Hi everyone,
In May this year, I had posted a request to Anwar sir for a Tech Topics board. This was seconded by none other than rcpilotacro. However, this may have missed getting the attention of Admin.

Be that as it may, I thought I'll start such a topic anyway.
The first one is 'How much power is needed?

From the posts it seems to me that many many members consider a motor's kv (rpm per volt) as the indicator of its power. This is a totally incorrect notion.

1. A motor's power is indicated by its max watts rating
2. Its ability to convert its power into thrust depends on the prop, primarily the prop dia
3. Its ability to fly the model at appropriate speeds depends on the prop, primarily the prop pitch

So the next questions are:
- How much thrust do you need?
- How many watts needed to provide the required thrust?
- What speed is required for the model to fly properly?
- What is the required prop pitch to deliver the required speed, given the rpm/kv of the motor?

We'll look at these questions in the next post...

How much thrust do you need?

Long accepted thumb rules, and personal experience (supporting data available) indicate that:
1. A trainer type model needs only 30-35% thrust to weight ratio to maintain level flight
2. 50% thrust to weight is enough for safe take off and climb out
3. 100% thrust to weight is enough for most aerobatics
4. 150% thrust to weight will give unlimited vertical
5. More than that may be required for 3D, but I have (so far) no experience to comment from

Contd from post #1
How many watts to produce the required thrust?

One thumb rule is 5gms thrust per watt. (Static thrust)

Small props / high pitch / high kv motors may be in 3-4 gms/watt range
Large props / low pitch / low kv motors may be in 6 gms/watt range

Take the example of a 40-48" span trainer weighing about 600 gms.
300 gms thrust should be enough to fly it safely
At 5 gms/watt, that needs 60watts.

This is a reasonable starting point for selecting a suitable motor. For the above example, a 100watt motor would have sufficient reserve.

But can it fly the model fast enough to avoid stalling?

Following thread sir
A tech section should surely be started. Many one off tips/techniques/thumb rules are lost over time.Similar to ancestral  knowledge being lost over time
keeping them stored in an active section where all the experts can share their thorough know how would be awesome. Then the repetitive questions will be reduced.

Sorry Sir for intruding and posting in between.

Thank you very much sir for starting such a thread.

that seem to me interesting  

go on sir

Contd from reply #2
How fast does it need to fly?

The answer is - sufficiently faster than its stalling speed
Thumb rule: Should be capable of 2.5-3 times the stall speed.

How to find the stall speed?
Thumb rule: 4-5 times the square root of the wing loading. (Haha, my favourite subject!)

Take a 40-48" span trainer. Wing chord say 6-8". Wing area say 2-2.5 sqft. All up weight say 550-650gms.
That will give a wing loading of about 9oz/sqft (say 250gms per sqft)
Stall speed estimate in mph is: 4-5 times square root of wing loading in oz/sqft
In this case 12-15 mph (18-22 ft/sec, 6-7 m/s, 20-25kmph)

Cruise capabilty required is therefore (2.5-3 times stall speed):
30-45 mph, or 45-60 ft/s, or 15-20 m/s, or 50-70 kmph.

Contd from reply #5

What pitch is required to get say 60 ft/sec speed?

Staying with our 40-48" span, 2-2.5 sqft, 550-650gm, 9oz/sqft model,
Let us consider 4" and 6" pitch (fairly typical)

A 6" pitch prop (ie, 0.5ft) needs to turn 120 times to go 60 ft.
So for 60 ft/sec speed, it has to turn at 120 rps or 7200 rpm.
A 4" pitch prop needs to turn 180 times to go 60 ft.
So for 60 ft/sec (about 65kmph) it has to run at 180 rps or 10,800rpm

Subscribed.!! keep it coming Iyer sir. 
It would be great if you could summarize all those concepts that you already talked about in this thread. like prop selection etc.

Ashok.P

Subscribed.
This is a superb start for all the beginners and others too. Very well explained.

Very nice method to calculate the speed. 

Contd from reply #6


On 2s, say 8v, 6" pitch needs 900kv (rpm per volt), 4" pitch needs 1350kv
On 3s, say 12v, 6" pitch needs 600kv, 4" pitch needs 900kv.

Since the motors that are generally available run at about 75% efficiency (actual rpm/v compared to theoretical kv), the above figures have to be divided by 0.75.

So,
On 2s, say 8v, 6" pitch needs 1200kv (rpm per volt), 4" pitch needs 1800kv
On 3s, say 12v, 6" pitch needs 800kv, 4" pitch needs 1200kv.

Now we know the pitch, we know the kv. What about dia?

Contd from reply #10

For prop dia I guess we'll just have to make up a thumb rule!

How does 15-25% of wing span sound?
Many of our 36" span models probably have 8" dia props (22%)
48" span models probably have 10" dia props (21%)

So let us assume (using 3s lipo), that we need:
1. Prop dia of 9"
2. Prop pitch of 5"
3. Motor of 1000kv
4. Capable of 100watts (or say 8-10 amps)

Will post examples shortly!

Excellent info sir.. my vote also goes for a dedicated tech section 

Contd from reply #11

Example: my electric trainer
Span 52", chord 8", area 2.9sqft, weight 500g w/o battery, 590g with 3s 100mah, 690g with 2200mah.
Wing loading 7.25/8.25 oz/sqft with 1000/2200mah battery.
45 degree climbout on 5amp, 60watts.
Maintains height on 2.5 amps, 30 watts.
Prop 9x5 ordinary orange, not EMP or APC.
Motor Turnigy D2830/11, 1000kv, rated at 210watts, thrust 890gms as per manual!
This motor is far too much for this model, as i never need over 30% of the available power!

Contd from reply #13

Example: my Spitfire
Span 34", avg chord 7", area 1.65 sqft, weight 360gms
Wing loading 7.7oz/sqft with 3s 1000mah battery
Prop 8x4 ordinary orange.
Motor L2210 1650kv
Flies loops and rolls at half throttle (72w) thrust about 280gms
Vertical rolls out of sight on full throttle (210w) thrust over 650gms (far too much!)


Edit: In case anyone is wondering what I'm talking about, kindly see from first post. Thanks.

subscribed sir !!
Looking forward to learn something new every day

subscribed for some amazing gyan   

  Iyer Sir

I'm sorry sir, but whereas this may apply frequently, it cannot be termed a rule. There are too many other factors involved. Some large diameter low KV combinations are frequently in excess of this

Great work Iyer sir....
Expecting a lot of good discussion here....  Subscribed!

Super Iyer sir! Just a suggestion that everything in a single post will be easy to read and digest than skim through all posts.

This is a very good idea sir. I have a suggestion to anwar sir though. I agree with Vishal ^, but have a better idea, if it can be implemented. Can we have an article section for stuff like this? It would keep all information posted by the Author in one place and then comments/discussion can be added at the bottom, rather than having comments mixed in between posts related to the topic. The author should also be allowed to add to his article any time he feels like.

Very Nice Iyer sir.
Sir are questions allowed.

Iyer sir one question if I have an 2836 1000kv and an 2822 1000kv motor both of them with a 10*4.7 prop. Basically both the motors should be producing the same thrust as they are of the same size and running the same prop but why does the bigger motor produce more thrust.

Good question Nitesh (although after asking 'can I ask a question' - (which was a question  ) you have without waiting for Iyer sir's reply asked a question, thus raising the question is it questionable to ask these questions  

What is good here is you have indirectly asked the 64 million $ question - how to predict the rpm (and therefore voltage)drop under load!

@rcrcnitesh,
Same size, or is one bigger? 2836 vs 2822

"From the posts it seems to me that many many members consider a motor's kv (rpm per volt) as the indicator of its power. This is a totally incorrect notion."

Both motors being 1000kv simply means that if you connect a 12v battery, they will both run at 12,000rpm. But this is without any load, ie without prop.

If you run both motors with the same prop at the same rpm (verified by a tachometer), the thrust will be the same.
But as you increase the throttle to full power, you will find that the smaller motor will not be able to match the rpm of the bigger motor with the same prop.
Result: The maximum thrust from the bigger motor will be more.
(Remember that the power consumed will also differ)

Edit: 'Are questions allowed?' was answered in the affirmative via pm yesterday

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.