Measuring pressure diffferences

This is where it gets tricky.

Force = Mass x Acceleration

Say I could measure the Force against a Piston supporting the outboard, and get say 1 Kilogram

If Force was 1 Kilogram, and the Speed was 2 Kilometer an hour, is that enough to calculate the Power ?

But - I have to use Newtons as the Force measurement in the previous formulae

eg.

Power = Force times speed

Say the force is 1 Newton, and the speed attained with that force is .5 meter/second. ( 2 klm/hr)


Then the power is what you get by multiplying the two: ,5 Newton meter per second = .5 Joule per second
= .5 Watt.

 

Power = Force times speed...indeed. Measure the force and speed and simply multiply the two. ??? What are you missing?

For example...if at 15 ft/sec velocity I'm measuring a thrust (or drag) force of 200 pounds...200 pounds times 15 ft/sec is 3000 ft-lb/sec, or 5.45 HP..or 4.07 kw..and so on.

Now of course that is the net power delivered to move the craft..not the total installed power required to deliver that net power.

 

Is it as simple as
"1 kg ( kilogram ) unit for a weight and mass measure equals = into 9.81 N ( newton earth ) "

so, in my formulae

"If Force was 1 Kilogram, and the Speed was 2 Kilometer an hour, is that enough to calculate the Power ?

But - I have to use Newtons as the Force measurement in the previous formulae

eg.

Power = Force times speed

So the Force is the force is 9.81 Newton, and the speed attained with that force is .5 meter/second. ( 2 klm/hr)


Then the power is what you get by multiplying the two:
.49 Newton meter per second
= ..49 Joule per second
= .49 Watt."

The mystery continues

 

So the Force is the force is 9.81 Newton, and the speed attained with that force is .5 meter/second. ( 2 klm/hr)


Then the power is what you get by multiplying the two:
.49 Newton meter per second
= ..49 Joule per second
= .49 Watt."

9.81 Newton * .5 meter/second = 4.9 Newton*meter/second (not .49)

Note this calculation is only valid for calculating power if the force and velocity are aligned. If they are not aligned then the appropriate cosine multiplier is needed.

 

Thanks David, I need a lot of checking on my math.

The vector calculation is a good point - I would have to do meaningful test runs on a straight course.

But overall - does the calculation sound correct. ?

My thoughts are that I need to calibrate some sort of Force meter using conventional weights ( which is using the gravitational acceleration), so using a Kilo Weight of Force could be pretty close to the 9.81 Newton as I see it.

Does that make sense ?

 

Pretty close?...as close as one can get.


1 watt = 1 newton*meter/second = 1 (kg/9.8)*meter/second ...and so on.

You simply need to convert as needed to maintain a consistent set of units.


In most of the tests we've done with outboard dynamometers, the correction for thrust angle and point of application between model setup and full-scale propulsion has been relatively small. But it is always taken in to account anyway. I think you misconstrued DCockey's comment; he was referring to the orientation and location of the thrust vectors in the vertical plane; nothing to do with steering/lateral vectors. It is always assumed that model tests conducted for measuring drag are performed in a straight line.

 

yes, a dynamometers on a real outboard is the best way to go, but on electric motors on small scale boats, the results would be meaningless.

It looks like I am going to have to manually calibrate an Analogue scale with weights, and observe the Force at different speeds.

Thats the next big challenge

 

?? I was simply referring to our typical setup..motor on a bracket reacted by a load cell. The term "dynamometer" covers a wide range of devices, including the simplest of thrust or load measurement setups.

 

Bingo. Exactly that. Otherwise known, since the 1700s as ...model testing.

 

I had the opportunity to visit the Swedish Naval Museum whilst working in Karlskrona some years ago. Some of the model testing innovations developed by Chapman, in the 1700s, are on display there...a very clever fellow he was.

 

For electric models we have been using the Eagletree V4 Datalogger for some time now, take a look;

http://www.eagletreesystems.com/index.php?route=product/product&product_id=54

 

Thanks for that Jim.
For $70, and only 100 grams, its a real contender on a scale model.

Its definitely less work than hassling with an Arduino systems, and having to 'roll your own'


The Pitot ( airspeed) addon could be a real bonus for the sailing component to detect wind speed.

Then, you could use it all again on the full size prototype for evaluation.

I wonder how you could record relative wind direction with this system ?





edit

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