Topic: rc car battery to power a mini-me?

[phatdats]

anyone using a 7.2v rc car battery to power thier mini me?  i just got a 4000mah pack.  anyone know how long it will last (without phantom)?  does the power draw change as the voltage increases (ie does the minime draw the same at 6v that it does at 7.2 or 9 or 12v?)

thanks!
steve

[caymanreview]

the mme's power consumption is 5.5W

I = P/V
5.5W/9.6V = 0.572A = 572mA

( Battery Capacity [mAH] / Current Draw [mA] ) * ( .80 )
3000mA / 572mA = 5.24h * 0.8 = 4.2 hours

this was the formula i used for the 9.6v pack and the 3000mah pack, just substitute what your using and you are good!

im not sure how much not using phantom makes for runtime, but this will give you a minimum runtime to go by

[caymanreview]

5.5W/7.2V = ~763mA

4000mah/ 764 = 5.25h * 0.8 = ~4.2 hours

 

[phatdats]

thanks dustin!! and thanks for helping me this afternoon!!!!

i'd give you some tix if i had any 

[leegeddy]

the mme's power consumption is 5.5W

I = P/V
5.5W/9.6V = 0.572A = 572mA

( Battery Capacity [mAH] / Current Draw [mA] ) * ( .80 )
3000mA / 572mA = 5.24h * 0.8 = 4.2 hours

this was the formula i used for the 9.6v pack and the 3000mah pack, just substitute what your using and you are good!

im not sure how much not using phantom makes for runtime, but this will give you a minimum runtime to go by

dustin;

the quoted power consumption should be divided by the Apogee's nominal operating voltage of the MME (whatever that may be).  we know that there's a range of voltage that the MME will accept, but you should calculate based on their specified operating voltage.  that's how i would approach it.

marc

[caymanreview]

well well well mark, always out to best me

its just a rough formula to follow, nothing scientific

my pleasure to help you out this afternoon steve

[Sean Gallemore]

the mme's power consumption is 5.5W

I = P/V
5.5W/9.6V = 0.572A = 572mA

( Battery Capacity [mAH] / Current Draw [mA] ) * ( .80 )
3000mA / 572mA = 5.24h * 0.8 = 4.2 hours

this was the formula i used for the 9.6v pack and the 3000mah pack, just substitute what your using and you are good!

im not sure how much not using phantom makes for runtime, but this will give you a minimum runtime to go by

dustin;

the quoted power consumption should be divided by the Apogee's nominal operating voltage of the MME (whatever that may be).  we know that there's a range of voltage that the MME will accept, but you should calculate based on their specified operating voltage.  that's how i would approach it.

marc

I tried doing that once and then people told me to use Dustin's method. I'm confuzzled

[leegeddy]

well well well mark, always out to best me

its just a rough formula to follow, nothing scientific

my pleasure to help you out this afternoon steve

hell, my other "count with fingers and toes and hope my battrees last" method still works for me. 

getting back to the subject at hand... there should only be a small time difference since we're dealing with small voltage difference.

marc

[Sean Gallemore]

i heart small deflections

[AT853rxwh]

I think what he is trying to say is this

If it is supposed to run on 12v you would get this:

5.5w/12v = .458a or 458ma



Personally I would always use the lower voltage current rating as it is always going to be higher, so worse case by using it your battery life will be longer than you expected, not shorter.

[caymanreview]

steve

check your pm's/email...

got a urgent ? for you!

Dustin