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« on: September 12, 2017, 04:46:36 PM »
sanjayrai55
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A friend asked this in a PM, and I choose to reply here, hoping it might be useful to others

"I have a doubt, please spare some time to answer the following:

1) 1000KV motor, 1045 propeller
2) 2200KV motor, 6045 propeller

Rest everything is same on a plane.

Which motor will make plane go faster? & Why?"

Many people do get confused here. I too, when I started out, had the same doubt.

Look at 1045. This designation (normally) is for a SF Prop, 10"diameter, and 4.5"pitch

The 10 inches will determine the thrust. Thrust is a factor of torque & RPM, and is defined as: Thrust is the force which moves an aircraft through the air. Thrust is used to overcome the drag of an airplane, and to overcome the weight of a rocket. Thrust is generated by the engines of the aircraft through some kind of propulsion system.

4.5" is the Pitch. Pitch: Is the displacement a propeller makes in a complete spin of 360 degrees. This means that if we have a propeller of 40 pitch it will advance 40 inches for every complete spin as long as this is made in a solid surface; in a liquid environment, the propeller will obviously slide with less displacement.

To illustrate

http://www.propellerpages.com/content/articles/images/tnl502.jpg
Props - Diameter and Pitch


So the speed of the aircraft directly depends on the pitch, and the Thrust on the Diameter, for the same RPM.

If the RPM increases, both these proportionately increase.

Let us consider the actual situation, using anyone of the popular thrust calculators

1045 prop 1000 KV @ 11 Volts
Estimated Static Thrust :    1671 grams
Supplied Power : 382Watts
Level Flight Speed :    90 Kmph


06045 Prop 2200 KV @ 11 V
Estimated Static Thrust :    1512 grams
Supplied Power : 535 Watts
Level Flight Speed : 200 Kmph

These figures are of course theoretical, under ideal conditions, but, they are fairly accurate in practice and very useful for comparison, thus also for prop selection! It is also enlightening to observe the power requirements in each case. The second case means a pretty large motor which can supply 535 Watts ++

So, the second case, the higher KV motor, will make the plane go faster

Incidentally, give some thought to the type of model too. If it's a trainer type model, go for the first. If it's a jet, the second. But always remember, your MODEL IS NOT ALWAYS FLYING AT FULL Throttle. It should have adequate thrust and fly above stall speed even at eg 40% throttle
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« Reply #1 on: September 13, 2017, 12:45:33 PM »
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Thank you so much for the info.  Great help.
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« Reply #2 on: September 13, 2017, 07:11:37 PM »
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I'm glad you found it useful

Hope it was useful to some others too
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« Reply #3 on: September 15, 2017, 01:01:38 PM »
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Yes Sir, it surely was.

can you please suggest a thrust calculator.

regds
Saurabh
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« Reply #4 on: September 15, 2017, 04:30:35 PM »
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This one is OK

http://rcplanes.000webhostapp.com/calc_thrust.htm

For Android, Rc Tools

For all of these, thrust depends on prop size, rpm. RPM should be measured by a tachometer

Most of us don't have a tacho!  Huh? Huh?

Use KV X Voltage (nominal) and calculate. Then cross check on Power consumed in watts

eg say I try a 9*6 prop at 1200 KV

I would need a 9*6 at 13920 rpm (11.6 X 1200)
Which yields: 2.4 Kg thrust, needing 680 W Power

Now, check how much is your motor capable of

eg Turnigy Aerodrive SK 3542-1185 Kv  https://hobbyking.com/en_us/turnigy-aerodrive-sk3-3542-1185kv-brushless-outrunner-motor.html

This can take max 49 A
So 49 X 11.6 = 568 Watts
Now that prop needs 680Watt to get the desired RPM and thrust

What will happen if I use it anyway?

The RPM will drop! The RPM will drop to the point which the motor can handle. Else, the motor will burn out!

PLEASE NOTE: The wattage in the thrust calculation is wattage available to Prop ie Motor Output, and the wattage drawn by the motor ie Motor Input, will always be higher because the motor efficiency is never 100%. A more realistic efficiency is 80%.

Further, Thrust so derived will be STATIC THRUST, ie with the model stationary. In the air, the thrust will be higher by anywhere between 10% to 20% depending on the type of model.

If one plays along with a thrust calculator, this selection will soon become nearly instinctive

eg I have a Sport Model weighing 1.6 Kg. I would (without calculation ) say I can use a 10*7 prop on a 3542-1200 KV motor with a 60 A ESC. Can someone tell me if it will work satisfactorily, or not. This is a Sport Model, not a Trainer, and I wish to be able to climb vertical and accelerate out of a loop or roll
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« Reply #5 on: September 16, 2017, 02:39:04 PM »
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I am trying to figure it out, can you explain the same stuff in simple language sir?

still here's what I think:
bigger prop means RPM will decrease,
higher pitch means A consumption will increase.

but what about thrust?

It is given that the model is 1.6KG so we need more than 1:1 thrust to weight ratio to be able to go vertical.

Thank you.
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« Reply #6 on: September 16, 2017, 05:02:00 PM »
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Bigger prop and higher pitch means bigger load on the power supply.

The power supply could be a glow or gas engine, an electric motor or even a wound up rubberband.

Even for the same pitch and prop dia you could have different loads depending on the number of blades in the prop.

eg., a 3 blade  10 X 7 prop will load the power source more than a 2 blade 10 X 7.

Most use 2 blade props, so let us for the moment look at that scenario for your electric motor.

Any increase in dia or pitch will demand more amps from your battery as the rpm of your motor is determined by its kV and the battery voltage.

You will need to size your prop based on desired thrust, current capacity of your ESC etc. Of course you will have to take care of other factors too - - eg a too large prop dia getting poor ground clearance.

There are multiple factors at play and  a thrustmeter and wattmeter are wonderful tools to have.

However, the simplest way is to arrive at a ballpark prop size. Then carry a few props of higher and lower dia as well as pitch close to your rough estimate, and try them out at the field. You will be amazed at how quickly you will hit the 'sweet spot'.

Cheers and happy landings !
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« Reply #7 on: September 16, 2017, 05:18:59 PM »
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Firstly you are correct that thrust needs to be more than 1.6 Kg. Point is, how much?


""
bigger prop means RPM will decrease,
higher pitch means A consumption will increase.

but what about thrust?
""
Not a correct reasoning although sometimes the answer may work out correct

Thrust is actually dictated by diameter (bigger prop)
Power consumption is a factor of Prop Diameter, RPM, and to some extent, Pitch
Power is literally Thrust * RPM (output of motor)
Power is also  V * A  (Input to motor)

Theoretically speaking, use a bigger diameter prop and you get more thrust
Use a higher pitch and get more speed
Use a higher RPM and you get more thrust and speed

But, in real life, as you do these, the Power Demanded of the motor will keep increasing! Can the motor keep giving this power? Obviously there is a limit

Due to the characteristics of the electric motor, the capacity is normally specified in Amps or Watts. If in Amps, the S rating of the battery is also specified (eg 3S = 11.1 V)
If you exceed that you will burn or otherwise damage the motor (Most ESCs have a safety cut-off if A exceeds a certain limit, some expensive motors too)
So you cannot keep increasing prop size or RPM without staying within the motor capacity

Battery -> Motor = Input Power
Motor -> Prop = Output Power
Motor has efficiency, which is never 100%; more likely 80%

Motor Input Power = Volts * amps

Now what you try is take any prop eg 9*6, and see the thrust calculator results (thrust and power, speed) at 2000, 5000 and 10,000 RPM
Next, change the prop to 9*7 and compare the results
Next change the prop to 10*7 and see the results

You will find this exercise enlightening

I'm sorry if the explanation is not as simple as one would wish. I graduated from IIT Bombay in 1979 and am a bit rusty!

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« Reply #8 on: September 18, 2017, 03:10:38 PM »
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So here's what I found:

Lets assume that V is 11.6 which is nominal voltage.

Motor KV 1200 so we get = 13920 RPM

Prop          RPM           Thrust          Watts          Observation
9*6           13920        2397            680             Control
9*7           13920        2397            793.4          Motor pulls more current, gives same amount of thrust. (but what will happen RPM?)
10*7         12528        2928            879.7          I considered reduction in RPM by 10% because of big prop, more thrust, more A consumed

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« Reply #9 on: September 18, 2017, 04:39:36 PM »
sanjayrai55
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Great, Swapnil

Actually I didn't want you to reduce RPM yet! I wanted you to observe and compare the Power drawn, and compare it with what the motor is capable of

If you do that, you will observe that the motor capacity is 568 watts, and the motor is therefore incapable of turning any of these props at the RPM indicated.

So, what will happen if you use these props anyway?

You will overstrain your motor
The Voltage across the motor will drop; the internal resistance being the same the current will increase
Voltage drop => RPM drop, as the KV is a fixed characteristic of the motor
Current Increase => motor temperature increases I2R , your motor is at risk of burn out, your ESC and LIPO may be stressed if they lack the capacity to handle the increased current. Also, the hotter the motor gets, the higher it's Resistance, leading to a progressively worsening situation
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« Reply #10 on: September 18, 2017, 04:42:55 PM »
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Further, you would have noticed as you increase the diameter of the Prop, the thrust increases, increasing the Pitch did not have this effect. If you had noted the speed in each case you would see the ground speed of the aircraft keeps increasing as you increase pitch


Which brings us to Prop/Motor Selection for a particular Aircraft
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« Reply #11 on: September 18, 2017, 04:58:23 PM »
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Yes the concept is clear now.

The same diameter prop with different pitch will give you difference in speed not in thrust.

Also I thought I have to assume that the above mentioned motor will handle those props and do the calculations.  In reality it will not work with those props.

Thank you so much Sir.

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« Reply #12 on: September 18, 2017, 06:02:02 PM »
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 Thumbs Up

Now try an exercise

Two models

Model 1: High winger trainer, AUW 1.1 Kg with 2200 3S LiPo
Model 2 : Low winger sport,  AUW 850 gm with 2200 3S LiPo

Question:
1. Select a motor prop combination for each
2. What performance and flying time would you expect?
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« Reply #13 on: September 19, 2017, 11:40:00 AM »
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Here's my submission:

Model 1: High winger trainer, AUW 1.1 Kg with 2200 3S LiPo
Model 2 : Low winger sport,  AUW 850 gm with 2200 3S LiPo

So the expected figures are:

Model 1: W:T >= 0.75 (I am not considering speed factor here)
Model 2: W:T >= 1.5 (which I feel will be better, also it should be faster)


Question:
1. Select a motor prop combination for each
Ans:  Model 1: 2212 1000KV motor with 1045 prop which gives me about 800-900 gram thrust @ 12A max i.e. T:W 0.72 - 0.81
        Model 2 : Sunnysky X 2216 1250KV with 1047 prop which gives me about 1350 grams of thrust @ 26.5A T:W 1.58 giving me ground
                      speed of about 100kmph

2. What performance and flying time would you expect?
Ans:  Model 1: considering I am using about 40% of throttle @ 4-5A consumption I must get at least 33 - 26.4 mins. If we go by rule of
                      draining upto max 80% battery the flight time should be around 26.4 - 21.2 Min.
         Model 2: considering I am using about 60-70% of throttle @ 10-15A consumption I must get at least 13.2 - 8.8 mins. If we go by rule of
                      draining upto max 80% battery the flight time should be around 10.56 - 7.04 Min.

Kindly do correct me.

Swapnil N.
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« Reply #14 on: September 19, 2017, 07:35:25 PM »
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Congratulations Swapnil! You got everything right!

There are of course various permutations & combinations of motor and prop

BTW, both of these are real models, of which you would be hearing more about in November!

http://www.rcindia.org/electric-planes/pilot-(ok-models)-junior-100-adapted-to-corro-electric/

http://www.rcindia.org/electric-planes/qb-10-in-corro-electric-for-2016-sweepstakes/

I personally used a 2836-1150 on both, with a 9*6 e-prop. Prefer the extra "punch"
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« Reply #15 on: September 19, 2017, 08:31:08 PM »
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@sanjayrai55,
Nice tutorial sirji!
Regards.
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« Reply #16 on: September 19, 2017, 10:04:23 PM »
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Congratulations Swapnil! You got everything right!

There are of course various permutations & combinations of motor and prop

BTW, both of these are real models, of which you would be hearing more about in November!

http://www.rcindia.org/electric-planes/pilot-(ok-models)-junior-100-adapted-to-corro-electric/

http://www.rcindia.org/electric-planes/qb-10-in-corro-electric-for-2016-sweepstakes/

I personally used a 2836-1150 on both, with a 9*6 e-prop. Prefer the extra "punch"

Thank you so much!  Would love to see them in Air.

Nice to see you back Iyer Sir.
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« Reply #17 on: September 20, 2017, 05:57:13 AM »
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@sanjayrai55,
Nice tutorial sirji!
Regards.

Iyer sir! Since you are not being active, I have to fill the void Wink Wink
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« Reply #18 on: September 20, 2017, 09:47:39 AM »
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I have another question.  Instead of starting new thread I thought, I will ask it here.

Planes are tuned by banking by deflecting ailerons in opposite direction.  In doing so one aileron moves down and other one moves up.  The one which moves down make more lift but the other one doesnt, also we lose little altitute and to compensate that we need to give a little bit up elevator.

Although this technique gives you more agility I feel that this is inefficient, as the aileron moving up is just creating mess.  So if we could program the radio to deflect only one aileron i.e. only in down direction will it be more efficient? and will it let plane hold the altitude?

So what I mean to say is that, if I want to turn right, I will deflect left aileron i.e. in down direction. But the right one stays in its place. (although creating more lift by deflecting left aileron down will cause more drag and plane might try to yaw left)

Kindly give it a try if resources and time permits.  I am curious to know what will happen.  Also what do you think will happen?

Swapnil N.
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« Reply #19 on: September 20, 2017, 09:56:45 AM »
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You are coming close to the concept of differential ailerons, which many of us use

Down movement is usually kept at 50% of Up movement. This reduces the chance of tip stall
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« Reply #20 on: September 20, 2017, 03:23:14 PM »
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You are coming close to the concept of differential ailerons, which many of us use

Down movement is usually kept at 50% of Up movement. This reduces the chance of tip stall

Ok. Thank you for sharing.

So the only aileron which is going down moves 50% w.r.t. the aileron moving up right?
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« Reply #21 on: September 20, 2017, 05:52:45 PM »
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Yes
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« Reply #22 on: September 20, 2017, 08:29:14 PM »
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@sanjayrai55,
Nice tutorial sirji!
Regards.

Iyer sir! Since you are not being active, I have to fill the void Wink Wink

Thanks for giving me an entry into this thread  Grin
So here I go...

@swapnilnimbalkar,
The first thing is that the ailerons do not make the plane turn. They make it roll.
Think of giving aileron input for a 360 degree roll. The direction of flight will not change, ie, no turn, in an axial roll. (Ignoring the barrelling effect for simplicity)

When you give left stick, the right aiieron goes down. This increases the angle of attack and hence the coeff of lift and the lift of the right wing. The reverse happens on the left aileron/wing. So the plane rolls left.
When the CofL of the right wing increases, so does its CofD, and its drag.
And the drag of the left wing decreases.
This results in adverse yaw, ie, right yaw on left aileron input.
As Sanjay sir said, this is corrected by having more up than down, ie aileron differential (say 50-75%)

If not ailerons, then what makes the plane turn?
Next post will explain.
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« Reply #23 on: September 20, 2017, 09:05:41 PM »
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So what makes it turn?
In level flight, there is a lift vector equal to the weight, pointing straight up.
When the ailerons are used, it starts to roll. When the desired degree of bank (say 30 deg) is reached, further rolling has to be stopped by neutralising the ailerons (maybe a bit of opposite aileron to overcome the rolling inertia)
Now the lift vector is tilted. It has a vertical component (LV), and a horizontal component (LH).
If LL is the lift in level flight (equal to weight), LV=LL x cos(30) and LH=LL x sin(30).

It is the horizontal component of lift that makes the plane change direction, ie, turn!

In the process, the vertical component LV has become less than LL, the lift required to maintain altitude.
This leads to a loss of height in a turn.
We usually compensate by pulling a little up elevator to increase the angle of attack, and hence the CofL and the lift force, without noticing that this results in a reduction in airspeed!

Imagine that you are flying north and want to turn south using a 90 degree bank. You can see that turn is entirely based on the horizontal component of the wing lift, and not on the ailerons at all!
Watch your altitude and airspeed when you try this!
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« Reply #24 on: September 20, 2017, 09:11:09 PM »
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Posts: 233
Join Date: Nov, 2016



That explains a lot.

Thank you so much for taking time and posting this info. 

I hope it will be helpful to others too.

 Hats Off
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