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« on: September 13, 2014, 04:35:20 PM »
K K Iyer
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Hi everyone,
In the 'Tech Talk - How much power' thread,
Rai saheb raised a 64 million $ question - how to predict the rpm (and therefore voltage)drop under load!

Could not answer right away for the simple reason that i didn't know!

However, after some head scratching, and actual testing, here is my attempt at an answer.
Grateful for scrutiny to see if the following analysis is reasonably correct.

Obviously, the voltage drop under load is a function of the current drawn and the internal resistance of the battery

To check this relationship, I connected a known load (55w 12v car halogen bulb) to 3 different 3s 2200mah lipos via the HK-010 Power Analyzer. Results were as follows:

Make       Watts   Amps   Volts      Volts       Volts   IR milliohm
                                     NoLoad  Loaded    Drop 
   
Haiyin        58.7    4.8     12.43     12.24      0.21      44
Turnigy1     54.1    4.7     12.04     11.78      0.26      55
Turnigy2     48.0    4.4     11.45     10.93      0.52     118

Haiyin from Vishal Rao of Kinetic Hobbies. About 3 months old, less than 10 cycles, only 1 flight.
Turnigy1 about 6 months old, about 20 cycles.
Turnigy2 about 12 months old, only 2-3 cycles, but bloated probably due to having been stored for several months on full charge.
All three rated 20-30c.

(Since this was only a preliminary investigation, please ignore for the time being, issues like all 3 batteries not being at the same starting voltage, the wattage not being exactly 55w etc)

Using the above IR figures, and assuming a 20amp draw, we can expect the following voltage under load:

Make          Volts      Amps     IR       Volts      Volts
                  NoLoad                         Drop      OnLoad
Haiyin         12.50       20        44      0.88      11.62
Turnigy1      12.50       20        55      1.10      11.40
Turnigy2      12.50       20       118     2.36      10.14 (ie, it can't take a 20amp load!)

Assuming
1. 1000kv
2. Actual rpm / kv x voltage of say 75% (Not a rule, only a thumb rule)

On a 20amp load, one could expect about:
8700 rpm with the Haiyin
8550 rpm with the Turnigy1
7600 rpm with the bloated Turnigy (before it dies...)

Will test to see if these estimates are any good, and whether there are any blatant errors in the analysis.

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« Reply #1 on: September 13, 2014, 05:03:49 PM »
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This is a very very rough extrapolation sir. It needs more precision. I think we have to first derive the "apparent load of the motor/prop"  ie an equivalent resistance connected in series over which there will be a potential drop. But I am not sure myself, that's why I asked Grin
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« Reply #2 on: September 13, 2014, 05:21:22 PM »
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Sir,
$64m ke liye kafi sir khujlana pada Head Scratching
The old grey cells ain't what they used to be.
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« Reply #3 on: September 13, 2014, 05:29:04 PM »
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Where are our young Electrical fellows? Busy overcharging? ROFL ROFL
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« Reply #4 on: September 13, 2014, 05:34:04 PM »
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Sir,
More importantly than precision (which I will refine),
I request an opinion on whether the approach to the problem is correct from the theory point of view.
Or is it funda gol?
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« Reply #5 on: September 13, 2014, 05:44:26 PM »
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With due deference sir, it is an incorrect approach.

One should be trying to identify the physical phenomena rather than looking for number patterns, ie Deductive rather than Inductive reasoning
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« Reply #6 on: September 14, 2014, 10:08:45 AM »
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Re:

Iyer sir...  Great thread...
I feel it's a good beginning... And if we could just add a Tachometer to the above setup...
We could get to know the output power... Which could be the equated with the input power...  We should be able to get the voltage drop ( ignoring power loss due to friction.. Heat etc) 
Sir please correct me if am wrong
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« Reply #7 on: September 14, 2014, 10:11:34 AM »
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Iyer sir...  Great thread...
I feel it's a good beginning... And if we could just add a Tachometer to the above setup...
We could get to know the output power... Which could be then equated with the input power...  We should be able to get the voltage drop ( ignoring power loss due to friction.. Heat etc) 
Sir please correct me if am wrong
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« Reply #8 on: September 14, 2014, 11:24:01 AM »
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Re:

How does the tacho help in detrmining power here?
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« Reply #9 on: September 14, 2014, 11:35:29 AM »
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Re:

The Power output can be calculated with a prop rotating at a know RPM...  Tachometer would help us get the actual RPM hence the power output... .. Correct me if am wrong
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« Reply #10 on: September 14, 2014, 01:30:36 PM »
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Re:

How? How will knowing rpm alone give you power? Power = torque x rpm. How will you measure torque?
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« Reply #11 on: September 14, 2014, 04:47:45 PM »
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We can get the power by...

Screenshot_2014-09-14-16-45-09.jpg
Re: Iyer's Tech Talk - RPM under load
* Screenshot_2014-09-14-16-45-09.jpg (44.04 KB, 800x450 - viewed 779 times.)
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« Reply #12 on: September 14, 2014, 04:52:27 PM »
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Re:

Or better way use one of the many webcalc available..
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« Reply #13 on: September 14, 2014, 05:08:01 PM »
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Re:

How will you measure Vo?
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« Reply #14 on: September 14, 2014, 05:09:01 PM »
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Re:

Doc, this is simply argument for the sake of it. Disappointing.
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« Reply #15 on: September 14, 2014, 05:18:04 PM »
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Re:

Sir... Please...  This formula is for dynamic thrust.. Label V0 as nil and you should get static thrust...
I really don't argue just for the heck..I know more then often  I am wrong... But I learn... And that's what am here for learning... And who else can be a better teacher than a person like you... Bow
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« Reply #16 on: September 14, 2014, 06:18:37 PM »
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Re:

This will give you thrust. How will you get power
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« Reply #17 on: September 14, 2014, 07:10:52 PM »
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Re:

Sir I sincerely don't know the formula but there are a few webcalc, that I routinely use...
Here's one..
http://adamone.rchomepage.com/calc_thrust.htm
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« Reply #18 on: September 14, 2014, 07:12:41 PM »
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Re:

Iyer Sir  ' SOS '
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« Reply #19 on: September 14, 2014, 07:12:52 PM »
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Re:

Sad
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« Reply #20 on: September 14, 2014, 08:44:09 PM »
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@docnayeem,
Sorry for delay. Was out flying!

Looks like i got too intrigued by Rai Saheb's $64m question, and too involved in trying to find an answer for something i didn't know.
In retrospect, i realise that i was on a wild goose chase, and barking up the wrong tree, in the process missing the woods for the trees!

The real issue is:
Why would i (or anyone) need to PREDICT the voltage drop/rpm loss under load?

I don't need to predict it, as i already know!
I just run the motor with a given prop (and different props) on my test rig, (which has a tach)
and it TELLS me:
- the voltage without load
- the voltage under different loads at different rpms
- the amperage under different loads at different rpms
- the rpm under different loads at different throttle positions
- the duration of the run
- the mah consumed in the duration of the run
- the thrust under different motor/ prop/ rpm/ power conditions

And i know the pitch speed from the rpm.
And the gms thrust per watt of power for different motor/prop combos at different rpms.

Can't think of anythink else i need.

Looks like i've utterly failed to understand the significance of the question!
As i said earlier, my old grey cells ain't what they used to be!

So the only outcome of this effort is that i could find out the internal resistance of some of my batteries.
That is something i didn't know earlier. Now i'll check all of them.
Regards
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« Reply #21 on: September 14, 2014, 09:06:19 PM »
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Re:

Iyer sir. Measurement imho is the only way. Just use a voltmeter Cheesy

But sir, for that many of us have test rigs Wink

The question was directed in the context of some issues that had come up, wherein one without a wattmeter etc. and without experience to make a reasonably accurate guesstimate, could predict, thus helping him or her to select a suitable prop for the given motor for the given model
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« Reply #22 on: September 14, 2014, 10:12:36 PM »
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The question was directed in the context of some issues that had come up, wherein one without a wattmeter etc. and without experience to make a reasonably accurate guesstimate, could predict, thus helping him or her to select a suitable prop for the given motor for the given model
Hope you agree that 'selecting a suitable prop' is a subset of 'How much power is required' and 'pitch required'

Precisely the questions i tried to address in my first Tech Talk - for precisely the same audience!

Glad we are on the same wavelength!
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« Reply #23 on: September 14, 2014, 11:08:57 PM »
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@docnayeem
Sir, I don't know whether you can calculate power or not, but even if you find the power, these motors have an efficiency of only 60-70%, Lower kv motors may go up to 80 or 85%, but even that is not small enough to neglect.
So your calculation of voltage drop will be incorrect. Correct me if I'm wrong.
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« Reply #24 on: September 14, 2014, 11:38:35 PM »
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@maahinberi,
@docnayeem
Taken as 75% in the first post.
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« Reply #25 on: September 15, 2014, 04:58:30 PM »
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Iyer sir. Measurement imho is the only way. Just use a voltmeter Cheesy

But sir, for that many of us have test rigs Wink

The question was directed in the context of some issues that had come up, wherein one without a wattmeter etc. and without experience to make a reasonably accurate guesstimate, could predict, thus helping him or her to select a suitable prop for the given motor for the given model
That's so true...
But most of them who do not have test rigs do rely on freely available thrust calculator.  And that's the only way out... Use specs and calculators
By the way Rai sir thanks for your inspiration I do have a test rig now...  Smiley
Will post pics soon
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« Reply #26 on: September 15, 2014, 06:06:11 PM »
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Hope you agree that 'selecting a suitable prop' is a subset of 'How much power is required' and 'pitch required'

Precisely the questions i tried to address in my first Tech Talk - for precisely the same audience!

Glad we are on the same wavelength!

If it makes you feel happy! Your happiness is paramount.

Otherwise, I don't. It certainly is not the same. In one case you talk about optimum prop selection. In the other, analysis of what is happening in prop rpm drop under load. If the first answered all questions, the second was never needed.
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« Reply #27 on: September 20, 2014, 05:14:51 PM »
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OK, here's an actual example of yesterday.

I had 2 old (in age, not use) 3300 mAh Turnigy Nanotech 30-50 C Lipos.

Just made a new model, aerobatic, 1.3 Kg. Wanted 350 Watts, and used a 3536-1250 KV with a 10*7 prop.

On testing with the old batteries, they gave 270 Watts @ 32 A, and 9.1 V

Then tested with a new battery. Got 560 Watts @44 A and 12.2 V

Massive difference

The old batteries are not puffed excessively, and all cells are in balance, and charge till 12.55-12.56 V

Just died from old age - however the important point is how the voltage can drop under load, especially on old batteries, and the importance of measurement (I did not measure thrust and RPM, did not need to in this case.)

And yes, the maiden was successful  Grin Grin  Will be posting that log soon  Wink Wink
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