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[boatmadman]

I have one of those big buhlers in my drifter, running on 12v it measured 0.2a off load, and 0.5a in the water with a 60mm prop, so looks like your guesses are about right, Andy.

Ian

[Telstar]

Hi
I had a quick look, maybe this info could be ueful

http://www.buehlermotor.org.uk/cgi-bin/sr.exe/standardmotorstartus
cheers Tom

[Bunkerbarge]

An interesting point to remember here is the old school days LAV (Length-Area-Volume) relationship.

If you take a 1" cube with a length of side of 1", a side area of 1"x1", or 1 square inch and a volume of 1"x1"x1" or 1 cubic inch and double the dimensions you end up with a side of 2", an area of 2"x2" or 4 square inches and a volume of 2"x2"x2" or 8 cubic inches.  This thinking follows on forever upwards of course.

Now relate this to a model of a statue of 1ft high with an outstretched arm, with say a contact area for the arm of 2 square inches and a volume of the arm of 5 cubic inches.  If you now make this into an exact scale replica of 10ft high the area of contact of the arm will have increased by 200 square inches but the volume of the arm has increased to 5000 cubic inches.  Consequently you can make the replica bigger and bigger until such point as the area of the arm becomes unable to support the weight of the arm and it falls of.

As above, if you talk about scales of models you are simply relating linear measurements but the volumetric differrence is according to the cube of the scale.  This means that if a 1/10th scale model has a mass of 10 lbs you would expect the mass of the real vessel to be in the region of 1000 lbs.  The power required to propel the two is therefore also in the same ratio so you have to apply the same thinking to the 1/72nd model.  You would expect the mass of the real vessel to be in the region of 373248 times the mass of the model.

I just thought that might help with some of thinking here.

[Telstar]

Hi Bunkerbarge
You're on the right track. There is at least one rather tedious book on the subject of scale modelling (mainly aimed at structural engineers and architects) that deals in convoluted detail with the subject
This still doesn't ensure they get it right (eg the millennium footbridge in London).
At least with our models its unlikely we'll kill someone

cheers

[malcolmfrary]

Now you're talking about professionals (i.e. people trained to have a set of good excuses for keeping your money when it goes pear shaped). 
Remember the saying that the Ark was built by an amateur, the Titanic was built by professionals?
The only noticable scale effect of water is due to surface tension - with smaller scale models you just dont get either a creamy bow wave or wake - the bubbles and drops of water are not scaled, but full size.

[Telstar]

Hi Malcom
(We are getting further from the original topic  ) Hope the moderator doesn't blue line us
On the subject of bow wave when I first started model boats I knew a then old modeller who had put small brass pins/tacks just about the waterline on the bow of his tug ( a bit like chin stubble) to break up the water and give a more realistic bow wave (as he called it a dogs bone)

cheers

[dreadnought72]

Hi Roy
I seem to remember reading that full size tankers could take several miles to stop from normal running speed

 Andy Thanks for the info. based on the 22,500HP and the scale factor of the model only (got to keep it simple ) I guesstimate the models power should be about 45W.
 The basis for this is no secret when dealing with displacement and power use the scale factor cubed
Cheers Tom

Hmmm.

I've taken a different tack to work out the power. Back to basics...

I've got 4*35mm diameter props, with a total surface area of 0.00385 sq m.
The props have a 45 degree pitch, so the advance per revolution is 0.11m.
Call it 0.055m with slippage, since a quoted 50% efficiency for props seems quite common.

Therefore, at 2000rpm, the water should be kicked out the back at 1.83 m/s.
This is above the scale speed of 1.3 m/s, which is essential, else it would never reach that velocity.

So, every second, 0.00385*1.83 = 0.007 cubic metres of water (7kg) is accelerated from zero to 1.83 m/s.
This means that the force required, in Newtons, is 7*1.83 = 12.8.

This is a force just under that of a 3lb weight resting in your hand.
If I were to pull the hull at the scale speed, that's the sort of drag I'd expect to feel...I think?

Back to The Force... the product of 12.8N and 1.83m/s gives me a value of 22W.

This is about half of that of Telstar's guesstimate - so are inefficiencies and shaft losses really 50% of the deal, even before slippage is taken in to account?

I think what I need to do here is check a motor running one of my props in water, with a greased shaft - both for current draw and rpm. (The hull is not ready for the water yet.) And if it's falling short of at least 6-8W and 2000rpm, then consider upping the voltage. A little.

Thanks for all the thought-provoking suggestions.

Andy

[Colin Bishop]

Small electric motors such as the ones we use are actually not very efficient. Of the top of my head I seem to remember that 70% would be considered very good.