Another way of looking at it is that at a given rpm the larger motor will swing a bigger prop.

  Iyer sir

But, now my question, repeated:

.............asked the 64 million $ question - how to predict the rpm (and therefore voltage)drop under load

thanks iyer sir you solved the 64 million $ question for me.

As Iyer sir has not replied, why don't you tell me Nitesh? In the case you have mentioned, what will be the voltage drop on the smaller motor?

Sir you didn't read my post properly.

Iyer Sir one more question won't the smaller motor burn if doing it this way.

@rcrcnitesh
Heat generated is propotional to the square of the current.
So if you exceed the max amps rating, you will soon get overheating.
Put another way, if you go on increasing the prop size beyond the motor's capacity, a smaller motor will burn before the larger one

Hi everyone,

This thread was only intended to help newcomers to RC Electric (fixed wing) models
1. To get a feel of how to estimate the power requirement
2. And not to think that a motor's kv is a sufficient indicator of the power.

The $64m question is beyond this scope.
Hence a new Tech Talk on RPM under load 
Hope you like it.
Regards.

Mr. Iyer, what is the value of maximum lift coefficient for getting 6 m/s stall speed or 20 m/s maximum speed??? And tell me the calculations because i am getting confused between maximum lift coefficient and lift coefficient.

@ashwini06,
Let me use my RCTLG glider as a reference.
It is 48"/120cm span, 11cm avg chord and weighs 200gms.
Video of test glide shows that it glides at about 5-6m/sec (with trim, ie, elevator set for hand launch test gliding)
Since it is gradually coming down, the lift must be slightly less than the weight. Let us assume that the lift is say 180gms.
This requires a CL of about 1 at 5m/sec, or about 0.7 at 6m/sec.
However if the trim was adjusted for much faster flight, far less CL is required.
At 20m/sec, the CL required is only about .065.

(Please note that my comments are based on practical observations and personal experience, and not on laboratory standard wind tunnel tests. Hence there is potentially a lot of room for error )

Have to go out now. Will post more details in the evening)

Contd...
If you look at the graph of CL vs angle of attack, for the airfoil used (AG03), you will see that it has:
CLmax of about 1 at about +8 degrees angle of attack.
At +4 degrees, CL is about 0.7.
And for CL of 0.65, it will be -1 degree (ie slight negative angle of attack)

(Clarification: this is at Reynolds number of 50,000)

Put differently, what this means is that:
1. When flown at 5m/s, the AoA is about 8deg, and the airfoil is already at its CLmax. If flown slower than this, a higher AoA would be required, resulting in a stall.
2. At 6m/s, the required CL of 0.7 is achieved at an AoA of 4deg.
3. At 20m/s, the CL needed is so low (<0.1) that the airfoil is actually flying at a slightly negative AoA!

Remember that the AoA is relative to the airflow, not the horizon! In flight even if the wing looks horizontal, it is actually at a positive AoA, because the flight path slopes downward.

So is there any point in all of this? Yes, there is.
If you look at the Coefficient of Drag CD vs AoA data, you will be surprised to find that (for this airfoil at this Reynolds number), the CD at 4deg AoA is only half of the CD at 8deg AoA.

So the model will be far more efficient when flown at 6m/s instead of at 5m/s (other things remaining unchanged)

The problem is how to know whether it is flying at 5m/s or 6m/s
Best wishes for whatever experiments you are doing...

Ok, i got it, we have to take the help of graph ultimately.... thank you so much.

What are the options to get max flight time from 1000 kv motors ?

make a rc helium blimp

K K Iyer Sir's valuable points + http://www.airfieldmodels.com/information_source/math_and_science_of_model_aircraft/formulas/index.htm = Many doubts cleared (We can now think about scratch building a plan)

Thank You Iyer Sir for such great valuable points.

This should be atleast a part of          "New to this forum or RC hobby ? Start here : The RC India Gems !" Space

Thank you Very much Iyer Sir. 

Hello Sir, since this topic is about power, here is the problem am going through. I am about to build a Hovercraft. I ll be using a 25mm Depron sheet. The length of the hovercraft is around 30 inches, and width is 18 inches. I am unable to decide what kv motor I should use. overall weight including battery is going to be around 1.5-2 kg. Please suggest me which motor I should use for lifting as well as pushing purpose. Thanks n Regards.

In addition to that I ve preciously build some hovercrafts but they were not fast enough. On one occasion I used sunnysky 980 kv motors both for lift and push. On another occasion I did used Avionic 1400 kv motors but in both of the cases the hovercrafts were not fast enough. Also they were good only on ceramic tiles and were struggling to move on concrete surface. This is the reason I want to choose the right motor combination. Thanks n Regards.

@miginstruments,
Not able to give you specific and useful answers due to lack of hovercraft experience.

However sharing some thoughts that may be of help:

1. Will it be 1.5-2kg? A 30"x18" body in depron (25mm???), skirt and rudder may not be over 200gms.
Lift Motor, thrust motor, props, ESCs, servos may be 3-400gms. Add 200gms for a 3s 2200mah lipo for lift and 100gms for propulsion lipo of 1000mah. AUW may not exceed 1kg.

2. I think for the lift it would be desirable to have the lowest kv, largest prop dia, and lowest pitch you can find. Eg, Hextronics DT700 (700kv) and a 12x3.8 prop. I remember having seen a test report that this would provide about 1300gms static thrust. (Looking through my own records, I found that my DT750 with 11x5.5 prop gave 950gms at 14.4amps/168watts on 3s lipo.

3. Once the weight is lifted, propulsion should not require too much power. I suggest 40-50mm EDF. This will help in keeping the CG as low as possible. And, if I remember correctly from my own tests a few years back, should give about 500gms static thrust.

4. Alternatively, use a high kv motor (2000/2500kv) that can take, say, a 6x4 prop (Eg eMax, I forget whether it is 2805 or 2215. Something like that)

5. Remember that kv is not a sufficient indicator! It is the power of the motor, and the dia of the prop that determine the thrust. The pitch determines the speed.

Hope this is of some use...
Regards

Dear Sir Thanks for the Suggestion. I was confused between the lift and the push motor. so far I was thinking that the push motor should have lower kv rating so that it should provide enough amount of thrust. Earlier I used a a bigger prop as well. This time I was thinking to use a 550kv motor for push. In fact I even ordered one. But I ll change the plan a little bit as per your suggestion, instead of using the 550kv motor for push I ll put it for lift and go with a higher kv motor and a relatively smaller prop for push. If you come across anything new then please do share the information with me. Thanks n Regards.