How do I work out how many Watts I'll get from my set-up...

[propogandhi]

Just a quick general question here. Its mostly likely been asked before and I'm most likely to have been told already. But my mind doesn't hold onto equations too well at the best of times. So, can somebody outline the basic maths behind this problem so I/we can have it for a point of reference in the future when I/we inevitably forget it again!

 

Lets say I have a 50", 5lbs model that I'm looking to power. I know roughly that I'd need a motor with a max rating (for 15sec) of 600watts to achieve 100+ w/lb, but how do I figure out how many watts I'd be getting constantly and not just at this Max peak?  I'd also like to know how I can figure out what amp rate speed controller I'd need to use, based on using a 3cell 3300mah lipo. I believe this is how I work out the basic amp draw :
 
600w ÷ 12.5v (3 cell lipo) = 48amps
 
But my issue is that this is only the stated max watts for the motor for 15seconds. Its the constant output I need to figure out, that way I can get a better estimate of my amp draw and thus what size speed controller to use. Of course prop size comes into it too, but how and at what point in the equation?
 

 Any clear help on this would be most welcome, maybe it can even become a sticky....unless there is already a sticky showing how to work out all these things (watts, amp draw, estimated duration etc) and I've just missed it.

 

prop.

 
 

[Colin Naylar]

Click on 'All Topics'

Scroll down to the 'New to Electric Flight' Folder

click!

 

You'll find all you need to know in there...

[Tim Mackey]

Colin.... you're da man

[Merlin spit]

The best way is to use a watt meter pretty much essential purchase for any one flying electric.This will give you measurments of total watts and amp draw and battery voltage under load which is usally around 10.5 v for 3S lipos

 

If you want to get 500W from a 3s lipo the maths would be watts /volts = amps 

so 500w/10.5v (battery under load)=47.6amps A 60a ESC would give some head room

 

now a 3300mah lipo if rated a 20c could be used at 66amps but would only last 3minutes so pulling 48amps would last a while longer and be better for the battery (ill have to look it up but theres a way of working out how much longer approx)

the first thing you need to is the right size prop to pull the amps and this depends on the motor as well what kv rating it is and what sort of model your flying jet warbird trainer?

[winchweight]

Ummm.... isn't a cell lipo 11.1v? Therefore 55 amps?

 

W = VxA

 

Recommend buying a Wattmeter, they're indispensible with leccy flight.

 

Summat like this

[propogandhi]

I do have one and its a great help. But I just need to get an idea of watts so I can pick the best esc. I was assuming that a 3 cell lipo being fully charged (12.5v) is what we work off. So its meant to be the nominal voltage right...11.1v?

 

prop.

Edited By propogandhi on 14/03/2010 19:31:31

[scott cuppello]

More like 10v under load after 30 seconds.....depends on how good the pack is...but I work off 10v for 3s.

[birdy]

Just get a big esc then - I would use 70 amp personally, then you don't need to worry about this kind of thing, and you have headroom as someone already said. I work off 10.5v by the way...

[Merlin spit]

use your watt meter it will tell you what voltage the battery is under load 10.5 v is average for 3s under load .if you were getting higher voltage you would need less amps to get 500watts ,a lot of 500-600watt setups use 4s lipos for this reason ie higher voltage less amps need to be drawn for same watts but your motor will need to be capble of using a 4s battery of course as the higher voltage will increase RPM

 

if your stick to a 3s battery and want 500watt you will need a 60amp esc as you are going to need to pull around 48-50 amps and it always wise to have head room for all parts of a electric setup.

watts=volts*amps

watts / volts=amps

 

 amps always goes up when either or both prop pitch or dia. are increased

  the higher kv rating of a motor the smaller the prop will be ie 2200kv good for ducted fan

1000kv good for warbird types 300kv motors will turn bigger props still 

[propogandhi]

Thanks for the input gents, I can figure out what Ii need from here now.

 

 

prop. 

[Wingman]

I use 10 volts for 3S cos the sums are easier 

[Simon UK]

I have found (at least with A123 cells) that working voltage is directly related to the amount of amps you are drawing from the cells. So the higher the amp draw the lower the voltage..... atleast this is my experience.

 

I also use use 10 volts for calculations of a 3 cel lipo pack, atleast this way you are getting - Working conditions - numbers and not theory.

 

 

[Ed Anderson]

Many motor manufacturers provide documentation that tells you what to expect from the motor.  For example the max pro motors have excellent documentation that includes charts that tell you what to expect based on the voltage of the pack and what prop you are using.  I have found them to be pretty good in predicting how the motors will behave.

 

For example, take a look at this motor:

HC3510-1100

 

The documentation says to use 3.3V as the voltage under load for each cell of your lipo.  So a 3 cell lipo is about 10V.   You can argue whether it should really be 10.5 or something else, but let's just follow their recommendtions.

 

So, if we use a 3 cell lipo and a 10X5 prop, it will draw about 24 amps. 

If we use a 10X7 prop, it will draw about 30 amps.

 

If we keep the 10X7 and change to a 2 cell lipo, it will draw about 15 amps.

 

Using one of these tables you can decide what prop to use with what voltage pack to acheive the desired wattage.  You can also see that certain combinations will over burden the motor.

 

 

Another is GWS.  If we look at this brushless motor we can see what to expect based on voltage and prop.

GWS chart

 

So you can see what to expect before you buy the motor.   There are many others who provide similar documentation.

 

 

Edited By Ed Anderson on 30/03/2010 23:23:08

[Shaunie]

The voltage on any battery will sag under load, the higher the current the more it sags.

The classic model is to consider it a perfect battery in series with a resistor (this being the internal series resistance of the battery). As the current increases so does the drop across this resistor and this of course follows Ohms law.

The higher the C rating the lower the internal resistance should be (as long as the manufacturer rates their battery honestly, not in fibs) and the better it will stand up under load. The internal resistance also increases as the battery runs flat, it's part of what stops it delivering energy forever. I am generalising a bit as different battery chemistries have different characteristics.

I have some 25C 3200 packs that drop 0.5V per cell at 35ish Amps and some 60C 2200 packs that drop 50mV (0.05V) at the same current, both sets being effectively new. Those figures from memory, the packs and data are at work.

Always use a Wattmeter and work with values under load, never use unloaded values.

Shaun