, not to mention the health of the family money tree, help suggest a best operating set up for you.

I suggest that you prop the motor for easy take offs from grass, fun aerobatics, not blistering speed nor unlimited vertical performance. 3D and pylon racing eat up batteries. Also charge the packs relatively slowly 0.5C to 1C. Dont use all the capacity before you recharge. Cool the pack before and during charging. And if lipos are the cell you like, get a balancer device or balancing charger.

I hope this helps, Happy Holidays.

fly high, brother -rjf

Originally Posted by cmulder

Am i right to think that if you use a low kv motor or higher gearing and a higer cellcount with lower mah you are running more effiecent then a higer amp setup like is common now?
And would it not also increase the amount of cycles a pack will do?

Gearing will make the power system more effiecent. Higher voltage compared to lower will also slightly improve effiecency as well - may or may not be significant (when you are taking about 2S-3S ) If you swing it way up to 5-8S, I yield to others for their input.

Gearing adds weight, as do the extra voltage - so if you look beyond the power system, things start to get a bit more complicated.

cmulder,

If you want a certain amount of watts to power you plane, and use higher voltage instead of more current. You will see longer batterylife.
Of course itīs possible to use xSxP packs, but only how much current a single cell se is important in this matter.

/Wheaz

Originally Posted by Wheazel

cmulder,

If you want a certain amount of watts to power you plane, and use higher voltage instead of more current. You will see longer batterylife.
Of course itīs possible to use xSxP packs, but only how much current a single cell se is important in this matter.

/Wheaz

So, two 3S 2100 mAh batteries connected in series for 22 volts running at 10 amps (220 watts) is better for battery life than two 3S 2100 mAh batteries connected in parralell running at 20 amps (220 watts).......

Danny

Originally Posted by cmulder

Am i right to think that if you use a low kv motor or higher gearing and a higer cellcount with lower mah you are running more effiecent then a higer amp setup like is common now?

No. If the motor is he same size and construction but simply has more turns on it, it wll simply be the same efficiency when you apply enough volts to get the same power input on the same prop.

And would it not also increase the amount of cycles a pack will do?

[/quote]
No, because typically you would use a smaller cell capacity to get the same weight.

Fiddilng with volatges and turns will not achieve greater effeicincy or pack life if the same wieght of pack and size of motor are employed. You do not get something for nothing.

DannyR, as I said, whats important is how much current the single cell se.

/Wheaz

are you assuming he is using an efficient motor vintage1? With cheaper built motors the increased amp draw of low wind motors could cause sagging effeciency from heat buildup.

I do see your point about keeping pack weight and can design constant though. What about the fact that lower turn motors pass more amperage? Isnt heat caused by amperage?

Originally Posted by DannyR

So, two 3S 2100 mAh batteries connected in series for 22 volts running at 10 amps (220 watts) is better for battery life than two 3S 2100 mAh batteries connected in parralell running at 20 amps (220 watts).......

Danny

220 watts on 6 cells is still 220 watts no matter how you hook them up. in both cases each cell will be pushing 10 amps. In case one it is obvious, but in case two the parallel packs will share the load and split the amperage. Even if you wired all six in parallel, they would still see 10 amps apeice.

Originally Posted by johnrobholmes

are you assuming he is using an efficient motor vintage1? With cheaper built motors the increased amp draw of low wind motors could cause sagging effeciency from heat buildup.

I do see your point about keeping pack weight and can design constant though. What about the fact that lower turn motors pass more amperage? Isnt heat caused by amperage?

Vintage. You need to get to bed earlier!

Typically, low turn motors can draw a higher current than high turn motors but this is because they use thicker wire. Because there are fewer turns, there is enough room in the motor to use a thicker gauge of wire. The thicker wire means that the resistance will be lower.

You are right that the effieciency decreases as the motor heats up. The heat comes from two types of losses: Copper losses are I x I x R and Iron losses are Io x V where Io is the no load current. A low turn motor has a low resistance, R but a high no load current, Io. A high turn motor has a higher resistance but a lower no load current. If you run both motors at the same rpmand the same load they will have about the same efficiency and consequently will get just as hot as each other. What will be different will be that the low turn motor will use low voltage and high current while the high turn motor will use higher voltage and lower current.

One more thing. For a given voltage, motors have a maximum efficiency current. Maximum efficiency occurs when copper losses equal iron losses. Sometimes, there are situations where you have chosen the load very specifically such as a scale sized prop or a particular ducted fan unit. In these cases, there may be a benefit in selecting a particular combination of motor and cells that will be operating closer to its maximum efficiency point than some other combination. Also, you may want to keep voltage and current within particular ranges because of the speed controller that you have or the actual cell capacities that are available.

In a perfect world, we would be able to select the voltage to exactly what we wanted. But we can't because cells come in multiples of either 1.2(ish) volts or 3.7(ish) volts.

So, it is more important to get the right setup that is operating efficiently but this could be either a low voltage/ high current solution or a high voltage/low curtrent solution (or anywhere in between).

I have always understood that higher voltage setups have less heat losses because the current flow causes excess heat build up. The reason that power transmission lines can be so thin is because they carry high voltage low current power that is converted to low voltage high current at the transformers. If they carried high current there would be much higher power losses. This has been my understanding up to now but perhaps I am about to be educated.

The really big gains from going low amps/high volts come when using cheap can motors.

These motors were designed for other uses than flying model planes, and their amps of best efficiency are so low that they'de be almost unusable as powerplants for airplanes when used at lower voltages unless we push way higher amps through them, and that pushes their efficiency very low, into the 40-50% range.

On these motors, big motor eff. gains can be had by using the same total watts through them by going much higher on voltage while holding amps much lower, ie: instead of using 12 amps at 7 volts, try 11 volts at 7.5 amps.

Both have about 83 watts, but the low volt one will likely be at 40% eff. while the high volts one will be perhaps 60%, so

low volts: 83 watts x .40% = 33.2 watts output
high volts 83 watts x .60%= 49.8 watts output

So by going the same exact input wattage but at higher volts you have a full 50% more power at the prop, all done through better efficiency.

All this becomes moot however with purpose built motors as they can set their amps of best eff. at almost any level they want, neatly bypassing the need to keep motor amps down to get good efficiency from them.

Batman,

If the losses just came from the winding resistance you would be right. But the other component of the losses is what is termed iron losses and is calculated from Io x V. Io is sometimes said to be a constant but if you measure it you will find that it actually increases with voltage. Anyway, the iron losses increase with voltage and maximum efficiency is where iron losses equal copper losses. So Io x V == I x I x R for maximum efficiency.

So this means that if you take a low turn motor with Io of 5 amps and put 10 volts on it with no prop, it will lose 50 watts (5 x 10) in heat! And it is not even doing any useful work yet!

The way to figure out how to best run a motor is to make an estimate of the heat losses that it can withstand. This can be tricky but with some experience can be estimated. Let's say a motor can withstand losing 100 watts without overheating. Divide that in two to get the iron losses and copper losses at maximum efficiency.

So, Io x V = 50 and I x I x R = 50. If you know Io and R, you can work out the ideal voltage and current to run at. You can then use that to choose the appropriate number of cells and prop.

Just in case you think that Io does not get very large, a Hacker B50 4L has an Io of 8.75 amps! The kV is over 6000 so you don't need many volts.

I have a robbe eolo heli with a kontronic 3700 kv twist and a jazz controller.

Curently i am flying with a old 8 cell sanyo 2200 mah pack wich gives about 6 to 7 min duration.

Since i already have updated my robbe infinety charger to be lipo compatable i am considering replacing the old pack with a lipo pack.

Its verry eazy to change gear ratio so i was thinking if i put a small pinion on the motor , and go with the highest cellcount the motor/ esc can handle (6) i will get a good duration with a decent headspeed for hovering while not stressing the pack / esc to mutch

Heli what? Sorry, I am not at all familiar with helis .

Messing around with Motocalc to do the calculations, it would seem that the maximum efficiency current on 8 NiCd cells is 30-35 amps giving about 250 watts. Are you running your pack this hard?

At this level (250 watts) the motor appears to be around 90% efficient with losses of about 23-25 watts.

Looking at other voltages to provide that same power, 3s lipo would draw 24 amps at 10.6 volts for a similar efficiency and 24 watts in losses. So a good setup but no more efficient than what you have.

2s, 4s, 5s and 6s solutions for the same power appear to be slightly less efficient with losses of 27-38 watts.

If you let me know what power level you are actually using (what current are you drawing at the moment? I guess it is hard to measure?) or would actually like to use I can run the numbers again.

One word of caution. Motocalc is not 100% accurate and I have input the motor parameters from the Kontronik web site for this motor. However, we should be in the ballpark.

great discussion.

one big question though, how can you truely measure losses? measure watts in and have a dyno measuring watts out?