Battery C rating question - Masters Pattern
#1
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Battery C rating question - Masters Pattern
I'm looking for facts about this problem I'm trying to solve. I'm flying the Masters Pattern with a 50 size plane on a 5S, 25C, 5000ma battery. The flight takes 2500ma (2.5A) +/- 200ma depending on wind. I'd like a bit more 'end of flight' power. If I increased the C rating to a 40C or 50C would this give me the desired more power at the end of the flight? The extra weight and size will be no problem.
Thanks for comments
Scott
(I have placed this in the Battery section also - but wanted to check here too)
Thanks for comments
Scott
(I have placed this in the Battery section also - but wanted to check here too)
#3
You left one letter off the battery size, an "h". It is very important. The battery is a 5000 mah battery and that defines the maximum amount of power available, not the current. You need to increase it to have more reserve power at the end of our flight.
#4
My Feedback: (3)
No, increasing C rating will not increase capacity (or minimize power consumption).
C rating is related to the output resistance (or internal resistance IR) of the battery pack.
A higher C rating correlates to the ability of the battery to deliver higher output current without significant internal temperature rise
(Power = Currrent ^2 * Resistance....Current is the output current and Resistance is the battery pack IR; higher internal power dissipation leads to higher internal temperatures, which will shorten the battery life)
Higher C rated packs will be heavier than lower C rated packs simply due to the differences in construction/materials needed to lower internal resistance.
You could try a larger battery (5400mAh packs) or try a ThrottleTech in the airplane.
C rating is related to the output resistance (or internal resistance IR) of the battery pack.
A higher C rating correlates to the ability of the battery to deliver higher output current without significant internal temperature rise
(Power = Currrent ^2 * Resistance....Current is the output current and Resistance is the battery pack IR; higher internal power dissipation leads to higher internal temperatures, which will shorten the battery life)
Higher C rated packs will be heavier than lower C rated packs simply due to the differences in construction/materials needed to lower internal resistance.
You could try a larger battery (5400mAh packs) or try a ThrottleTech in the airplane.
#5
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If the internal resistance is lower, with a higher C rated battery, won't this equate to having more voltage throughout the flight? There would be less battery IR drop giving me more power to the motor - is that correct? Put another way, as the battery IR drop approaches zero won't I get more power at the motor?
#6
My Feedback: (12)
if you are going to buy a new battery, I would increase the mah. F3A has these Power Unlimited-X 5400 mAh 5S 18.5 Volts 25C Lipo battery JST/XH (World's Lightest packs)! FREE SHIPPING ON 2 or more packs!!!! and will give you the most bang for your buck. I would also agree with the throttle tec suggestion. In this situation, there's the science and then there's the reality.
Ken
Ken
#7
If the internal resistance is lower, with a higher C rated battery, won't this equate to having more voltage throughout the flight? There would be less battery IR drop giving me more power to the motor - is that correct? Put another way, as the battery IR drop approaches zero won't I get more power at the motor?
Last edited by big_G; 07-01-2016 at 07:40 AM.
#8
Senior Member
Ohm's Law
Amps = Volts / Ohms
EXAMPLE:
Fully charged, 5000mAh, 5s LiPo:
Amps = 21Volts (5s) / 1.1 Ohms (40C battery): 19.09 Amps
Amps = 21Volts (5s) / 1.3 Ohms (25C battery): 16.15 Amps
Power (Watts) = Volts x Amps
21 Volts x 19.09 Amps = 400 Watts (40C battery)
21 Volts x 16.15 Amps = 339 Watts (25C battery)
During a flight, the battery consumed 2500mAh and the 5s voltage is now 20 Volts. If you increase the mAh to 5400, the battery WILL have a higher voltage.
Amps = 20 Volts / 1.1 Ohms: 18.18 Amps (40C battery)
Amps = 20Volts / 1.3 Ohms: 15.4 Amps (25C battery)
20 Volts x 18.18 Amps = 363 Watts available using 40C battery
20 Volts x 15.4 Amps = 308 Watts available using 25C battery
Note the voltage is resting state and not under load
If the "C" rating is increased (lower resistance), the amount of energy available at the end of flight is increased.
If the mAh rating is increased (higher capacity), the amount of energy available at the end of flight is increased.
Go get the 5400/40C !!!!
Clear Skies and Happy Pattern Flying
-PD
EXAMPLE:
Fully charged, 5000mAh, 5s LiPo:
Amps = 21Volts (5s) / 1.1 Ohms (40C battery): 19.09 Amps
Amps = 21Volts (5s) / 1.3 Ohms (25C battery): 16.15 Amps
Power (Watts) = Volts x Amps
21 Volts x 19.09 Amps = 400 Watts (40C battery)
21 Volts x 16.15 Amps = 339 Watts (25C battery)
During a flight, the battery consumed 2500mAh and the 5s voltage is now 20 Volts. If you increase the mAh to 5400, the battery WILL have a higher voltage.
Amps = 20 Volts / 1.1 Ohms: 18.18 Amps (40C battery)
Amps = 20Volts / 1.3 Ohms: 15.4 Amps (25C battery)
20 Volts x 18.18 Amps = 363 Watts available using 40C battery
20 Volts x 15.4 Amps = 308 Watts available using 25C battery
Note the voltage is resting state and not under load
If the "C" rating is increased (lower resistance), the amount of energy available at the end of flight is increased.
If the mAh rating is increased (higher capacity), the amount of energy available at the end of flight is increased.
Go get the 5400/40C !!!!
Clear Skies and Happy Pattern Flying
-PD
#9
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Thanks for the comments. I would agree with PD on this also..
If the "C" rating is increased (lower resistance), the amount of energy available at the end of flight is increased.
If the mAh rating is increased (higher capacity), the amount of energy available at the end of flight is increased.
If the "C" rating is increased (lower resistance), the amount of energy available at the end of flight is increased.
If the mAh rating is increased (higher capacity), the amount of energy available at the end of flight is increased.
#10
Being as you are currently only using approx. 50% of your battery capacity, get the Throttle Tech. You will still see a drop off in voltage even with higher capacity packs without a compensating device like the Throttle tech. While you don't mind the extra weight of a 40-50C pack, seldom does adding weight improve flying characteristics. JMHO.
Gary
Gary
#11
My Feedback: (90)
Without a compensating device, you may also consider select a more aggressive throttle curve at some time into the flight.
From this discharge curve for 5000mah (http://www.rcgroups.com/forums/attac...mentid=6727159), it seems such a time is at 0.5ah. Hard to translate that point into time into the flight without telemetry. But my guess would be after turnaround and right before entering the box.
From this discharge curve for 5000mah (http://www.rcgroups.com/forums/attac...mentid=6727159), it seems such a time is at 0.5ah. Hard to translate that point into time into the flight without telemetry. But my guess would be after turnaround and right before entering the box.
#12
My Feedback: (3)
Amps = Volts / Ohms
EXAMPLE:
Fully charged, 5000mAh, 5s LiPo:
Amps = 21Volts (5s) / 1.1 Ohms (40C battery): 19.09 Amps
Amps = 21Volts (5s) / 1.3 Ohms (25C battery): 16.15 Amps
Power (Watts) = Volts x Amps
21 Volts x 19.09 Amps = 400 Watts (40C battery)
21 Volts x 16.15 Amps = 339 Watts (25C battery)
During a flight, the battery consumed 2500mAh and the 5s voltage is now 20 Volts. If you increase the mAh to 5400, the battery WILL have a higher voltage.
Amps = 20 Volts / 1.1 Ohms: 18.18 Amps (40C battery)
Amps = 20Volts / 1.3 Ohms: 15.4 Amps (25C battery)
20 Volts x 18.18 Amps = 363 Watts available using 40C battery
20 Volts x 15.4 Amps = 308 Watts available using 25C battery
Note the voltage is resting state and not under load
If the "C" rating is increased (lower resistance), the amount of energy available at the end of flight is increased.
If the mAh rating is increased (higher capacity), the amount of energy available at the end of flight is increased.
Go get the 5400/40C !!!!
Clear Skies and Happy Pattern Flying
-PD
EXAMPLE:
Fully charged, 5000mAh, 5s LiPo:
Amps = 21Volts (5s) / 1.1 Ohms (40C battery): 19.09 Amps
Amps = 21Volts (5s) / 1.3 Ohms (25C battery): 16.15 Amps
Power (Watts) = Volts x Amps
21 Volts x 19.09 Amps = 400 Watts (40C battery)
21 Volts x 16.15 Amps = 339 Watts (25C battery)
During a flight, the battery consumed 2500mAh and the 5s voltage is now 20 Volts. If you increase the mAh to 5400, the battery WILL have a higher voltage.
Amps = 20 Volts / 1.1 Ohms: 18.18 Amps (40C battery)
Amps = 20Volts / 1.3 Ohms: 15.4 Amps (25C battery)
20 Volts x 18.18 Amps = 363 Watts available using 40C battery
20 Volts x 15.4 Amps = 308 Watts available using 25C battery
Note the voltage is resting state and not under load
If the "C" rating is increased (lower resistance), the amount of energy available at the end of flight is increased.
If the mAh rating is increased (higher capacity), the amount of energy available at the end of flight is increased.
Go get the 5400/40C !!!!
Clear Skies and Happy Pattern Flying
-PD
#13
My Feedback: (10)
. C rating is the measure of the batteries ability to hold rated voltage at a current. You are not drawing anywhere near the current that your existing batteries are capable of so you wont see a significant voltage increase. Higher c rated packs are generally more fragile and wont take as much abuse either. Also c rating has nothing to do with the batteries ability to transfer heat away as it discharges. So if you are operating at the higher current levels of a high c pack you will probably melt it down and ruin it in a few flights unless you have some good cooling air flow. And i mean alot!!!
#14
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#16
If I read this tread correctly, This is what I got out of this.
Scott asked about how to get more power at the end of the flight. Assuming everything else stays the same, a Throttle Tech will do that, at the cost of power at the beginning of the flight.
Changing the C rating will give you a higher voltage through out the flight resulting in a higher RPM. (Power will therefore be higher as described by the equations provided by others above)
Changing the mah rating will help also by slowing the rate that the voltage drops off. Thus having a higher voltage at the end of the flight.
Summary:
Each of the above solutions will work. But each has their own trade off's
Increasing C rating and mah's will both result in higher weights that will consume some of the increases you are achieving.
The Throttle Tech will cost him power at the beginning of the flight.
Did I miss anything??
Scott asked about how to get more power at the end of the flight. Assuming everything else stays the same, a Throttle Tech will do that, at the cost of power at the beginning of the flight.
Changing the C rating will give you a higher voltage through out the flight resulting in a higher RPM. (Power will therefore be higher as described by the equations provided by others above)
Changing the mah rating will help also by slowing the rate that the voltage drops off. Thus having a higher voltage at the end of the flight.
Summary:
Each of the above solutions will work. But each has their own trade off's
Increasing C rating and mah's will both result in higher weights that will consume some of the increases you are achieving.
The Throttle Tech will cost him power at the beginning of the flight.
Did I miss anything??
#17
If I read this tread correctly, This is what I got out of this.
Scott asked about how to get more power at the end of the flight. Assuming everything else stays the same, a Throttle Tech will do that, at the cost of power at the beginning of the flight.
Changing the C rating will give you a higher voltage through out the flight resulting in a higher RPM. (Power will therefore be higher as described by the equations provided by others above)
Changing the mah rating will help also by slowing the rate that the voltage drops off. Thus having a higher voltage at the end of the flight.
Summary:
Each of the above solutions will work. But each has their own trade off's
Increasing C rating and mah's will both result in higher weights that will consume some of the increases you are achieving.
The Throttle Tech will cost him power at the beginning of the flight.
Did I miss anything??
Scott asked about how to get more power at the end of the flight. Assuming everything else stays the same, a Throttle Tech will do that, at the cost of power at the beginning of the flight.
Changing the C rating will give you a higher voltage through out the flight resulting in a higher RPM. (Power will therefore be higher as described by the equations provided by others above)
Changing the mah rating will help also by slowing the rate that the voltage drops off. Thus having a higher voltage at the end of the flight.
Summary:
Each of the above solutions will work. But each has their own trade off's
Increasing C rating and mah's will both result in higher weights that will consume some of the increases you are achieving.
The Throttle Tech will cost him power at the beginning of the flight.
Did I miss anything??
The Throttle Tech gives a more linear feel to the flight from start to finish. Most of our set-ups have an abundance of power at the beginning of the flight anyways...so giving up a little up front to gain at the back end seems like a win. In my case it probably helps conserve energy as I occasionally use too much power on the up-lines.
#19
Amps = Volts / Ohms
EXAMPLE:
Fully charged, 5000mAh, 5s LiPo:
Amps = 21Volts (5s) / 1.1 Ohms (40C battery): 19.09 Amps
Amps = 21Volts (5s) / 1.3 Ohms (25C battery): 16.15 Amps
Power (Watts) = Volts x Amps
21 Volts x 19.09 Amps = 400 Watts (40C battery)
21 Volts x 16.15 Amps = 339 Watts (25C battery)
During a flight, the battery consumed 2500mAh and the 5s voltage is now 20 Volts. If you increase the mAh to 5400, the battery WILL have a higher voltage.
Amps = 20 Volts / 1.1 Ohms: 18.18 Amps (40C battery)
Amps = 20Volts / 1.3 Ohms: 15.4 Amps (25C battery)
20 Volts x 18.18 Amps = 363 Watts available using 40C battery
20 Volts x 15.4 Amps = 308 Watts available using 25C battery
Note the voltage is resting state and not under load
If the "C" rating is increased (lower resistance), the amount of energy available at the end of flight is increased.
If the mAh rating is increased (higher capacity), the amount of energy available at the end of flight is increased.
Go get the 5400/40C !!!!
Clear Skies and Happy Pattern Flying
-PD
EXAMPLE:
Fully charged, 5000mAh, 5s LiPo:
Amps = 21Volts (5s) / 1.1 Ohms (40C battery): 19.09 Amps
Amps = 21Volts (5s) / 1.3 Ohms (25C battery): 16.15 Amps
Power (Watts) = Volts x Amps
21 Volts x 19.09 Amps = 400 Watts (40C battery)
21 Volts x 16.15 Amps = 339 Watts (25C battery)
During a flight, the battery consumed 2500mAh and the 5s voltage is now 20 Volts. If you increase the mAh to 5400, the battery WILL have a higher voltage.
Amps = 20 Volts / 1.1 Ohms: 18.18 Amps (40C battery)
Amps = 20Volts / 1.3 Ohms: 15.4 Amps (25C battery)
20 Volts x 18.18 Amps = 363 Watts available using 40C battery
20 Volts x 15.4 Amps = 308 Watts available using 25C battery
Note the voltage is resting state and not under load
If the "C" rating is increased (lower resistance), the amount of energy available at the end of flight is increased.
If the mAh rating is increased (higher capacity), the amount of energy available at the end of flight is increased.
Go get the 5400/40C !!!!
Clear Skies and Happy Pattern Flying
-PD
Jim O
#20
Senior Member
Not sure what you are asserting here. It would appear you believe the 25C pack has 200 milliOhms more internal resistance than the 40C. I'm pretty sure that is not the case as I consistently get values in the neighborhood of 2 milliOhms per cell on 25C and 35C packs which would yield 10 milliOhms total for a five cell pack. If there is a difference it is in the range of 0.1 milliOhms per cell or half a milliOhm for 5 cells. At 20 Amps the difference in the voltage drop due to the internal resistance would be 0.01 volts. (20 x .0005). 25C packs are good enough!
Jim O
Jim O
#21
My Feedback: (3)
Please look again at what Jim stated above.
You can also run a fairly simple experiment. Take a 25C and 50C 5000mAH battery, both charged to the same termination voltage (say 4.1 or 4.2 ... your preference)
Connect the battery to an electronic load Do this at a reasonable discharge current (say 5C - 25A). If you don't have an electronic load, then a power resistor should suffice (sized and cooled appropriately)
Monitor/record the capacity discharged when you reach your desired termination voltage (say 3.7VDC). You can use an inline wattmeter to do this as well.
Do this for both batteries and compare results.
If your application of Ohm's law is correct, then a 50C battery will always have a significant increase in (discharge) capacity compared to a 25C battery (20% in your example math in your post), independent of discharge rate/output current.
#22
To the op....buy a 40 or 50C pack and try it. All our guessing and numbers crunching can't hold a candle to real data. Worst case is your out 2 batteries. They will make great practice packs anyways! Let us know how you made out.
#24
I'm looking for facts about this problem I'm trying to solve. I'm flying the Masters Pattern with a 50 size plane on a 5S, 25C, 5000ma battery. The flight takes 2500ma (2.5A) +/- 200ma depending on wind. I'd like a bit more 'end of flight' power. If I increased the C rating to a 40C or 50C would this give me the desired more power at the end of the flight? The extra weight and size will be no problem.
Thanks for comments
Scott
(I have placed this in the Battery section also - but wanted to check here too)
Thanks for comments
Scott
(I have placed this in the Battery section also - but wanted to check here too)
1) Scott's objective is to have more end of flight power.
2) He has more than enough ENERGY in his battery pack.
3) The reason he wants more power at the end of flight is because battery voltage drops as it is discharged. Lower voltage means lower RPM.
4) The way to solve the problem is to take the battery variability out of the equation.
My experience with packs providing 60 to 70 Amps is that the C rating is insignificant and I believe there is no theory that says it will solve his problem. So my answer is no, going to 40C or 50C will not give him the desired more power.
Because he is not energy limited, the solution is straightforward. Get a motor with a higher Kv that is rated at the desired power or go to a 6 cell pack, and then add a Throttle-Tech to regulate the voltage. He can have more power and have it from the beginning to the end of flight and set it to his desire.
I use Throttle-Techs in all my pattern planes. Make sense?
I suggested to Jeti that they build the TT concept into their ESCs. Got no answer.
Jim O