Not blown away with increasing voltage

[RogerD]

My board - 6S SK 6364 245Kv. 12000Mah (4x3000Mah 6s packs). Tested range 7 to 7.5 miles.
Now upgraded to 24000Mah (2 x 12000Mah Multistar packs). Good for 12 to 15 miles - done 10 so far with lots left.

Upgraded mate's board to 10S 6374 195Kv, 8000Mah (2 x 4000Mah 10s turnigy) - range tested around 6 miles. 3.67 volts per cell at 5.5 miles.

I though going up in voltage would offer greater returns?

 

[Simonvtr]

Let me see if I can understand this

1st setup 6s 3000mah lipo x4 in parallel?
6s @ 4.2v = 25v x (3000mah x 4 = 12000mah (12ah)) = 25v x 12ah = 300watt hours

2nd setup 6s? 12000mah multistar lipo x2 in parallel?
6s @ 4.2 = 25v x (12000mah x2 = 24000mah (24ah)) = 25v x 24ah = 600watt hours

3rd setup
Now here is where I get confused by your statement.
If its a 10s setup then 10s x 4.2 = 42v x 8ah = 336watt hours
What is (2 x 4000mah 6s)? is it a booster pack on the existing 10s?

If we only look at the watt hours on each setup then the one with 600watt hour will give you more range.
Increase voltage but reducing ah will only increase your speed.
Also the motor will eat more juice to maintain the higher speed which will also reduce your range.

 

[bobfandango]

My board - 6S SK 6364 245Kv. 12000Mah (4x3000Mah 6s packs). Tested range 7 to 7.5 miles.
Now upgraded to 24000Mah (2 x 12000Mah Multistar packs). Good for 12 to 15 miles - done 10 so far with lots left.

Upgraded mate's board to 10S 6374 195Kv, 8000Mah (2 x 4000Mah 6s turnigy) - range tested around 6 miles. 3.67 volts per cell at 5.5 miles.

I though going up in voltage would offer greater returns?

I'm fairly confused. How are you measuring range? By my math, 6S @ 12Ah ~ 250Wh which rule of thumb says should be good for roughly 15 miles. How do you get 10S from two 6S packs? 3.67 volts is probably close to the average voltage across discharge, not the voltage when the batteries are dead. That is, that pack is roughly half charged at 3.67 volts... A 3.3 volt cutoff is more typical.

 

[RogerD]

I made an error and have edited it. The 10S setup is 2 x 10S 4000Mah packs in parallel

Let me see if I can understand this

1st setup 6s 3000mah lipo x4 in parallel?
6s @ 4.2v = 25v x (3000mah x 4 = 12000mah (12ah)) = 25v x 12ah = 300watt hours

2nd setup 6s? 12000mah multistar lipo x2 in parallel?
6s @ 4.2 = 25v x (12000mah x2 = 24000mah (24ah)) = 25v x 24ah = 600watt hours

3rd setup
Now here is where I get confused by your statement.
If its a 10s setup then 10s x 4.2 = 42v x 8ah = 336watt hours
What is (2 x 4000mah 6s)? is it a booster pack on the existing 10s?

If we only look at the watt hours on each setup then the one with 600watt hour will give you more range.
Increase voltage but reducing ah will only increase your speed.
Also the motor will eat more juice to maintain the higher speed which will also reduce your range.

 

[RogerD]

My board - 6S SK 6364 245Kv. 12000Mah (4x3000Mah 6s packs). Tested range 7 to 7.5 miles.
Now upgraded to 24000Mah (2 x 12000Mah Multistar packs). Good for 12 to 15 miles - done 10 so far with lots left.

Upgraded mate's board to 10S 6374 195Kv, 8000Mah (2 x 4000Mah 6s turnigy) - range tested around 6 miles. 3.67 volts per cell at 5.5 miles.

I though going up in voltage would offer greater returns?

I'm fairly confused. How are you measuring range? By my math, 6S @ 12Ah ~ 250Wh which rule of thumb says should be good for roughly 15 miles. How do you get 10S from two 6S packs? 3.67 volts is probably close to the average voltage across discharge, not the voltage when the batteries are dead. That is, that pack is roughly half charged at 3.67 volts... A 3.3 volt cutoff is more typical.

I tyoed wrong 8000Mah 10s = 4000Mah 10s x 2 in parallel.

My measurements are gps logged on rides until LVC on my Car ESC (on my 6s setup) (which leaves my packs requiring an 80% charge up - 9600Mah into a 12000Mah pack), which is bang on). I've never gone past 7.5 miles on that setup. I'm 90Kg. I'm assuming my new setup will give me double-ish - around 15 miles. Seems to stack up based on today's 10 mile ride. Bear in mind it's a converted Boom Boarda, which uses 90mm "rubbery" tyres that don't roll anywhere near as well as proper skate wheels, but they are "comfy"

 

[RogerD]

After correction, the 10S setup:

42 volts x 8Ah = 336Wh

So he should get more or similar to my older setup of around 7.5 miles (exactly the same boards and wheels etc).

Looks like I've set the LVC too high on his Vesc. Will play with it.

Any recommendations? Hard to know what sag to allow for.

 

[Simonvtr]

That makes alot more sense ^^

Ok so 1st setup with a 6s 12ah = 300watt
Doing the math you get 7 miles on that setup so you burn 41 watt per mile.
300watt / 41watt per mile = 7.3 miles

2nd setup 6s 24ah = 600watt
600watt /41 watt per mile = 14.6 miles

3rd setup 10s 8ah = 336 watt
336 watt /41 watt per mile = 8.1 miles

Looking at the numbers your multistar setup will give you the most range while the 10s setup will get you there faster but with less range.
Distance is not voltage but the total wattage which is volt x ah.

 

[RogerD]

Nice one, well explained - thanks

Getting my mate's 10S board tomorrow and am going to check how many Mah I can put back in, then change the LVC settings.

P.s. If I remember, I'll put my watt meter on and see how close your numbers are (based on my numbers!) - will be fun if nothing else

 

[bobfandango]

Distance is not voltage but the total wattage which is volt x ah.

Not wrong, but higher voltage *can* lead to more distance depending on the setup... A higher voltage buys you lower current which in turn reduces electrical losses... Reduced electrical losses mean extra power to put into making distance (overcoming rolling resistance, air resistance etc). Electrical losses, however, are only part of the story. If your overall load, mechanical and electrical both, is dominated by mechanical losses (e.g. friction from whatever source... tires, drive train, air resistance), then decreased electrical losses won't buy you much perhaps in terms of distance. It probably does, however, buy you reliability and longevity since the electrical components should run cooler.

 

[liveforphysics]

Distance is not voltage but the total wattage which is volt x ah.

Not wrong, but higher voltage *can* lead to more distance depending on the setup... A higher voltage buys you lower current which in turn reduces electrical losses... Reduced electrical losses mean extra power to put into making distance (overcoming rolling resistance, air resistance etc). Electrical losses, however, are only part of the story. If your overall load, mechanical and electrical both, is dominated by mechanical losses (e.g. friction from whatever source... tires, drive train, air resistance), then decreased electrical losses won't buy you much perhaps in terms of distance. It probably does, however, buy you reliability and longevity since the electrical components should run cooler.

Common misconception.

 

[bobfandango]

Distance is not voltage but the total wattage which is volt x ah.

Not wrong, but higher voltage *can* lead to more distance depending on the setup... A higher voltage buys you lower current which in turn reduces electrical losses... Reduced electrical losses mean extra power to put into making distance (overcoming rolling resistance, air resistance etc). Electrical losses, however, are only part of the story. If your overall load, mechanical and electrical both, is dominated by mechanical losses (e.g. friction from whatever source... tires, drive train, air resistance), then decreased electrical losses won't buy you much perhaps in terms of distance. It probably does, however, buy you reliability and longevity since the electrical components should run cooler.

Common misconception.

Which part?

 

[RogerD]

I charged his 8000mah 10s pack today, from 3.7 volts per cell to full, with a brick style charger. I put a hobbyking watt meter in line and I put 7000mah in, which seems odd. Far more than I expected. It's looking like he'll only get around 6 miles on his 10s 8000mah setup. Was better at 6s 12000mah with around 7 miles.

Top speeds are similar. Perhaps it's the gearing. Or the vesc? ( I geared him down from 16 :44 to 14:44 (90mm wheels)

I did advise against going 10s but he was keen...( none of my decent chargers are 10s capable)

 

[Ohbse]

Not wrong, but higher voltage *can* lead to more distance depending on the setup... A higher voltage buys you lower current which in turn reduces electrical losses... Reduced electrical losses mean extra power to put into making distance (overcoming rolling resistance, air resistance etc). Electrical losses, however, are only part of the story. If your overall load, mechanical and electrical both, is dominated by mechanical losses (e.g. friction from whatever source... tires, drive train, air resistance), then decreased electrical losses won't buy you much perhaps in terms of distance. It probably does, however, buy you reliability and longevity since the electrical components should run cooler.

Common misconception.

Which part?

Your first sentence - motor doesn't deal with pack voltage, it deals with phase current. Higher pack voltage will deliver exactly the same phase current to deliver same torque on same motor. Losses are dominated by copper resistance in motor, this doesn't care about pack voltage. Generally you'll lose on higher pack V with same motor as you'll go faster. Speed is inefficient (but fun)

 

[bobfandango]

Your first sentence - motor doesn't deal with pack voltage, it deals with phase current. Higher pack voltage will deliver exactly the same phase current to deliver same torque on same motor. Losses are dominated by copper resistance in motor, this doesn't care about pack voltage. Generally you'll lose on higher pack V with same motor as you'll go faster. Speed is inefficient (but fun)

Maybe there is something I don't understand... I always thought of it as power in must equal power out. So, Tw=IV. If your mechanical load is constant, then Tw is constant and to keep IV equal to Tw in turn requires halving current when you double voltage. So, current through the battery is halved which means less waste heat which in turn means more power to make distance. But you are saying, I take it, that copper losses in the motor dwarf ohmic losses in the battery and so higher voltage really doesn't buy you much at all....

 

[Ohbse]

Your first sentence - motor doesn't deal with pack voltage, it deals with phase current. Higher pack voltage will deliver exactly the same phase current to deliver same torque on same motor. Losses are dominated by copper resistance in motor, this doesn't care about pack voltage. Generally you'll lose on higher pack V with same motor as you'll go faster. Speed is inefficient (but fun)

Maybe there is something I don't understand... I always thought of it as power in must equal power out. So, Tw=IV. If your mechanical load is constant, then Tw is constant and to keep IV equal to Tw in turn requires halving current when you double voltage. So, current through the battery is halved which means less waste heat which in turn means more power to make distance. But you are saying, I take it, that copper losses in the motor dwarf ohmic losses in the battery and so higher voltage really doesn't buy you much at all....

You got it If your connections between battery and controller are suitably sized this area of loss should be minimal. There are other advantages of higher voltage, the ability to spin the motor faster for instance can be useful, but once again that's only with the same motor. If you have control of the selection of motor KV then again pack voltage becomes relatively unimportant. There are certainly 'sweet spots' in terms of power output from the controller, current MOSFET technology this seems to be around 80v with another sweet spot around 120v but this is only really due to currently available switching hardware, as technology evolves this will not necessarily be the case. There is no inherent requirement for performance to equal higher voltage is what I'm getting at.

 

[liveforphysics]

Your first sentence - motor doesn't deal with pack voltage, it deals with phase current. Higher pack voltage will deliver exactly the same phase current to deliver same torque on same motor. Losses are dominated by copper resistance in motor, this doesn't care about pack voltage. Generally you'll lose on higher pack V with same motor as you'll go faster. Speed is inefficient (but fun)

Maybe there is something I don't understand... I always thought of it as power in must equal power out. So, Tw=IV. If your mechanical load is constant, then Tw is constant and to keep IV equal to Tw in turn requires halving current when you double voltage. So, current through the battery is halved which means less waste heat which in turn means more power to make distance. But you are saying, I take it, that copper losses in the motor dwarf ohmic losses in the battery and so higher voltage really doesn't buy you much at all....

You got it If your connections between battery and controller are suitably sized this area of loss should be minimal. There are other advantages of higher voltage, the ability to spin the motor faster for instance can be useful, but once again that's only with the same motor. If you have control of the selection of motor KV then again pack voltage becomes relatively unimportant. There are certainly 'sweet spots' in terms of power output from the controller, current MOSFET technology this seems to be around 80v with another sweet spot around 120v but this is only really due to currently available switching hardware, as technology evolves this will not necessarily be the case. There is no inherent requirement for performance to equal higher voltage is what I'm getting at.

It is so refreshing to be on ES. Thank you for that excellent post Ohbse.

 

[chuttney1]

I'm finding it difficult to follow 10S 8000 mAh battery gets less mileage vs 6S 12000 mAh battery; besides physics saying Watts = Voltage x Current. I'm thinking from this small test a third unaccounted variable influencing the results is how fast one guns the throttle with the usual saying "your mileage will vary."

 

[RogerD]

It is odd. When I tested it - I rode with the board's owner. We went the same route, similar speeds. He's definitely heavier on the throttle than me.

 

[Simonvtr]

I'm finding it difficult to follow 10S 8000 mAh battery gets less mileage vs 6S 12000 mAh battery; besides physics saying Watts = Voltage x Current. I'm thinking from this small test a third unaccounted variable influencing the results is how fast one guns the throttle with the usual saying "your mileage will vary."

10s 8ah will give you just about the same distance as 6s 12ah. The 10s setup will give you a bit less distance depending on the road condition and how the equipment is setup. I am just throwing theory here because I am no engineer. If both boards are setup the same way and both riders are of the same size the 10s will give a bit less distance compare to the 6s setup. 10s being a faster setup will create more wind resistance when moving thru the air which in turn will cause more friction on the drive system turning that extra energy into heat.

 

[Vanarian]

If both boards are setup the same way and both riders are of the same size the 10s will give a bit less distance compare to the 6s setup. 10s being a faster setup will create more wind resistance when moving thru the air which in turn will cause more friction on the drive system turning that extra energy into heat.

Kinda wrong, well that's same issue for both 6S and 10S since what you're saying is determined by speed of the rider.

Watt Hours are Watt hours, no matter 6S or 10S. What changes is how you use these watt hours.

Motors are driven by phase amps and voltage is oddly the mean to carry these phase amps eases with more voltage, apart from spinning speed you draw amps more easily. That's why high voltage setups pull like train from start.

Still that's odd, you should not get less distance, add watt hours get more range!
What should also help efficiency and range with higher voltage is : if triggering power becomes easier, you spend less time pressing the throttle to get it. So for the same result you should abuse less of your throttle. That's what should make for better range IMHO.

Isn't there a problem with the setup or a random loss somewhere (wheels? Belt? Broken bearings? Damaged wiring on batteries?)?

 

[RogerD]

We did a full to flat (both 3.3v-ish per cell) test today - lovely sunny day

6s 24Ah (cheapo multistar packs) - I got 12 miles till flat (3.45 per cell at rest)

10s 8Ah (Zippy compact packs) - he got 8 miles till flat/lvc kick-in (around 3.3v per cell when measured at rest - prob climbed to around 3.45 at rest after 10 mins, same as mine)

Much better than 1st test. 8 miles is OK in my books for his 10S setup.

Loving my 12 mile range (and I have 12Ah of spare 6s packs I can carry with me for a further 6 odd miles range)

I'm going to stick with 6s. Also found today his 10s setup, even with a reduced KV motor and smaller pulley was tripping the VESC heat protection. Took the cover off his ESC case and the problem immediately went away.

We are both geared for the same top speed so wind resistance doesn't come into it.

Reassuring to see 10s outperforming 6s in range terms as I'd hoped it would.

 

[saul]

so about 50 whr/mile on 6s
and 40 whr/mile on 10s.

That is pretty significant improvement.

 

[whitepony]

too many variables around this test, dont think you can really tell anything about the efficiency of these setups. would be cool to have a real bench with 2 different batteries made of the same cells, 1 vesc, 1 motor and a proper way to measure the mechanical work.

in these real life tests, alone wheel choices and belt tightness, possibly bearings (or well, generally rolling resistance) will affect this probably significantly. even more dominant (depending on speed) is also the riders surface and cv value. :lol:

I remember someone was doing an interesting experiment to measure efficiency of motors by using one motor to drive another motor, both powered by vescs with one driving and the other recuperating. the losses of incoming vs. outcoming power are the losses of the whole chain, so vesc 1, motor1, motor2, vesc2. ah, found it: http://vedder.se/forums/viewtopic.php?f=6&t=82

this setup with constant or well defined varying motor RPMs with changing battery voltages can at least provide a quite interesting insight in how different battery voltages compare relative to each other.

 

[RogerD]

Well it's as close as you can get without being scientific.

We both weigh the same.
Same boards,
Same bearings, wheels, trucks.
Same route, and speeds. Same surface.
Almost the same motors - 6364 on 6s vs 6374 on 10s
Zippy compacts on one, Multstar on the other (so higsly difference C ratings for sure (20c vs 10c) but you never even get close to the C rating in use at the capacities we are running)