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Feb 01, 2008, 07:41 PM
Texas Buzzard
Texas Buzzard's Avatar
Originally Posted by jrb
Missed this in my copy & paste; from one of my previous postings:

"Many folks remember F=MA; but that is not what Newton actually derived.

Newton actually stated that a force is equal to the time rate of change of momentum (mass times velocity) = d/dt mass * velocity – this difference is very important with respect to propulsion.

In fundamental units, force is mass * length divided by time squared = M*L/T/T.

Velocity = L/T

And, then Power = v * T = L/T * M*L/T/T = M*L*L/T/T/T!

Just as shown below:

So force divided by power leaves you with a parameter that has the fundamental units of T/L. = Velocity!

Efficiency, like Reynolds is a parameter w/o units; its makes no sense if units are when T/W!"
Thanks, All very true jrb. Can youi help with actual calculations in solving this little problem OR are you trying to say that the problem can't be solved using only watts / watts out?
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Feb 01, 2008, 08:02 PM
Texas Buzzard
Texas Buzzard's Avatar

Charging batteries

Looks like a beautiful day tomorrow. Maybe the people who frequent this thread are tired of a bit of the real-world approach.

Some are throwing in factors that we can't measure. How can we measure thrust while the plane is flying.

My friend from GERMANY should be specific since he says his H.S. teacher would punish him for what I wrote. BE SPECIFIC in critique please.

This started out as an honest question, no one would repond with calculations so I finally did as a courtesy to all interested parties.

Vampire67's post reminds me of the day I asked a question in a Seminar Class at the Natl Labratory at White Sands, NM. There were several PhD there in the class who I thought were competent. In answering my question the prof used the chalk board for a time, then turned aroud and said, "There are 9 physistists her and I think each of the 9 would have a slightly different comment".

I say that to emphasise that your training might be different fom others thus, your explanation would be different but still would arrive at a common answer.

The German commented on the use of Imperial or English Units and that the CGS and the MKS was superior. DID HE NOTICE that I converted to MKS as soon as I could? If he had attended major universities in the U.S. he would know that in Physics we haven't used the English System ( slugs,feet & seconds) since WW II. But, the engineers still use what's called the Engineering System which actually the English System. Of course MKS os more convenient to use.
Feb 01, 2008, 11:44 PM
Registered User
T B -
You started your first post of this two-part thread with a deception; you implied ignorance.

You asked for the opinions of others and you got their opinions, cased closed? no.

It is not a great feat to mathmatics to follow energy input from the battery, through the ESC, motor, and fan and finally as thrust out the tailpipe. But, it is fruitless to do so, when a bathroom scale is available.

As a number of participants in this thread have accurately noted; (and I paraphrase) static thrust means diddily-squat, the proof of this pudding is in flight.

So, was there a real question in there somewhere, or was it just a demonstration of mathmatical agility?

Feb 02, 2008, 12:17 AM
The blade numbers go up to 11
stumax's Avatar
Originally Posted by Texas Buzzard

No Kong, you were NOT confusing at all. You would make a good H.S. teacher. Why? You talk on a level that is understandable to the mentally handicaped like me - HA HA HA just joking. But your post on effeciency of two fans w/ the same motor and same watts is good for me!
.................................................. ........
When anyone solves a problem is is common courtesy to show your calculation for two reasons.
1. If a novice is reading then he might glean more by seeing the progression.
2. If a reader disagrees with what is written then that reader can be specific in his critique, therefore I wrote as if I was being informative & courteous.

Watts is easy, but to measure thrust isn't. Back in my glow days I had a dial type spring balance ( a thing to measue the mass of an object ) since it was calibrated in kg. Kg is a mass unit. 1.0kg = 1000 gms. Max. scaling was 10 kg or since 1.0 kg = 2.2 lbs it went up to 22.0 lbs.

1.So first I have to find it, hang it from ceiling and hang the fan/motor unit with electronics facing so the thrust is aimed downward - right? The difference in the reading before fan is turned on AND after it is turned on will give the actual measured thrust.
2. Since effeciency is nondeminsional (as you said) Watts IN / Watts out will give a fraction lessss tham unity ( 1.00 ) - right? So multiply the quotent by 100 to get percent effeciency....don't you agree?
3. But there is a small problem, how to convert thrust into Watts OR Watts into Force units since Thrust is measured in Force units - right?
4. We can say that the fan is pumping air out of the tailpipe. Essencially this fan is doing WORK on the air. The fan is applying a Force on the air causing the air to be moved through a distance. The rate (speed) at which the air is moving is it's velocity since we know the direction the air is moving. O.K.,(I am thinking out loud now!) Since Work is F=ma or force x distance and Power is the rate of doing the Work or in English Syst: pounds feet per sec (ft lb/sec). We are getting closer to the conversion now.
5. Volts x Amps = Watts - right? And Force x Distance = Work - O.K.?
6. We have to do something with the thrust we measured to change it to Watts so we can divide Watts IN by Watts OUT.

7. When we use a spring balance to measure the thrust we probably are using an English balance calibrated in ounces. Lets say the measured thrust was 10.0 ounces. Since one ounce is equavalent to ABOUT 29 gms of mass we can say that 10 oz x 29 gms/ounce = 290 grams - but that is Not a force unit! 290 gms x the acceleration of gravity (980 cm/sec^2) gives us the conversion of Ounces into Metric Units or 290 gms x 980 cm/sec^2. This gives us 284,200 dynes. The dyne is a unit of Force in the MKS system.. 1.0 gm resting on your table pushes down with a force of 980 dynes.
So our thrust being converted from ounces to dynes . That is a big number. Lets do better. 10 oz x 29 gms/oz = 290 gms thrust. Can we change it to JOULES which is more convenient? Yes I think so. A Joule is a unit of work JUST LIKE WATTS IS A UNIT OF WORK.
1 Joule = 1.0 newton meter ( The force of 1 newton moving thru' one meter. Like ft. pounds is 1 pound of force moved thru' 1 ft /sec- this is 1 ft lb/sec ... 550 ft lbs/sec = 1 H.P.
Going back to the 290 gms of thrust... it happens to be = to 0.290 Kg! When using Kgs we use the MKS system or 9.8 m/sec^2. So multiply acceleration of gravity 9.8m/sec^2 x 0.290 kg , it's = 2.842 kg m/ sec^2
This is a force unit. POWER EQUALS FORCE x DISTANCE/TIME OR THE RATE OF DOING WORK. So if you push a trunk across the floor with a force of 550 pounds and move it 1.0 foot in 1.0 sec then you exhibilted 1.0 H.P. In metric 1.0 Joule of Work done per second = 1.0 Watt. So think that a WATT IS EQUAVALENT TO H.P. or just Power.

1 joule = 1 newton meter .... a newton is a force unit having the unit kg m/sec^2. If some of you are engineers you may want to use the engineering system and stick to pounds, slugs and feet etc. I don't
want to do that.

** Work being done by the fan = 2.84 kg m/ sec^2

Since 1 newton meter = 1 Joule we are in business.

1 Joule/sec = 1 watt That is: work done per sec = power

So the 2.84 Joules/sec has to = 2.84 watts being done per sec on the air by the fan unit.

If watts IN was 100 watts and watts OUT was 2.84 watts

100 watts/2.84 watts = 35.2 or 35.2% effecient

.................................................. .........................

Everyone except two have said that watts IN/watts OUT is the way to go. It does simplify it for us and for all practical purposes the slight differences in Temp, Hunidity and air Pressure doesn't make a significant difference. The NASA engineer at Langley in the wind tunnel needs more than we can give him with our spring balance and watt meter. Nuff said.

PLEASE - ANYONE, if you find an error make me aware of it by referencing the line in which the error appeared. 35.2% seems to be a low effeciency but after all it was a hypothetical question.

8. Remember the thrust was 10 ounces. .[U] We just measured the thrust didn't we? The amt. of work done by the fan on the air was ~ 2.84 Joules ( force x distance OR newton meters) What about the time? It is taken care of since when we measured WATTS IN we have the Amt. of Work being done per second by the battery. Ha Ha , we just converted OUNCES TO WORK BEING DONE ON THE AIR! iT WAS DONE IN JOULES OF WORK ! A WATT IS A UNIT OF WORK/sec or what we call POWER. Looks like we are just about there![/U]

By definition the Joule is a unit of work done/sec .

1 H.P. = 746 Joules = 746 Joules /sec
again - - - -- 1 joule/sec = 1 watt.

9. Watts divided by Watts is unitless! You CAN COMPARE TWO DIFFERENT FAN UNITS WHEN YOU KNOW WATTS IN and THRUST produced .

OHG! There's so much totally wrong in there! I don't have time to teach highschool physics - I'm sure there's a wiki page for it - but I'll fix up a few fundamental errors which may lead people astray.

1. work is NOT ma, force = ma or mass x acceleration
2. force is force, work is force x distance, don't confuse them
3. the Joule is a unit measure of energy or work, not a measure of force, you cannot change thrust into Joules!!!!!
4.Watts IS NOT A UNIT OF WORK!!!!!
5. Here's the biggy: IF POWER IN WAS 100WATTS AND POWER OUT WAS 2.84WATTS, EFFICIENCY IS 2.84%!!!!! ie 2.84 / 100 . Unfortunately, this is the efficiency of some people's brains.

Fan efficiency is nothing more than the ratio of actual thrust divided by the theoretical thrust available with a given amount of shaft power. Note that I said shaft power, motor efficiency doesn't come into it. Unfortunately the subject is far too involved to get into here. If you really need to know, get some propulsion textbooks and study real hard.

Feb 02, 2008, 12:32 PM
jrb's Avatar

There are dozens of threads around that speak to the physics; note it works for either EDFs or Props!

Here’s a few that might be of interest:

Any unit of measure can converted from one system to another via a multiplication factor: feet to meters, pounds to grams, pounds to Newtons, etc.

So, in the end simply measure the thrust, and velocity to get Propulsion Output Power. Compare that to Watts In and you’ll have great measure of efficiency!

You can scale it via a multiplier for easy use for display, or if wanted to convert it to its actual magnitude.

Here’s the above cleaned up just a little:

In fundamental units, force is mass * length divided by time squared = M*L/T/T.

Velocity = L/T

Power is M*L*L/T/T/T

Velocity times Thrust as fundamental units = L/T * M*L/T/T;

Which = M*L*L/T/T/T -- Power!

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