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calculating lipo wattage

mokeife

Bronze Contributor
Member For 3 Years
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to calculate wattage i normally pluged in 16.8 volts since is a 4s lipo and then plugged my resistance in in this case its .3 and i come up with 940 watts. i have my potentiometer set at 50 percent so i take 50 percent of 940 and come up with 470 easy enough right? well ive come to notice that the potentiometer isnt linier using an inline volt meter i figured this out. so i asked my builder and he said this "The formula is squareroot of % duty cycle multiplied by battery voltage after sag". now when i try that my wattage is way to low. what i did was take the square root of 50 percent which is 7 and multiplied it by my voltage after sag which was 15.6 volts and came up with 110 watts. Now i know i am nowhere near that im closer to 500 watts so what am i doing wrong? 50 percent pot setting with a .3 build on a 4s lipo sure isnt 110 watts
 

Iliketurtles

Silver Contributor
Member For 4 Years
PWM guesstimates are voltage across load x %duty cycle. But you won't know what the %duty cycle or actual voltage across the load is without scoping the output, which could be a bit equipment explodey. Sounds like he was talking about sine waves, PWM isn't a sine wave.
 

mokeife

Bronze Contributor
Member For 3 Years
Member For 2 Years
Member For 1 Year
PWM guesstimates are voltage across load x %duty cycle. But you won't know what the %duty cycle or actual voltage across the load is without scoping the output, which could be a bit equipment explodey. Sounds like he was talking about sine waves, PWM isn't a sine wave.
so how exactly do i calculate my wattage since the percentage method doesnt work since the pot isnt linier thanks for your help
 

Iliketurtles

Silver Contributor
Member For 4 Years
You can't without scoping the output. Even if the pot was linear the 50 on the pot does not represent a 50% duty cycle its usually the 0 on the pot that represents around a 50% duty cycle. The voltage you actually get across the load will also depend on how the electronics were put together. There is no x on the pot = y watts out equation unless somebody scopes the device under load and records the data for it.
 

mokeife

Bronze Contributor
Member For 3 Years
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nevermind i figured it out he wanted me to take the square root of 50 which is 7.07 and multiply that by my sag which well put at 16 volts so you multiply 7.07x16 which equals 113 so my volts are 11.3 plug that in to an ohms law calculator as well as my resistance which is .3 and you come up with 425 watts bingo my wattage with sag
 

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