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Old Nov 26, 2012, 03:39 PM
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Originally Posted by green_flyer View Post
I did a static thrust test of the Turnigy 5000kv inrunner. Sadly, its a bit dissapointing, I had higher hopes. Efficiency is low, the motor is essentially an amp hog. Almost 22A, 238W for only 343g static.
Here is a comparison of the Turnigy motor with the Lander motor. I wouldnt want to eat up 7A more for only 20g more thrust. My search for a high efficiency low weight motor continues.
Its true some things were not constant, different airframes, different ESCs, but still I was expecting more out of the Turnigy.

Lander Rotor, JPower housing/JPower copy housing, Nanotech 3s 850mah 45C, 25A ESC (Phoenix/Turnigy plush)

Motor--------------------Airframe----------------Peak Thrust----------Peak A----Peak V---Peak W----Efficiency
Lander 4600kv-------F35-----------------------323g---------------------14.98------11.36------170-----------1.9g/W
Turnigy 5000kv------F22-----------------------343g---------------------21.77------10.96-------238----------1.44g/W

http://youtu.be/e4Qn0Yu5WUs
Put that same set up in a different airframe with better ducting and it will be much better. The F/A-18 is really the only twin exhaust jet in the series to not be bothered by the dual outlet. I bet a squashed thrust tube would be better than the twin outlet in that plane...
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Old Nov 26, 2012, 03:42 PM
we can take off without that
green_flyer's Avatar
London, UK
Joined Nov 2008
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good poing about the F22 ducting, since its the same housing I will swap it into my F35, hopefully better results.
The strange thing is the stock Jpower fan still makes 265g in the F35 and the F22. Maybe thats its magic default thrust in any airframe
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Old Nov 26, 2012, 03:56 PM
chuck
santa barbara, CA
Joined May 2009
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grams / watt is not a linear measurement , u are aware of that correct?
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Old Nov 26, 2012, 04:14 PM
we can take off without that
green_flyer's Avatar
London, UK
Joined Nov 2008
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I'm not sure what you mean? its not meant to be linear, its a data point. Its a comparison of 2 data points. Like if you were looking a 2 different motors, and one had 80% efficiency and another had 75% efficiency.
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Old Nov 26, 2012, 04:16 PM
chuck
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its not comparable at 2 different thrust levels.
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Old Nov 26, 2012, 04:19 PM
we can take off without that
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Joined Nov 2008
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Thats true for a single system, you are thinking of the grap plot. But these are singular data points of 2 different graph plots. Both are at the peak of each graph for each plot. So it is a valid comparison of a system, where all you change is the motor (although not strictly true in this case because I used different airframes and ESCs).Essentially it lets you decide which motor to use.

here is a quick example. You cant expect to compare at the same thrust, because at full thottle the two set ups cant acheive the same thrust. Also, ESCs are most efficient at full throttle, so it would be invalid to reduce throttle of one system to match thrust.

Essentially you are correct, but its not thrust that has to be kept constant, its voltage. I would have to test both setups at the same voltage to get a fairer efficiency comparison. However this is impractical as in real life the system drawing more amps will sag the battery more ( I cant fly with a power supply ). Also, I can guesstimate that if this motor only acheives 1.44g/W at 11V, then it highly unlikely to have much higher efficiency at 11.36V at which the other motor was tested.
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Old Nov 26, 2012, 04:38 PM
chuck
santa barbara, CA
Joined May 2009
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lets try it this way ; 100g at 100w is not the same efficiency as 200 g at 200w .

it just does not work this way as your efflux speed changes much more than static thrust as power goes up.

your thrust #'s where close so it's not a huge deal in your test , just wanted to make sure u dont make this common mistake in g/w comparison. the test is only valid in the same system at the same thrust level.
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Old Nov 26, 2012, 04:55 PM
we can take off without that
green_flyer's Avatar
London, UK
Joined Nov 2008
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I disagree, you can only compare the efficiency of 2 different motors at the same voltage. If you wanted to match thrust, the two motors would end up being at different voltages, therefore not a valid efficiency comparison. Also, note that whilst trust is not far off in these two tests, the Amps are very different, resulting in a very revealing result.
Also power (watts) is not directly connected to efflux speed or static or dynamic thrust. It all depends on system efficiency. You could have a very inefficient system pulling huge watts but producing little efflux speed and static thrust. Also, static thrust is proportional to efflux speed, although not linearly.

edit- also, in your example that is the same efficiency
watts is energy expended, grams is work done, so in both examples the efficiency is the same, you are expending 1 Watt to produce 1g of work (thrust).
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Old Nov 26, 2012, 08:38 PM
chuck
santa barbara, CA
Joined May 2009
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wrong again. in my example, the 200watt system is much more efficient. u can also look at it this way, watts to efficiency is not linear. or , thrust is not linear to watts. to double thrust the wattage will be more than double, period. and again this is within the same fan/duct combo.

another example that is not linear is power to speed , it takes ~8 times the power to double the speed.

if u dont believe me , look it up.

chuck.
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Old Nov 26, 2012, 08:47 PM
Augermeister
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Is it just me or is the paint/decals on the Katana a POS ? Completely flaking off and I even coated her with Minwax.
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Old Nov 26, 2012, 09:26 PM
Wonderfully Wicked
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Willoughby, Ohio
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Grams per watt is actually not a valid way to compare power systems. The only way you can compare that unit is if the watts on both systems are identical and in that case the direct thrust measurement is all you need. The reason it does not work is that thrust and watts are not linear and this is due to the physics involved. If you have 100 grams of thrust at 100 watts, then 200 watts will never equal to 200 grams of thrust in the same power system. Gram/watt was a marketing concept made popular by GWS (best I can find) way back when as way to "compare" power systems but the underlying assumption of the linear relationship is incorrect so the number has no real meaning.

Here is the correct way to compare power systems.
http://www.rcgroups.com/forums/showp...65&postcount=4

At the bottom of that post you will see a chart showing how the same power system at the same efficiency appears to be less efficient as the watts increase (if you use grams/watt method) when in reality the performance falls on the curve following the formula.


So for this case:
Lander setup:
K= 323/170^.666= 10.52

Turnigy setup:
K= 343/238^.666= 8.93

So the Lander setup is better as the K factor is higher, and yes the grams/watts figure is also higher but it is a mathematical coincidence, partly due to the lower watts of the Lander setup. When using grams/watt, the number ratio will always decrease as the watts go up due to the exponential nature of how thrust and watts are related.

If you use this "K" method then you will always have apples to apples on power systems and it allows you to predict thrust and watts at other power levels, compare different fans and motor combos etc.

Don
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Old Nov 26, 2012, 10:02 PM
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gooniac33's Avatar
Sunnyvale
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Quote:
Originally Posted by dahawk47 View Post
Is it just me or is the paint/decals on the Katana a POS ? Completely flaking off and I even coated her with Minwax.
Nothing but crap!!! I love the plane but want to take all the decals off and paint it myself!! AGH!!! I flew mine this weekend in formation with F/A-18's and it was just as fast but more stable...Great flying plane! I can't imagine it with about 500 watts!! May not be all that much faster but it will climb better! But its already amazing as it is!!

Quote:
Originally Posted by The Don View Post
Grams per watt is actually not a valid way to compare power systems. The only way you can compare that unit is if the watts on both systems are identical and in that case the direct thrust measurement is all you need. The reason it does not work is that thrust and watts are not linear and this is due to the physics involved. If you have 100 grams of thrust at 100 watts, then 200 watts will never equal to 200 grams of thrust in the same power system. Gram/watt was a marketing concept made popular by GWS (best I can find) way back when as way to "compare" power systems but the underlying assumption of the linear relationship is incorrect so the number has no real meaning.

Here is the correct way to compare power systems.
http://www.rcgroups.com/forums/showp...65&postcount=4

At the bottom of that post you will see a chart showing how the same power system at the same efficiency appears to be less efficient as the watts increase (if you use grams/watt method) when in reality the performance falls on the curve following the formula.


So for this case:
Lander setup:
K= 323/170^.666= 10.52

Turnigy setup:
K= 343/238^.666= 8.93

So the Lander setup is better as the K factor is higher, and yes the grams/watts figure is also higher but it is a mathematical coincidence, partly due to the lower watts of the Lander setup. When using grams/watt, the number ratio will always decrease as the watts go up due to the exponential nature of how thrust and watts are related.

If you use this "K" method then you will always have apples to apples on power systems and it allows you to predict thrust and watts at other power levels, compare different fans and motor combos etc.

Don
Couldn't have said it....no really I couldn't have...LOL But it makes a ton of sense!
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Old Nov 27, 2012, 01:52 AM
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Vienna, Austria
Joined Apr 2007
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Simple phyics behind thrust and efflux

What's happening when a fan accelerates air?

First, in a nutshell:
thrust = force = mass * acceleration. Twice the mass accelerated, twice the thrust. AND the relation is also linear with regards to the acceleration: Twice the acceleration, twice the thrust.
But the kinetic energy imparted on the accelerated air rises linearly only with the mass, but NOT with the velocity:
It's mass times velocity squared, divided by 2.
Therefore, you need to put in four times the energy to accelerate a mass to twice the speed (both times from zero).

This is the reason why props should be as big as possible in order to accelerate a lot of air by just the amount demanded by the airspeed. Also, bigger fans will give more thrust with the same Watts and are preferred AS LONG as you get enough efflux for the desired speed of the model.
The whole business can easily be influenced by thrust tubes - forcing the air through a narrow exhaust speeds it up - at the expense of static thrust (once going below 85% FSA or so, depending on the fan geometry).

Let's do the figures for two systems, each 5N thrust (some 500g):

System 1 (low efflux, e.g., no thrust tube) achieves 5N thrust by accelerating 250g of air to gain 20 m/s in speed (approx 45 mph) in one second:
Force = mass * acceleration, in this case F = 0.25 * 20 = 5N
Looking at the static case for simplicity, the energy in the moving air was zero at the start, and is now:
E(kin) = m * v^2 / 2 = 0.25 * 20^2 / 2 = 50 Joule or Wattseconds, and since this happens during one second
we have put in 50W (that's NOT Watts in, that would be a lot more due to motor and fan inefficiency)

System 2 (high efflux, e.g., smaller fan or same fan with narrow hrust tube) achieves 5N thrust by accelerating, say, 200g of air to gain 25 m/s in speed (approx 56 mph) in one second: F = 0.2 * 25 = 5N
Energy imparted on the air now E(kin) = 0.2 * 25^2 / 2 = 62.5 Joule or Wattseconds, and it takes 62.5 Watts of power to impart that energy on the air in one second (power is just energy per period of time).

This means that systems should only be directly compared with the same exhaust diameter. You can do two things:
- look at the Watts needed to get the same thrust in both systems, or
- put in the same power (Watts) and look at the difference in thrust
EDIT: or you use the formulas provided by the DON which take into account the relationship between momentum and energy

cheers
Clemens
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Old Nov 27, 2012, 01:57 AM
we can take off without that
green_flyer's Avatar
London, UK
Joined Nov 2008
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Quote:
Originally Posted by The Don View Post
So for this case:
Lander setup:
K= 323/170^.666= 10.52

Turnigy setup:
K= 343/238^.666= 8.93

Don
Thanks Don, thats useful to know. So the Turnigy should be making over 406g static to be comparable to the Lander.

p.s. - because we are comparing different systems, it is possible to have more thrust for less watts. Neutron 4900kv for example, in the same fan, produces 333g peak for 168W.
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Old Nov 27, 2012, 06:54 AM
Wonderfully Wicked
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Willoughby, Ohio
Joined Jan 2002
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Quote:
Originally Posted by green_flyer View Post
Thanks Don, thats useful to know. So the Turnigy should be making over 406g static to be comparable to the Lander.
Correct. and since you use dthe same motor, and only changed the motor (if I read the post correctly) then it says the Turnigy motor is not converting the power as efficiently as the Lander motor.

Quote:
Originally Posted by green_flyer View Post
p.s. - because we are comparing different systems, it is possible to have more thrust for less watts. Neutron 4900kv for example, in the same fan, produces 333g peak for 168W.
This setup has a K of 10.93 and since you are using the same fan unit, it is pretty good at energy conversion, about the same as the Lander motor (slightly better actually). If you are trying to compare two different rotors, say stock fan to Lander, you would want a few data points on the stock fan to see what its K value is coming in at. If you have only one data point and it is a lower K value, you do not know if it is low due to motor being inefficient or the rotor design is bad, unless you are running similar watts in the same motor on two different rotors so can assume the motor is running about the same efficiency. What I am getting at is the K factor is really a combination of the motor and fan efficiency and is a representation of the overall efficiency of the combination of the two. Generally the fan factor stays about the same over the power ranges we use, it is the motor that varies the most, generally dropping efficiency as watts increase.
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