2S/3S & Motors?
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- battery motor performance
quite a bit differently.
With the jump in voltage you get proportional increase in power for the same current load, a nearly proprtional increase in RPM (if unloaded, it is proprtional), and typically a need to drop the prop size to maintain the same current load.
For a given prop you'll see in increase in pull, current/power draw and speed, and if your prop was properly sized for 2s, an dramatic increase in smoke generation capabiliy
Effectively, a motor's greatest limit is current -- if you can keep that down to a safe level, it wont' overheat and burn out. by increasing the voltage you can boost the power output, and by spinning a smaller prop faster, you can maintain the same cooling, but have a dramatic boost in thrust and speed. This can generally be done until the bearings become the limting factor.
For a "D x P" prop, power drawn by the prop is proportional to D x D x P. If you increase the voltage, you need to change your prop's diameter or pitch to maintain that same ratio to keep the current roughly the same . . . For example:
2S w/ 9x6 prop -> 3S w/ 8x???
- get the ratio of the voltage change: 3S/2S = 1.5
- get the prop's power loading: 9x6 -> 9 x 9 x 6 = 486
- reduce the prop's power factor by the voltage ratio: 486 / 1.5 = 324
- find the desired prop size (some might not be available, but make a guess): 324 = 8 x 8 x ??? -> ??? = 5
the proper prop would be an 8x5" prop on 3S . . . or a 9x4", or 7x6.6"
This all assumes the two props are of the same make and model (don't compare nylon with glass filled with CF props -- won't work).
General Rule of thumb is to drop an inch on each number, which for the sizes we deal with, works fairly well.
With the jump in voltage you get proportional increase in power for the same current load, a nearly proprtional increase in RPM (if unloaded, it is proprtional), and typically a need to drop the prop size to maintain the same current load.
For a given prop you'll see in increase in pull, current/power draw and speed, and if your prop was properly sized for 2s, an dramatic increase in smoke generation capabiliy
Effectively, a motor's greatest limit is current -- if you can keep that down to a safe level, it wont' overheat and burn out. by increasing the voltage you can boost the power output, and by spinning a smaller prop faster, you can maintain the same cooling, but have a dramatic boost in thrust and speed. This can generally be done until the bearings become the limting factor.
For a "D x P" prop, power drawn by the prop is proportional to D x D x P. If you increase the voltage, you need to change your prop's diameter or pitch to maintain that same ratio to keep the current roughly the same . . . For example:
2S w/ 9x6 prop -> 3S w/ 8x???
- get the ratio of the voltage change: 3S/2S = 1.5
- get the prop's power loading: 9x6 -> 9 x 9 x 6 = 486
- reduce the prop's power factor by the voltage ratio: 486 / 1.5 = 324
- find the desired prop size (some might not be available, but make a guess): 324 = 8 x 8 x ??? -> ??? = 5
the proper prop would be an 8x5" prop on 3S . . . or a 9x4", or 7x6.6"
This all assumes the two props are of the same make and model (don't compare nylon with glass filled with CF props -- won't work).
General Rule of thumb is to drop an inch on each number, which for the sizes we deal with, works fairly well.
. . . Ok . . . have you heard the proverb, "you don't realy know something until you can explain it to someone else" . . . .
I started to write out an explanation and found out I was pulling some slight of hand that I know works empircally, but the reasoning doesn't hold . . . pointing out to me I'd made a few assumptions too far, but lets hit the points I know work, and I'll ponder and others can call me crazy and point to the holes in my logic
The key word is "Proportional".
To calculate power from the properties of the power system is a pretty nasty mess, but if I don't care about the actual wattage demand, I can make some comparisons, knowing what will happen if I change one thing or another, becasue of the proportionality.
- Power is proportional to the pitch of the prop. keep most everything the same, and make the prop a little steeper, and a 4" pitch will demand half as much as an 8" pitch.
- Power is also proportional to the area the prop spins in (the "prop disk"), which is proprtional to the square of the diameter. If the prop diameter goes from 7 to 8", keeping everything else the same, the power will increase around 30% ((8²/7²) -1) for a 14% increase in length.
Pitch and length, because they're independant dimisions are independant variables, so we can say, for a DxP prop (like a 9x6"):
Power ∝ D² x P
So what does power equal? Dunno. Too many other factors contribute and most of those factors I don't care about -- All I care about is how two different props compare in an apples-to-apples comparison.
So if I have two props of the same manufacture on the same power system (motor/ESC/battery), I can compare the numbers . . . so which draws more current? a 9x6" or an 8x8"?
9x6" -> 9² x 6 = 486
8x8" -> 8² x 8 = 512 <- We have a winner!
In this case, the shorter prop will draw more power to run. I can also use this number to predict which shorter/faster prop draws the same power as a longer/slower prop.
****** (hold on, it's about to get bumpy)
Now where I'm getting tripped up is bending the comparison by changing the voltage . . . a simple change . . . but one that has subtlties that I'm pondering through.
If voltage increases, power goes up (P = I x V -- the equality is easy), as does RPM (RPM = kV x V), but from expeince, Keeping the prop the same, but boosting the voltage causes a dramatic increase in current, somewhat proprtional to the increase in voltage. So experience says if you keep current the same and increase voltage by, say, 50% percent, the D² x P of the prop needs to go down by about that amount . . . . but running out the math (assuming all the other factors in the proprtion and independant of voltage) means the prop size would need to go up to hold the proportion . . . not down, which we know is the right answer . . .
The hole as I see it:
See above where I say "most of those factors I dont' care about" . . . well, maybe I should. the relationship with RPM to voltage plays heavily, and is likley the missing sauce. I haven't played enough with it to know in detail how RPM relates to power in more than a general side.
VVVVV Shear Conjecture VVVVV (still with me?)
My guess I'd say P ∝ RPM². I need to divide out the voltage, not multiply, and since RPM = kV x V, if RPM² is on the right, it will fix this, but I have no mechanism I feel solidly supports this . . . soemthing vauge in the past about wind resistance being proprtional to the square of velocity seems to ring a bit, but that class was a long time ago . . .
If that were so, then:
Power = I x V ∝ RPM² x D ² x P = (kv² x V² x D² x P)
. . . KV can drop into the hidden constants in the proprtion . . . and
I/V ∝D² x P
So if I stays the same and V goes up, D²xP needs to go down by that amount.
</Conjecture>
As I said before, this matches what I've seen IRL, but the numbers are a little squishy and the explanation involves waiving hands at a specific speed . . . I suppose I may be failing in that proverb . . .
(Sorry for the math dump. Parametric scaling is a particularly nasty subject, but handy when it works)
I started to write out an explanation and found out I was pulling some slight of hand that I know works empircally, but the reasoning doesn't hold . . . pointing out to me I'd made a few assumptions too far, but lets hit the points I know work, and I'll ponder and others can call me crazy and point to the holes in my logic
The key word is "Proportional".
To calculate power from the properties of the power system is a pretty nasty mess, but if I don't care about the actual wattage demand, I can make some comparisons, knowing what will happen if I change one thing or another, becasue of the proportionality.
- Power is proportional to the pitch of the prop. keep most everything the same, and make the prop a little steeper, and a 4" pitch will demand half as much as an 8" pitch.
- Power is also proportional to the area the prop spins in (the "prop disk"), which is proprtional to the square of the diameter. If the prop diameter goes from 7 to 8", keeping everything else the same, the power will increase around 30% ((8²/7²) -1) for a 14% increase in length.
Pitch and length, because they're independant dimisions are independant variables, so we can say, for a DxP prop (like a 9x6"):
Power ∝ D² x P
So what does power equal? Dunno. Too many other factors contribute and most of those factors I don't care about -- All I care about is how two different props compare in an apples-to-apples comparison.
So if I have two props of the same manufacture on the same power system (motor/ESC/battery), I can compare the numbers . . . so which draws more current? a 9x6" or an 8x8"?
9x6" -> 9² x 6 = 486
8x8" -> 8² x 8 = 512 <- We have a winner!
In this case, the shorter prop will draw more power to run. I can also use this number to predict which shorter/faster prop draws the same power as a longer/slower prop.
****** (hold on, it's about to get bumpy)
Now where I'm getting tripped up is bending the comparison by changing the voltage . . . a simple change . . . but one that has subtlties that I'm pondering through.
If voltage increases, power goes up (P = I x V -- the equality is easy), as does RPM (RPM = kV x V), but from expeince, Keeping the prop the same, but boosting the voltage causes a dramatic increase in current, somewhat proprtional to the increase in voltage. So experience says if you keep current the same and increase voltage by, say, 50% percent, the D² x P of the prop needs to go down by about that amount . . . . but running out the math (assuming all the other factors in the proprtion and independant of voltage) means the prop size would need to go up to hold the proportion . . . not down, which we know is the right answer . . .
The hole as I see it:
See above where I say "most of those factors I dont' care about" . . . well, maybe I should. the relationship with RPM to voltage plays heavily, and is likley the missing sauce. I haven't played enough with it to know in detail how RPM relates to power in more than a general side.
VVVVV Shear Conjecture VVVVV (still with me?)
My guess I'd say P ∝ RPM². I need to divide out the voltage, not multiply, and since RPM = kV x V, if RPM² is on the right, it will fix this, but I have no mechanism I feel solidly supports this . . . soemthing vauge in the past about wind resistance being proprtional to the square of velocity seems to ring a bit, but that class was a long time ago . . .
If that were so, then:
Power = I x V ∝ RPM² x D ² x P = (kv² x V² x D² x P)
. . . KV can drop into the hidden constants in the proprtion . . . and
I/V ∝D² x P
So if I stays the same and V goes up, D²xP needs to go down by that amount.
</Conjecture>
As I said before, this matches what I've seen IRL, but the numbers are a little squishy and the explanation involves waiving hands at a specific speed . . . I suppose I may be failing in that proverb . . .
(Sorry for the math dump. Parametric scaling is a particularly nasty subject, but handy when it works)
Gramaphobe
Member
I know this is an old thread but I think it has answered a question I have.
I cant find any info for a nrm propdrive 1200kv motor for 2s operation. The suggested prop is 8x6e on 3s and it runs great. I want to build a ft3d and use a 2s setup for more low end grunt for prop hanging and such. Crunching the numbers in this formula tells me I could try a 10x3.8 at 3s though...
Any thoughts on a good 2s prop? Ir should I even bother?
I cant find any info for a nrm propdrive 1200kv motor for 2s operation. The suggested prop is 8x6e on 3s and it runs great. I want to build a ft3d and use a 2s setup for more low end grunt for prop hanging and such. Crunching the numbers in this formula tells me I could try a 10x3.8 at 3s though...
Any thoughts on a good 2s prop? Ir should I even bother?
Gramaphobe
Member
Ok... 8x8x6 is 384 x 1.5 is 576, so 12x12x4 is 576 exactly....hmm... ill give that a try...
Hey Gramaphobe,
One thing to keep in mind, this balance isn't to keep the power output constant between swapping props & batteries, but to keep the *current* constant. In reality, the power consumed will drop, and with it, the thrust generated . . . pretty much the opposite of what you want.
When you stick to the same battery, matching this power factor prop to prop will keep *both* power and current relitively constant, since voltage is fixed.
Looking at your setup, an 8x6 prop is a fairly fast prop -- there are faster, but at 1200kv, it has a pitch speed (the speed at which the prop is most efficient) of 75.7mph. That's by no means slow.
Best way to get that low end thrust is to slow down the pitch speed on the prop.
If you were to switch over to a 9x4.7 (common slow-fly and multirotor prop size), you'll find it matches out pretty close in power, but the pitch speed is down to 59.3mph. Down further to a 10x3.8 (a harder to find prop size, but still standard), you're looking at 48mph.
. . . alterntiavely . . .
If you were to INCREASE the battery voltage, the pitch speed would increase *BUT* so would the power.
moving up to 4S and changing to a 8x4.5 (also a common size) you'd still keep the 75mph pitch speed, but you'd have FAR more power behind it. you'll still be weak in the low end, compared to the high, but becasue the whole power curve shifts up by 30%, you'll have that much more power on the low end. Trading pitch for length again, the matching 9" prop for 4S is a 9x3.5 (once again a rare but standard size)
One thing to keep in mind, this balance isn't to keep the power output constant between swapping props & batteries, but to keep the *current* constant. In reality, the power consumed will drop, and with it, the thrust generated . . . pretty much the opposite of what you want.
When you stick to the same battery, matching this power factor prop to prop will keep *both* power and current relitively constant, since voltage is fixed.
Looking at your setup, an 8x6 prop is a fairly fast prop -- there are faster, but at 1200kv, it has a pitch speed (the speed at which the prop is most efficient) of 75.7mph. That's by no means slow.
Best way to get that low end thrust is to slow down the pitch speed on the prop.
If you were to switch over to a 9x4.7 (common slow-fly and multirotor prop size), you'll find it matches out pretty close in power, but the pitch speed is down to 59.3mph. Down further to a 10x3.8 (a harder to find prop size, but still standard), you're looking at 48mph.
. . . alterntiavely . . .
If you were to INCREASE the battery voltage, the pitch speed would increase *BUT* so would the power.
moving up to 4S and changing to a 8x4.5 (also a common size) you'd still keep the 75mph pitch speed, but you'd have FAR more power behind it. you'll still be weak in the low end, compared to the high, but becasue the whole power curve shifts up by 30%, you'll have that much more power on the low end. Trading pitch for length again, the matching 9" prop for 4S is a 9x3.5 (once again a rare but standard size)
Gramaphobe
Member
I noticed in the build thread for the ft3d it suggest the exact motor I have with a 8x4 prop. That should be really low amp draw on an already efficent motor so I should be able to get away with a low mah pack, and being an 8 inch prop it should spin up quick.
I can always try the 9x4 and 10x4 on 3s.
Now I just need to finish building...
I can always try the 9x4 and 10x4 on 3s.
Now I just need to finish building...
Jacob Shore
New member
I have a question. I'm building the ft blunt nose versa in pusher configuration for fpv and have had some problems in the past because a lack of power. I really want to get it right this time. So can someone give me some advice on a beef equivalent motor but from hobby king? Any help will be great.
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