Watts is watts?

Hey guys,

Like many of you, I'm eagerly awaiting and anticipating the release of the Mamba Monster Max. I'm currently running a few Mamba Max escs in my 1/8 scale conversions and, for the most part, really like them. The only reason that I'm awaiting the MMM is so that I can run more than 4S lipo.

I think it is an agreed upon thing when I say that voltage is your friend and it's better to run a higher voltage and draw fewer amps to perform a task.

Please forgive me if this gets overly simplistic, but I am relatively new to having to be even worried about all of this.

So...just to keep things simple, let's just say that I'm running a 4S 4Ah pack, which is capable of delivering around 59.2Wh of power (A*V, is that right?). Now, let's just assume that I want to have the same power on tap, but in a 6S configuration. I would divide 59.2 by the voltage of the 6S pack (22.2v) to find that I would need a 2.7Ah pack to have the same power on tap.

At this point, it appears to me that to run a more efficient system, you don't necessarily need to run a physically larger battery pack. You just need more, but smaller cells to get the same amount of stored energy.

To apply that to actually running a system in a buggy/truggy, let's just assume that I'm running a 7XL, with a target max rpm of 30K, on 4S Lipo and it's max current draw is "X." If the above is true, and I can make a correlation, then...with the above example, the 6S pack only has to have about 68% of the capacity of the 4S pack to have the same stored energy. Is this to say that (assuming that the batteries being used can meet the current demands of the system in which they are running) running a system, with a target rpm of 30K on 6S lipo will only require 68% of the current that the 4S system would need. If this is true, then the advantages of running higher voltage would be at least two fold.

First, you would have the advantage of the motor running more efficiently, requiring less current.

Second (and this is the biggie for me), you wouldn't necessarily have to bother with getting the latest and greatest cells that can deliver 100+ amps of current without significant voltage drop. If the above is true, in the instance where I would need a good 20C pack for the 4S setup, the 6S setup would only require a good 15C pack, which is much less expensive.

Is this right or am I off of my rocker?

[skellyo]

This thread should answer that for you:
http://www.rc-monster.com/forum/showthread.php?t=5977

[Serum]

You mix different things up.

Power isn't the same as capacity.

If you want the same capacity; (assuming you mean to achieve the same runtime) you can go with 60V 10Ah batteries or 6V 100Ah.
But i understand where you want to go;

I would prefer a 6S setup 3200 over a 4S 5000. Even though the 4S setup has got more stored energy. Most likely, assuming both motors on both setups run 30000 rpm at WOT (different KV motors) , the 6S setup will run cooler (more efficient)

You could play with the EPA on your transmitter and use a higher KV motor on a higher voltage and tame it down by limiting the throttle to lets say 60 percent.

OK, I'll go read the other post. I missed it somehow.

Serum,

Thanks for taking the time. Just to discuss the points that you made above for a sec. If I understand correctly, the 5K pack will have more stored energy (= mAh?) than the 3.2Ah pack, but the higher voltage system will expend LESS energy doing the same job as the 4S system?

If that is the case, then my point should still be valid. You also need "better" battery packs with a 4S system than you would with a 6S system. Notwithstanding the price of a controller to handle 6S, getting into a 6S system would make for sense for the more budget consious than getting into a 4S system.

[Serum]

It's quite simple;

P=UXI

1000watts = 14.8VX67.5A
1000watts = 22.2VX45A

[nativepaul]

You are perfectly correct for the first part a 4s 4AH pack has the same energy as a 6s 2.7AH and your battery will weigh the same, so if you gear you 7XL way down you will get the same input power, and maybe more power out due to greater efficiency at lower amp draws, however your motor will be spining half as fast again, 45k instead of 30k on 4s and this may lose you some of the efficiency that you just gained. It would be better to fit a 10xl instead of the 7xl on the original gearing, to spin the same 30k rpm, the 10xl has its max efficiency at lower currents than the 7xl so its you lose none of the efficiency that you gained with higher volts.

The second benefit isn't realy there, if you where to carry on using 4AH packs in 6s you could use 15c instead of 20c, but your battery would weigh half as much again as the 4s pack.
If you use a 2.7AH 6s pack like you sugested to keep the weight the same, you will be pulling less amps but have smaller cells to pull it from so it is still the same 20c rate.

If you want to use cheaper lower C cells you will have to install larger cells which isnt much cheaper per AMP you need, but is alot cheaper per AH capacity, but you will pay for it with weight, for example (using made up prices) a flightpower evo20 6s 2500mah costs around £100 and weighs 400g
a cheaper lower C lipo would have to have 4000mah to handle the same amps and a cheap 4AH pack will likely cost the same £100 as the evo 2.5AH, that seems like an extra 1500mah worth of runtime for free, however nothing is free, it's just that you don't pay with cash you pay with weight instead by hauling another 200g of battery around with you.

lol you got loads of replys while I was writeing this, I type too slow.

[Finnster]

You are on the right track, and your general conclusions are correct, but you are messing up the units/terminology a bit. Power would just be the wattage, but you are calculating the Energy of the batt, either Watt-hours or Joules (=> ~213 kJ.) Ie Energy is Power over time.

You will be expending the same amt of E & P regardless, P is what moves the truck, its just you will be building P with V instead of A. You know this tho, as you are trying to find a batt with the same Power (output potential) and Energy (work capacity) but higher V.

First, you would have the advantage of the motor running more efficiently, requiring less current

One note tho, I don't think its important to make a big deal of the efficiencies. Both motors (of the same line) can be highly eff, but the lo kv motor will better by a ~couple %. This is not going to make dramatic differences is heat generation. What's going to make the greatest diff is the amp reduction, as heat gen is directly related to current. Look at the data @ Lehner comparing a 2000XL @ 22V vs a 4200XL @ 11V. This would be comparable to converting a 3S system to a 6S system. The power output is about the same, the rpms are ~same, and even the eff are only down a bit across the board. The huge diff is the amp draw. Its half, thus heat gen would be ~half.

Nor would I assume a motor running at hi rpm is going to be in-efficient or automatically undesireable. Again the charts. Also see the one for the 1950 6T/ high amp. Its over 94% peak eff @ 68K rpm. Its actually more eff than a 1950/11 turn running at same voltage and ~1/2 rpm. However, for a given output in our range (~1500W) the 11T is more eff, but the current draw is the same. Tq is ~half go gearing should about follow. The end result looks like they should both be about the same speed wise, but the 11t run a bit cooler.

=> New high current/ high voltage controllers should make motor choices interesting. :)

[BrianG]

If you can use an equivalent HV solution, one of the side benefits is the use of thinner gauge wire and connectors because there are less losses with lower current. Also, the ESC doesn't have to be so robust. It just becomes a matter of finding an ESC that will handle HV.

[Serum]

There is a contradiction in your post finn..

I don't think its important to make a big deal of the efficiencies.
and
heat gen is directly related to current

This is not true.

a 150A lurking motor on 10V can be more efficient than a 150V motor on 10A.

It IS all about efficiency. they higher the voltage, the less 'important' the resistance becomes But like i said, it's not just 'with high amps comes high temperature'

[Finnster]

Originally Posted by Serum

There is a contradiction in your post finn..

I don't think its important to make a big deal of the efficiencies.
and
heat gen is directly related to current

This is not true.

a 150A lurking motor on 10V can be more efficient than a 150V motor on 10A.

It IS all about efficiency. they higher the voltage, the less 'important' the resistance becomes But like i said, it's not just 'with high amps comes high temperature'

I was waiting for your post to that. ;) That is part of the point I was demonstrating with the 1950 6/HA, tho I had to cut the post off somewhere, so it got a bit lost. (My point assumes they have to be reasonable eff to begin with, if you have a 50% eff motor you have a prob.)

So unless I forgot my physics or I missing some other point, heat is a function of resist and current.

The motor is consuming a whopping 200A and remaining 94% eff. Impressive. But since that current is so high, heat gen is still signifigant. :dft002: But at 6400W @ 64V/100A, the motor could even be same or less eff and put out less heat. Thus you could run a poorer quality motor, but still have a cool system.
If the eff % are just changing a couple %, I don't see that the being the critical factor over the current you are designing to run thru the system.

[zeropointbug]

Finnster, are you sure you have your understanding or efficiency down?

You said: "But at 6400W @ 64V/100A, the motor could even be same or less eff and put out less heat. Thus you could run a poorer quality motor, but still have a cool system."

Heat output does not have to do with current directly, it's how much power you are putting in, how well the motor transforms that energy, and how much mechanical power is put out at the shaft. It doesn't matter if a motor has 94% eff. at 100v/5A, or 94% at 50v/10A... you will still have the same heat output for a given power input.

The correct assumption is that for a given motor, it will run more efficiently the higher the voltage. And as you up the voltage, the smart thing to do is usually run a higher turn motor (thats with same gearing). The amount of heat that is generated from current will go down proportionally as you increase voltage, and increase turns. This is NOT total heat output however, there are other losses too that contribute.

I think these larger RC cars such as truggies, we need to increase the voltage to say 36V, but that's just IMO. That would be using something like a 16-18 turn XL size motor. Or better yet a Neu, or LMT. Efficiency would go up several % points, and that makes a HUGE difference as far as heat output is concerned.

Zero

[Finnster]

Originally Posted by zeropointbug

Finnster, are you sure you have your understanding or efficiency down?

You said: "But at 6400W @ 64V/100A, the motor could even be same or less eff and put out less heat. Thus you could run a poorer quality motor, but still have a cool system."

Heat output does not have to do with current directly, it's how much power you are putting in, how well the motor transforms that energy, and how much mechanical power is put out at the shaft. It doesn't matter if a motor has 94% eff. at 100v/5A, or 94% at 50v/10A... you will still have the same heat output for a given power input.

The correct assumption is that for a given motor, it will run more efficiently the higher the voltage. And as you up the voltage, the smart thing to do is usually run a higher turn motor (thats with same gearing). The amount of heat that is generated from current will go down proportionally as you increase voltage, and increase turns. This is NOT total heat output however, there are other losses too that contribute.

I think these larger RC cars such as truggies, we need to increase the voltage to say 36V, but that's just IMO. That would be using something like a 16-18 turn XL size motor. Or better yet a Neu, or LMT. Efficiency would go up several % points, and that makes a HUGE difference as far as heat output is concerned.

Zero

You know, I was thinking about this on the way home (forgive me, I've had a long crappy day and feeling a bit off.) I made that statement, but either way that leaves ~400W of wasted wattage. "Where would that go?" Then realized that it has nowhere to go other than as heat. I think I have been too stuck on the resist/amp relationship, treating this as a simple circuit and its effects on the rest of the system, but its more involved than that.

Yes I do see where then 1 or 2% would make a big diff when talking about the large amts of power we are trying to generate. 2% of 1500W is 30W, which is a sig amt of heat to try and dissipate. The eff differences seem more dramatic at "non-ideal" loads, where the eff diff can be several %, so this would be more dramatic in totality.

My thinking was along the lines of: would I prefer a 1500W system on 4S@ 92% eff, or a 1500W system on 6S@90% eff? (just a thought experiment, say comparing a 4s LMT vs a 6S XL) thinking the diff in eff would be more than made up for by the lower current demands. Its now apparent this choice isn't so obvious....

Yup, this is all correct from reading over it quickly. One benefit of running a more efficient HV setup is that you can reduce the size and weight of your packs by using lower capacity cells to give a lower overall power capacity while still retaining max power output and run time.

So you gain efficiency by reducing weight, reducing electrical losses, and reducing motor losses.

[Serum]

Assuming you mean total setup efficiency, the 4S setup would run cooler, since it's 2 percent more efficient.

[Scoob]

I'm a little confused but I have some questions.

Regarding what finnster is saying with equal power setups, the HV/LC (high volt low current) setup will cause less heat to the ESC, wires, connections ect. from less current flowing through them. You could even get away with a less amp capable ESC, wire of higher resistance, ect. This is correct right?

However, with the motor the total wattage that is being put to the motor, whether it be by HV/LC or LV/HC, is not flowing through the motor but being converted to mechanical power. Therefore the efficiency at which it converts the wattage to mechanical power is the diciding factor for heat. Is this correct?

The higher turn motor has higher resistance and that is why, even though it pulls less current, it will not create less heat at same power level unless it is more efficient. Is this correct?

Trying to understand this better.

Thanks.