3s vs 4s Comments please

[Ashnook]

I am about 50% IC and 50% Electric now and leaning more towards electric. I have converted some planes and kits to Electric with, I think, good success. I have decided to primarily use 4s batteries instead of what seems to be a general preference to 3s, here are my reasons, am I wrong or am I missing something?

The mail reason is one of current, that is for the same power you need less current with 4S than with 3S. This effects both the speed controller and the motor in a positive way. Almost all motors and speed controllers are rated up to 4s and above.

There is more battery power available for what seems like little extra weight. Also I have found in a couple of cases that the whole electric setup (battery, motor, leads, mount, esc) are generally lighter than the equivalent IC setup (motor, mount, tank, servo, receiver battery) and the extra 4s battery weight goes a long way to compensate for this, why add lead instead of extra battery power.

So, am I OK in this way of thinking or am I missing something (important?)

Cheers, Brian

[Dave Hopkin]

I think its a case of "horses for courses" really, depends on the size of models, the type of flight you want, not sure there is a generic recipe for what suits..... I have ended up with a mix of 3S and 4S to suit the type of models I have plus some planes advertised as 3 to 4S need a 4, the Seagulls E-Pioneer for instance, with a Eflite Power 32 in it flies (almost) like a dim witted slug in concrete on a 3S but shove a 4S in it and it transforms into perorming like the IC boomerang it was spawned from

[Andy Green]

I'm no expert Brian, but your thinking is sound, higher voltage means lower current to give you the same power, and current is the killer in electric setups. If weight isn't an issue then go 4S

It goes without saying to use a power meter to check the system, and change prop if necessary.

Andy

[Frank Skilbeck]

As long as you are selecting the motors for a 4s battery then the logic is sound, if you use a model/motor combo that's been designed around a 3s setup then you'll need a smaller prop to compensate and the max rpm will be approx 30% higher.

I have a mixture of 3s and 4s batteries and on one plane use 2 x 4s in series to make an 8s, just ordered some 6s batteries for a new model..........................

[Ashnook]

Hi Dave, yes a friend flying on Monday on 3s tried a 4s pack in his high wing sport plane and was amazed at the extra power it gave him. He also said it flew much better with the added weight up front - another 4s convertee

Yes Andy, a power meter was the first bit of 'kit' I bought and I do several prop combination tests before I decide what prop to go for.

Monday was very cold but clear sunny skies here and we both agreed that messing around with IC on a day like that would not be so much fun. Plug-and-fly seemed a better plan. I still like IC for the better feeling of real power and the noise of a 4-stroke!

Brian

[Ashnook]

Hi Frank, I built and OLD wot-4 kit and converted it to electric using a 5055 1KW motor! I use 2 4s-2200ma batteries in parallel to give me 4s-4400ma. I did not need to add any weight to the tail and it flies like a rocket (if you want) for 8-10 minutes.

Brian

[Dave Hopkin]

I dont think a increase of voltage will decrease current! Quite the opposite I find

Increase in voltage will increase the total power output and the only way power can increase is by drawing more current from the LiPo......

[Ashnook]

Hi Dave, according to the power calculation, watts = current times volts for the SAME power (that's the key) the current will reduce as the voltage increases - got to!

Brian

[Dickw]

If you keep the same motor and prop, increasing the voltage by changing from 3s to 4s will increase both the current and power.

If you increase the voltage but want to keep the same power you need to change the motor/prop combination.

Dick

[Peter Beeney]

I think you have to consider the principles on which an electric motor operates to understand the power consumption of said motor.

With regard to 3S and 4S packs, for cells of identical capacity, say 2.2Ah, a 3S will contain 2.2Ah x 11.1 volts = 24.42 watt hours, a 4S becomes 2.2Ah x 14.8 volts = 32.56 watt hours; thus a 25% increase.

Now it simply depends on the operator how he wishes to use the extra power.

PB

[Dickw]

I am probably being very pedantic here, but "Watt hours" is a unit of energy not power. The unit of power is Watts.

So it is the energy available that is being increased in changing tfrom 3s to 4s with the same sized cells.

Agreed it is the operators choice how the extra available enegy is used, but unless something else changes the rate at which the energy is used (i.e power) wil be increased if the voltage is increased.

Dick

[Peter Beeney]

Dick,

With the greatest respect, but I think that watt hours are indeed a measurement of power, not energy. It’s a convenient way of stating what power is available, or what has been used. For instance, your electricity bill is measured in kilowatt hours, it’s how much power you’ve used, but you might not have used it at the rate of one kilowatt per hour.

I’m now going flying for a while, back later this afternoon to continue, perhaps…

PB

[Willyuk]

I am new to electric and a question I have is why you use a smaller prop with a 4s as opposed to a 3s setup? A mystery to me.

thanks

[Dave Hopkin]

Posted by Ashnook on 31/12/2014 10:12:31:

Hi Dave, according to the power calculation, watts = current times volts for the SAME power (that's the key) the current will reduce as the voltage increases - got to!

Brian

You are assuming nothing else in the equation changes, however a brushless motor will try to spin at its KV rating, which is RPM * Volts, therefore the RPM will increase roughly 25%, the work the motor has to do (ie the power expended) will therefore increase to spin the prop - therefore the current will INCREASE to provide that power - which is why reducing the prop load reduces current flow and an increase in prop load will increase the current flow (note I use the term "prop load" as a combined diam & pitch as the two parameters operate in conjunction to create the interia the motor must overcome to spin it as X RPM)

[Dave Hopkin]

Posted by Willyuk on 31/12/2014 11:41:47:

I am new to electric and a question I have is why you use a smaller prop with a 4s as opposed to a 3s setup? A mystery to me.

thanks

Brushless Motors have a designed RPM speed (defined by the turns and poles) that RPM is usually stated as a "KV" number (confusingly) that translates to "Revolutions per Volt applied" - so the actual theoretical RPM of a given moter will change depending on the Voltage (3S roughly 11v, 4S roughly 14V) - so a motor that spins at 1000KV would want to do 11,0000 rpm on a 3S and 14,000 rpm on a 4S - the loading of the motor is a combination of the size and pitch of the prop - so to prevemt a increase in current draw you reduce the load on the motor by a smaller or finer prop

Or you could keep the 11S prop and extract more power from the set up at the cost of higher current draw - so duration of flight will reduce (or magic smoke if the ESC/Motor is not suitable for the increased voltage)

[Ashnook]

Yes Dave I agree with all you are saying. As well as taking the KV into account (I work on 800-1000kv for my type of aircraft) I use the watt-meter as my bible in terms of atcheiving the desired power-to-weight balance for my setup. I haven't just swapped a 3s for a 4s but rather decided I want, say, 500W for this model and used the watt-meter to get that with the right prop, the only thing I don't check, and perhaps I should, is the RPM of my setup. I do know that for my 500W the current flowing through the setup is lower than for the similar 3s setup and that makes me feel better and posibly be able to use a smaller ESC.

All this discussion is useful to me as I still am learning about Electric power as I am like many 40 year IC guys trying to get to grips with this new capability.

[Dave Hopkin]

Ashnook, I agree you will lower the current on a higher voltage by adjusting the prop load to get the wattage you want, but dont assume that increasing voltage will reduce current (if nothing else changes) it wont, on the same prop current will increase

Its a triangular balance, voltage/current./prop load - where we can influence only two factors, voltage and load

Increase either of them and current and wattage will increase...

Be wary of going too low on your ESC I think its better to have plenty of current headroom on an ESC rather than take it closer to its limits - I think they will probably last longer that way

[Dickw]

Posted by Peter Beeney on 31/12/2014 11:35:42:

Dick,

With the greatest respect, but I think that watt hours are indeed a measurement of power, not energy. It’s a convenient way of stating what power is available, or what has been used. For instance, your electricity bill is measured in kilowatt hours, it’s how much power you’ve used, but you might not have used it at the rate of one kilowatt per hour.

I’m now going flying for a while, back later this afternoon to continue, perhaps…

PB

No - but a common missunderstanding.

"Watt hours" is a unit of energy used or work done, My electricity meters measures the energy I have used over a period of time in Watt hours.

Power is the rate at whcih work is done or energy consumed, and is measured in Watts. The power output of my car engine is quoted in kW.

If an electric fire has a power of 2kW and is run it for an hour it has consumed 2kW hours of energy. If it has a power of 1kW and is run for 2 hours it also consumes 2kW hours of energy. Stand in front of them and you will feel the difference in power.

None of this is that important when playing with our toy planes, but if it is going to be quoted it is best to get it right to avoid confusion later.

Dick

 

Edited By Dickw on 31/12/2014 12:22:36

Edited By Dickw on 31/12/2014 12:23:11

[Ashnook]

Posted by Dave Hopkin on 31/12/2014 12:11:51:

Ashnook, I agree you will lower the current on a higher voltage by adjusting the prop load to get the wattage you want, but dont assume that increasing voltage will reduce current (if nothing else changes) it wont, on the same prop current will increase

Its a triangular balance, voltage/current./prop load - where we can influence only two factors, voltage and load

Increase either of them and current and wattage will increase...

Be wary of going too low on your ESC I think its better to have plenty of current headroom on an ESC rather than take it closer to its limits - I think they will probably last longer that way

All points well taken Dave. In most cases (except the experiment with a friend on Monday I am designing the setup from scratch using the desired power as the starting point. I have read where one should allow an excess of 15-20% on the esc.

[Cuban8]

If I recall my school physics lessons correctly, the letter k (lowercase) represents a mathematical constant, which in our case is the theoretical rpm/volt (unloaded) of a given motor hence kV. Nothing to do with kilovolts, thankfully.

[Phil Green]

Posted by Peter Beeney on 31/12/2014 11:35:42:

Dick,

With the greatest respect, but I think that watt hours are indeed a measurement of power, not energy.

PB

No, Peter. The watt-hour is an energy unit. Dick is absolutely right 
Watts are joules per second (ie time is a divisor) but watt-hours are watts times hours - so in simple terms the divisor and multiplier cancel, leaving watts-hours as a pure energy measurement (in joules for example - time has been 'cancelled' from the calculation)

Happy New Year everyone

Phil

 PS isnt it normal to choose a model, then a suitable power setup?

 

Edited By Phil Green on 31/12/2014 17:12:44

[Dave Hopkin]

Posted by Cuban8 on 31/12/2014 12:30:30:

If I recall my school physics lessons correctly, the letter k (lowercase) represents a mathematical constant, which in our case is the theoretical rpm/volt (unloaded) of a given motor hence kV. Nothing to do with kilovolts, thankfully.

I find it confusing having a radio and line transmission morphing into an IT backgroud, in both the use of the character "k" represents 1000 (or 1024 in IT) a following character represents the the unit, so kv would be Kilo Volts etc - with Brushless its rpm per volt.... why they decided that was the way to denote it and not RPV god knows!

[Bob Cotsford]

could it be because it's less characters to save wearing out keyboards?  Or K as used for a constant

As for 'power' - is it defined as a measure of anything? The ones I remember are pressure, force, work and the like.

 

Edited By Bob Cotsford on 31/12/2014 15:51:21

[Pete B]

Posted by Dave Hopkin on 31/12/2014 15:42:29:

with Brushless its rpm per volt.... why they decided that was the way to denote it and not RPV god knows!

Ah, we would confuse that with a Remote Piloted Vehicle then - unless we called them a UAS, UAV, RPA or drone -oh, we do....

Pete

[Dave Hopkin]

Bob - Power is the "rate of doing work" and is measured in Joules per second

Pete - we could call them TBT (Those Blasted Things)