Scaling power?

[Saqa]

The other item that needs to be addressed is model speed vs big boat speed

Found this discussion
Scale speed

I take it to mean that with my 1:10 ratio, 3 knots with the model will mean 30 knots at full size. Am I reading this correctly?

 

[Ad Hoc]

.... For a planing type hull the proper scaling factor for speed I do not know....

Doesn't matter. Scale is scale, no matter the hull type.
However, planning boats tend to be scaled based on various ways/methods .. each to their own, depending upon what aspect one appears to be interested in the most.
Such as on volume, or beam etc... but the method scaling is constant.

 

[Saqa]

Found this discussion
Scale speed

I take it to mean that with my 1:10 ratio, 3 knots with the model will mean 30 knots at full size. Am I reading this correctly?

Anyone?

 

[mc_rash]

L(realsize)/L(model) = scalefactor (lambda) for geometrical dimensions,
surfaces are scaled by lambda^2
volumes (and displacements) are scaled by lambda^3

you'll get the modelspeed Vm for a given realsizespeed Vs by making use of Froudenumber which is actually a model law and a way to compare boats of different lengths

Fn = v/((g*L)^1/2)
with v = speed, g = 9,81m/s² en L = Length

since model and realsize have the same Froudenumber you can say:
Vm/((g*Lm)^1/2) = Vs/((g*Ls)^1/2)

which you can work out to
Vm = Vs*((g*Lm)^1/2)/((g*Ls)^1/2)

now you can determine the right speed on modelscale for a given realsize speed

Edit:
I would use:
speed in m/s
L in m

 

[mc_rash]

You could work the formula out to:
Vm = Vs*((Lm/Ls)^1/2)

 

[TANSL]

Post #19 means knowing what you are talking about. This is where the tests in towing tanks are based.

 

[fredrosse]

"I take it to mean that with my 1:10 ratio, 3 knots with the model will mean 30 knots at full size. Am I reading this correctly?
Anyone?"

That is incorrect, the proper answer for displacement type hulls is that the big boat and the model are equivalent only if they have the same "speed-length ratio" .

For displacement type hulls this is NOT a linear relationship, but rather it is based on the square root of waterline length. For your big boat to model parameters, the proper ratio is scale ratio is squareroot of (1/10) which is 0.316. Thus if you are going to simulate a big boat speed of 10 knots, then the 10:1 scale model hull needs to be driven at 0.316 * 10 = 3.16 knots.

As it turns out for your particular parameters the scale model speed to represent the big boat speed of 30 knots would be having the scale model boat traveling at 0.316 x 30 knots = 9.5 knots for the model.

 

[tonipu]

please can somebody help

i made two scale model testing 1:10 and 1:20 and power requirement mech more formula (for scale 1:10)
ehp= hp / (10x10x10)

than
EHP = hp x (1/10)^3.5 =

 

[tonipu]

the boat is in planing mode

 

[tonipu]

froude number 1.8

 

[TANSL]

please can somebody help

i made two scale model testing 1:10 and 1:20 and power requirement mech more formula (for scale 1:10)
ehp= hp / (10x10x10)

than
EHP = hp x (1/10)^3.5 =

So EHP = ehp/10^0.5

 

[tonipu]

so original power requirement is 540kw

i do 2 test 1:20 and 1:10 rc model

1.test
so my power match 1:20 540kw20:20:20)=0.0675kw = 65 w thats whta i mesure more or less and dont match

P_model = P_real/(20^(7/2)). The power on model scale is now = 530 / (20^(7/2)) = 0,0148 kW and dont match 14W

2.test
my power match 1:10 540kw10:10:10)=0.540kw=540w thats what i mesure more or less and dont match
P_model = P_real/(20^(7/2)). The power on model scale is now = 530 / (10^(7/2)) = 0,170 kW and dont match 170W

but i will do more precise power measuruing
today my first 1:10 test

 

[tonipu]

maybe i need find a dc motor effiecency curve, to see a shaft power, as i expect 90% efficency but in high load dc motor 755 are less efficient

 

[Ad Hoc]

please can somebody help..

If you can post the data clearly and consistently for each model that you have or would like.
What you have noted above, to me, is clear as mud...just numbers trying to get an answer...