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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 



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.




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. 


santa barbara, CA
Joined May 2009
4,424 Posts

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. 



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). 


santa barbara, CA
Joined May 2009
4,424 Posts

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. 


Willoughby, Ohio
Joined Jan 2002
12,001 Posts

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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Vienna, Austria
Joined Apr 2007
2,925 Posts

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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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. 



Willoughby, Ohio
Joined Jan 2002
12,001 Posts

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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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