Motor "Sweet Spot" - Page 2 - RC-Monster Forums

neweuser · 06.30.2006, 03:46 PM

It is you! I didn't want to say and be wrong! LOL. I look dumb enough asking all the questions i do! I'll try it now, thanks!

coolhandcountry · 06.30.2006, 04:13 PM

I think the sweet spot varies just a little with vehicle motor and few other factors. The 35k is a good shooting point though.

neweuser · 06.30.2006, 04:18 PM

I'm getting, if i did it right, about 46mph? wasn't sure about the draw and that but i did what i knew and could figure out. I also figured that I'm getting about 33,779.2 rpms, that's calculating with 1.40 x 16 = 22.4 i used this as an average since there are variations. and after what brian said, you can't really go by 1.2 per cell. and my cells are (with no load, fully charged, an avg of 1.40). Does this sound right?

squeeforever · 06.30.2006, 05:18 PM

Try them at around 1.3-1.35 volts per cell. They will certainly drop some.

BrianG · 10:46 AM

I agree with Squee; they will drop. Just think about it for a minute and it will make sense. At no load, the batteries will have their highest voltage. When you take off, the motor is drawing the most current so the battery voltage will drop the most. Then, as the truck reaches the speed the motor wants to be at, the current draw will decrease and the battery voltage will rise again. It won't rise to its unloaded value since there is still some current draw, but much less than take-off current. At the motor's "sweet spot", it will be the most efficient. This does not mean it will draw the least current then, it just means that most of the battery power will be on the motor rather than dissipated as useless heat on the ESC.

This is probably a bit off-topic and outside the scope of this discussion, but a battery is really a constant voltage source (meaning voltage stays constant no matter how much current is drawn) in series with a small resistor. This resistor is not an actual component, but can be represented that way for this example. When you hook a load up to it, that load is another resistor. This creates a voltage divider. Basically, all that means is that these resistances create a current. This current drops a voltage on each resistor, and this drop is proportional to the resistance value.

Example; Let's say we have a GP3300 sub-C battery. It's unloaded voltage is 1.4v. Now, let's say it has an internal resistance of .01 ohms. Doesn't sound like a lot does it? Now, let's say we are drawing 50A of current from the ESC. This 50A will drop 0.5v (vdrop=amps*resistance, or 50A*0.01ohms) across the internal resistance of the battery. This means the battery will be 0.5v less than it's unloaded value of 1.4v, or, 0.9v.

Less current draw = less voltage drop and more voltage at the battery terminal.
More current draw = more voltage drop and less voltage at the battery terminal.

I've attached a pic that visually shows what I am talking about above. Hopefully this made sense.

neweuser · 06.30.2006, 10:54 PM

Ok, in english it means, when there is no load or "draw" the volts will read more in the batt, when there is a "draw" the volts will drop more and less will be in the batt...did i get that all right? I don't know what language Brian was speaking, but it sure wasn't troy!

BrianG · 06.30.2006, 11:52 PM

Quote:

Originally Posted by neweuser

Ok, in english it means, when there is no load or "draw" the volts will read more in the batt, when there is a "draw" the volts will drop more and less will be in the batt...did i get that all right? I don't know what language Brian was speaking, but it sure wasn't troy!

Yeah, in "English" that's what it means. :)

Sorry, I tried to make it easy to understand but obviously failed. :dft001:

coolhandcountry · 07.01.2006, 06:50 AM

Quote:

Originally Posted by BrianG

Yeah, in "English" that's what it means. :)

Sorry, I tried to make it easy to understand but obviously failed. :dft001:

You made it simple for people that is in the electrical field.
I was lost. I unserstand it some what but not all the way.:007:
How much does the batteries voltage really come up when under less of a load? At a 30 amp draw the nimhs if good can hold about 1.2 or so. give or take a .05 volts.

neweuser · 07.01.2006, 10:05 AM

I thought you did great, it's long so i had to go over it twice. And the neat little illustration did well! You should be a teacher! : - )

BrianG · 07.01.2006, 01:45 PM

CHC: Sorry, it's hard for me to explain things sometimes. I either go too deep into detail and ramble on and one, or not deep enough. :dft001: The explanation was greatly simplified for the sake of simply explaining the voltage drop under load. Batteries generate their power via chemical reaction so there are other factors that come into play. The discharge curves are not linear, so the voltage drop will vary more or less than a simple resistor equivalent would show. The best way to see how a specific battery will react is to look at the specific spec sheet with various plots and graphs.

neweuser: LOL, I would NOT be a good teacher - I have far too little patience/tolerance. Plus, I don't know enough. I'm glad if it helped you at all though. :027:

coolhandcountry · 07.01.2006, 02:06 PM

I was wondering though like if you draw 50 amps out the battery and it falls to like1.05 volts. then as the rc accels it would become less draw so if it fell back to 30 amp draw wouldthe voltage come back up to 1.2 like if it was a consistant 30 amp draw.

BrianG · 07.01.2006, 02:36 PM

Yes, it will come back up somewhat.