


wot motor should i go for on a CM pro discus? i was thinking 500650watt motor with about 1000kv? thanks austin






Hello,
Quick question for the experts. I am trying to understand power systems and I have reached a small roadblock: Q: Looking at this theoretical setup: Battery: 3s 1300 mah, 30C Motor/components at full load/speed pulls: 8A How do you determine flight time? Preston 





Quote:
https://www.rcgroups.com/forums/showthread.php?t=748302 






There is the calculation, as stated in the other thread, which is:
Capacity in AH/amp draw X 60 = minutes of run time. 1.3 AH/8 amps = .1625 hours .1625 X 60 = 9.75 minutes at 8 amps. This assumes you use up all the battery capacity. It also assumes that the battery can actually deliver its total rated capacity before the LVC, low voltage cutoff, kicks in to keep you from running it too low. In your case you are not pushing that battery hard at all so you would probably get this full run time at full throttle. Normally you don't run at full throttle all the time. For mixed flying that is probably more like 15 minutes. Above is the more precise way to do this and it is correct. However I usually use this quick estimate method. If the battery can delvier 1.3 amps for one hour then it can deliver 13 amps for 1/10 of an hour ( 6 minutes ) You are drawing about 2/3 of that ( 8 amps) , so your run time will be longer, about 9 minutes. Just a quick estimate method I use. Not exact, just a quick approximation. Your actual run time will vary by battery quality, the LVC setting on the ESC and how much time you spend at what throttle setting. 





I would like to add a idea concerning the selection of propellers. I use a good fish scale and a watt meter, fit a prop of your choice, hold the model in place with the fish scale, power up and write down the thrust ( fish scale ) and watts, continue with other choice propellers, and use the one that gets the most thrust for least amps if you are looking for performance, and least amps and close to 1:1 thrust for sports. Thrust will give you good idea about performance, and amps will give good idea about flight times.
This way you will know exactly if your prop choice is the best. 





Good info for beginner. New thing learn from this article. Cheer...
Inrunners high rpm can be converted into torque by using a gearbox (divide the kv of the motor by the ratio of the gearbox, it's a 5070kv motor but using a gearbox with a ratio of 6.7:1 you get a kv of 5070/6.7 = 757) Pitch speed in mph = ( RPM X pitch ) / 1056 A higher pitch speed means lower thrust > longer take off > high landing speed. As a general rule 1" pitch relates to 1" of diameter, if you step up 1" in pitch you need to step down 1" in diameter to keep the same amp draw. With more normal kind of planes we usually use a prop with the proportion of 1:2 i.e. 10x5, 11x5.5, 12x6 and so on as it is most effective (from what I heard). 

Last edited by scng; Jul 25, 2010 at 10:18 AM.




Hey,
I'm having trouble making out the wattage of a brushed motor form its name... for example, given the motor "370 Motor & Gearbox (V2),5.33:1 .4 Module from Eflite" how can i tell what the wattage/voltage/current draw is? Sometimes the vendor's website only tells me the voltage... for example in this case http://www.efliterc.com/Products/De...ProdID=EFLM108. Cheers, Jawad. 





GREAT POST!! Here's something I can use!!!






Quote:
First the Voltage: The guide didn't really mention the voltage per cell. Does a 1 cell have 3.7v and and a 3 cell 11.1? If so why does a 4 have 14.4 instead of 14.8? Second the duration: Intuitively I understand that a 6000 mah last longer than a 2100 mah. Both support the 58amp draw because of the 12C vs 30C rating respectively. My question is what is the math that calculates the duration 1012 minutes vs 22.5 minutes? Is 10 minutes because you are drawing 10x the hourly rating? In that case shouldn't it be more like 6 minutes? Great hobby but I never expected the depth of math LOL 






Quote:
2. You aren't necessarily drawing the full current available from the C rating of the pack. The current drawn governs the duration of your pack, within the limits of the battery rating. In general the more power you extract, the greater current you will need and the faster your battery will be consumed. Think of your battery as the water butt with a tap, open the tap a little bit and the flow will be low and the tank take ages to empty, open the tap full and the flow is greater, but the tank empties faster. Clear now? sparks 






Conceptually I get it, it the math that is throwing me off.
I'm not sure why one cell is 3.7 and two is 7.2 (where did the other .2 go?) but 3 is 11.1, (3 x 3.7 which makes sense) but 4 now is double the 2 cell (now .4 is missing). I'll just accept it though! I think I understand your tap/tank analogy and the current drawn. Putting it in context of the example a 30C rating will allow you to drain the 2100 mah at a rate supporting the required 58amp draw (as will the 6000 mah at 10C) but the 2100 mah tank is just plain smaller than the 6000 mah tank? 





I think that the declared voltage levels are rounded. 3.6  3.7 is about the lowest useful voltage level. A fully charged cell is supposed to maintain around 4.2V , on no load, but will rapidly drop to 4V and lower when in use.
The analogy for the tank example is to think about the elevation of the tank and the level inside. A full tank will squirt water out of the open tap at high pressure, this will gradually reduce to a dribble when the tank is nearly empty. correct, the 6000mAh tank is bigger. sparks 





Quote:
Quote:
Take a look at the specs in this example.... http://www.hobbyking.com/hobbyking/s..._20C_Lipo_Pack Spec. Minimum Capacity: 1000mAh Configuration: 2S1P / 7.4v / 2Cell Constant Discharge: 20C Peak Discharge (10sec): 30C Pack Weight: 62g Pack Size: 70 x 35 x 13mm Charge Plug: JSTXH Seems to me.... most lipo sellers/makers say 3.7V per cell....nominal voltage..... 1s = 3.7 2s = 7.4 3s = 11.1 4s = 14.8 ......etc..... 



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