Converting Pressure to Power in a Scale Model Boat

[Dnomyar]

I have built a model boat, ( http://schoolroad.weebly.com/rpgmodel_11.html )

and am trying to find out the amount of Power - maybe in Watts, that it takes to propel it at different speeds while it is running on the water. I have a rough sketch of the mechanism principle attached.
When the trimming arm is stretched by the motor propulsion, it is converted into a value on a manually calibrated set of marks.


My first thoughts are that if I can measure the pressure applied to the hull by the motor,
-say the Propeller is say, a square centimeter in area, and the pressure produced was .1 kilograms,

does equate to
.1 Kilograms per square centimeter,
that converts to a Force measurement of 0.980665 Newtons per square centimeter (http://www.allconversions.com/)

If I can apply that Force Unit x the Velocity of the boat , does that give me the Power required measurement ?

i.e.
Power = Force x Speed

eg Force is 1 Newton,
the speed attained with that force is 1 meter/second.

Then the power would be:
1 Newton meter per second
= 1 Joule per second
= 1 Watt.



Any and all all clarifications welcome, thanks

 

[Dnomyar]

Perhaps this Simplified Diagram is a better way to start.


The actual details of the mechanism can be considered later.

Basically , if an Outboard Motor creates a force of 1 Kilogram at 2 Kilometers per hour

can we calculate what Power it is producing in say Watts.

 

[Baluncore]

Water pressure on the hull is irrelevant.

When a boat hull moves through water it suffers a drag force due to water movement from in front of the hull to behind the hull. The wetted surface of the hull also causes drag. Total drag is usually proportional to the square of the water speed, but at some limiting speed the hull drag begins to rise very rapidly with water speed. The attitude of the hull will also change at higher speeds.

The speed of a boat is therefore determined by an equilibrium or balance between the total of the drag forces and the propulsion force of the motor – propeller combination.

 

[Dnomyar]

Yes, granted. But it doesn't solve the problem

How much Power (Watts) is the engine having to produce to attain the speed ?

is its a s simple as

"1 kg ( kilogram ) unit for a weight and mass measure equals = into 9.81 N ( Newton Earth ) "

so, in s formulae

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

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,
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."

 

[billy_joule]

Yes, granted. But it doesn't solve the problem

How much Power (Watts) is the engine having to produce to attain the speed ?

is its a s simple as

"1 kg ( kilogram ) unit for a weight and mass measure equals = into 9.81 N ( Newton Earth ) "

so, in s formulae

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

Yes, P=Fv (power = force * velocity) is one way to calculate power.

However, getting a value for the force in your situation is difficult for the reasons Baluncore mention.

You could drag the boat with a spring scale to get a force value but the boats attitude in the water will not be the same as it would if prop driven so will not be accurate. If the water is choppy so too will the force value.

Another point to consider is that the efficiency at which shaft power is converted to thrust force will, at a guess, vary with water speed. In other words propeller efficiency varies with speed. Your motors efficiency will also vary with speed. Ideally you want to cruise near your motors peak efficiency point and get a prop with good efficiency at that point.

Judging by your boat design you are probably interested in range? That's why you want to find power consumption?

 

[Dnomyar]

[Yes, P=Fv (power = force * velocity) is one way to calculate power.]

Any other ways you can suggest ?



Its the conversion of the Force component as Weight, being converted to the Newton figure that I am most uncertain about. ie. 1 Kilo of Force = .981 Newtons,

I have had other confirmation that this approach has some logic to it, so its looking gooder.

I don't see a problem of calculating the Force as a weight under experimental conditions, and the method of calculation doesn't really care *why* that amount of Force is required. Its a function of water resistance mostly, certainly wave and wind action, hull friction etc etc.

The important thing is to calculate that the amount of Force required for the designed speed is not excessive.

Say I built a hull with a flat boxlike bow. The amount of Engine Power to go 5 kph could be 5 times the amount if I put a pointy bow on it. Likewise, with a new hull design, will the amount of Force required to drive the boat be less if I modify the boat design slightly, or more.

Likewise, if I load the boat heavily, what does it do to the Force component with more immersed hull ?.

Naturally, the greater the Force needed for propulsion, the more fuel used, and shorter range. However, the real aim is to find the most efficient Hull shape to get target speed with lower Force.

The design is not a radical one, similar full size boats will require similar size of motors ( 50 to 70 Hp, 52 Kilowatts), but if small hull shape changes reduce the power requirements, that's great, and that's what I what to find out before I build the full size one.

 

[jim hardy]

Your basic physics is right,

work = force X distance
and power is rate of doing work, work/time
so power is force X distance/time = force X speed .
60hp is a big motor to me. What kind of boat are you building?Elco and Grebe yachts from 1920's used what they called "Canoe" hulls to achieve best speed with the modest powerplants available then.

http://www.boneyardboats.com/Archives/0030_Winter_2006/

Planing hulls are a lot different from displacement. Once it climbs up over its bow wave and rides on only the aft quarter of the hull , ordinary viscous drag is significant ... so you want to make lift to reduce wetted area by tricks like trapping air under the hull (as in hydroplanes) and deflecting downward the water that's moving sideways out from under the hull . Turned down chines and lifting strakes as on Richard Bertram's famous "Moppie" hulls are examples of the latter.

http://www.powerandmotoryacht.com/design/understanding-hull-design
http://www.navaldesign.co.za/articles/Planing Hulls 07 Oct 2006.pdfVery old "Rudder" magazines might interest you, as will Ralph Munro's "Commodore's Story" which popularized his "Sharpie" hull.

Interesting project.

http://www.huntdesigns.com/deepv-questions-answers.htm

see the lifting strakes ?
[URL=http://s232.photobucket.com/user/oldjimh/media/bravemoppie-lg_zps030b57a5.jpg.html][PLAIN]http://i232.photobucket.com/albums/ee289/oldjimh/bravemoppie-lg_zps030b57a5.jpg[/URL][/PLAIN]

bigger photo here:
http://forum.woodenboat.com/showthread.php?94771-Do-Wood-Cigarettes-Exist

old jim

 

[billy_joule]

[Yes, P=Fv (power = force * velocity) is one way to calculate power.]

Any other ways you can suggest ?

I would use a data logger to record voltage and current while driving the boat at various speeds/conditions and calculate power from P=IV.

Its the conversion of the Force component as Weight, being converted to the Newton figure that I am most uncertain about. ie. 1 Kilo of Force = .981 Newtons,

I have had other confirmation that this approach has some logic to it, so its looking gooder.

I don't see a problem of calculating the Force as a weight under experimental conditions, and the method of calculation doesn't really care *why* that amount of Force is required. Its a function of water resistance mostly, certainly wave and wind action, hull friction etc etc.

I don't understand why you are talking about mass (kg)? Force is measured in Newtons. The drag force acts horizontally, the force due to gravity acts vertically. The mass of the boat only indirectly affects The drag force and doesn't need to be considered when measuring drag force.

 

[Dnomyar]

"60hp is a big motor to me. What kind of boat are you building?"

Check out http://tattooyachts.com/under-power/


"I would use a data logger to record voltage and current while driving the boat at various speeds/conditions and calculate power from P=IV."

Unfortunately, the Power calculated that way includes losses due to friction, electric motor inefficiency etc - and not just the Power used to Drive the boat.


"I don't understand why you are talking about mass (kg)? Force is measured in Newtons. The drag force acts horizontally, the force due to gravity acts vertically.."

I have attached a diagram of how I plan to calibrate the 'measuring device'. Its simply standing the boat on its nose, and marking a 'meter' or similar as different weights are applied. These are gravitationally created Weights, that have a known equivalent as Forces.

When the equivalent Weight is achieved under Horizontal power, I know how much Force is required to achieve the speed. Its essentially a horizontal Weighing Machine

 

[jack action]

These tools might help you:

Boat speed calculator
http://www.cyberiad.net/michlet.htm
http://www.delftship.net/DELFTship/index.php/delftship/delftship-free

 

[Dnomyar]

Thanks for the links Jack. They probably won't be much help as I have had the hull designed with all those tools.

The big unknown is not the speeds attained, as I can measure those with the model. And even the Power required isn't totally unknown, just unproven.

What I am most interested in is the effect of minor hull changes on Power requirements, and to verify computer maximum power requirements for a range of speeds.

In effect, physically verifying the theoretical calculations already done.

 

[jack action]

Thanks for the links Jack. They probably won't be much help as I have had the hull designed with all those tools.

The big unknown is not the speeds attained, as I can measure those with the model. And even the Power required isn't totally unknown, just unproven.

What I am most interested in is the effect of minor hull changes on Power requirements, and to verify computer maximum power requirements for a range of speeds.

In effect, physically verifying the theoretical calculations already done.

I don't understand. I went quickly over the links, but the boat speed calculator gives the power needed according to speed in a nice graph and the other two say «calculates the total resistance» and «Perform basic hydrostatic and resistance calculations» which I thought meant evaluating the drag force of the hull. If that is the case, all you need is to multiply this drag force by the boat's velocity and you get the power needed. It would be that simple.

 

[Dnomyar]

I don't understand. I went quickly over the links, but the boat speed calculator gives the power needed according to speed in a nice graph and the other two say «calculates the total resistance» and «Perform basic hydrostatic and resistance calculations» which I thought meant evaluating the drag force of the hull. If that is the case, all you need is to multiply this drag force by the boat's velocity and you get the power needed. It would be that simple.

Yes indeed. The theoretical drag x velocity would give a good approximation of engine power required. I have subjected the model to a number of calculations that give me much more accurate predictions ( see attached) , but the difference between theory and practice can be considerable.

For example, at some speeds, I expect the hull to plane. The online power calculator only allows rough 'typing' of hulls, and is not sophisticated enough to allow for subtle design changes. How would it handle adding a slightly deeper bow stem with a slight bulb for example ?

I am hoping I can detect slight changes in performance while adding trim tabs, adding running strakes and the like. All of these are extremely difficult to evaluate with software.

 

[jack action]

I re-read the OP and now I understand better what you are trying to achieve.

If you measure the force acting on the boat by the motor (in N) and multiply it by the velocity of the boat (in m/s), it will indeed give you the power required by the boat (in W). Just to make sure we are on the same page, if you measure a pressure between the engine and the boat, then the force (in N) is the product of that pressure (in Pa) and the area (in m²) where that pressure is applied by the engine.

However, you will be missing the power required by the engine itself. I'm no expert in hydrodynamics, but I'm sure that if your propeller is closer to the water surface or is at a different angle, it will affect its own drag (which is added to the drag of the hull). I'm telling you this because you seem to look for some very small differences.

 

[Dnomyar]

Thanks Jack - you have indeed struck the crux of the question.

The relationship between the Pressure ( Pascals ) and Force (Newtons), as you can tell from the title of the thread, was what I thought I would need to establish for measurements. By taking the Pressure at the motor/boat boundaries and convert it into Power via the Newtons. Your mention of converting Pa to N was the first one I had come across, and I hadn't understood previously that it was straight calculation process. I even found a ready reckoner on the web to help me.

http://www.translatorscafe.com/cafe/units-converter/pressure/calculator/pascal-[pa]-to-Newton-per-square-millimeter-[n/mm%5E2]/

The actual measuring of Pressure accurately though, is hard at such small scale. The mechanics and/or electronics available are not very accurate.


As a result, however, I have tried to simplify the process a little. I am envisaging setting up a 'Horizontal Scale' arrangement, where I simply measure the Grams of Force ( calibrated previously from fixed weights) generated by the motor , and convert that to Newtons ( 1 Kilo Weight = .98 Newtons ), to simplify the process. I am hoping that a simple mechanical device will accurately measure the Weight being applied to the boat hull.

I have had a lot of thought about your comment on the effect of the Prop and Submerged part of the Motor on the degree the Power required to move the boat at different speeds. I can't think of a way of isolating the effect, so I will just have to hope that either the drag from the motor is similar to the drag from a real outboard, or that I can minimize the effect with making sure the prop is operating efficiently. In worst case, I will have to be satisfied with the degree of change, and not rely on it for the scaled up values :-(


I will do a sample sum for my own mental exercise, please check my logic
If I find, for example, that I can get 6 Kph out of say, 800 Grams of 'Weight' or Thrust from the motor

Weight in Grams 800
Equivalent in Kilos 0.8
(A) Equivalent in Newtons 0.7848 ( * .981)

Kilometers Per HOUR 6
= (B) Metres Per Second 0.066666667

= Newton Meter Per Second 0.05232 (A * B )
= Joule Per Second 0.05232
= Watts 0.05232


I might expect that the Full Size Boat ( Model is 1:5 Scale )
For speed of
13.4 Kph, (6*(5^0.5) (Ref 1)
I would require
6.54 Watts of Power ( 0.05232 * (5*5*5) )

I hope I have done the math properly.


______________________________________________________


Ref(1)
Scale speed is calculated :
vM= vO/((M)^0,5)
vO= speed of original
vM= speed of model
M= scale

 

[jack action]

First, I don't know what happened between post #4 and #15, but your conversion are wrong:

• 6 km/h is 1.6667 m/s;
• 0.8 kgf is 7.848 N.

There is a nice wikipedia page about conversion of units and you have this great online unit converter to help you if you need it.

Second, I did not realize that you are measuring on a scale model. I don't know where you found the equation written in your (ref 1) and how you calculated your power, but this has to be done based on what is called a dimensionless analysis. Again, I'm no expert in boat design, but I know that when you scale your model, not all quantities are necessarily scaled equally. You have to know what you are doing.

Here, you have a list of dimensionless quantities. The ones related to fluid dynamics or fluid mechanics may need to be considered in your conversion from one size to another (The Froude number is certainly one of them, but I'm sure that there are others as well).

 

[Dnomyar]

Sorry, it was a transposition error from my spreadsheet. Thanks for the checking.


Weight in Grams 800
Equivalent in Kilos 0.8
Equivalent in Newtons 0.7848

Kilometres Per HOUR 6
= Metres Per Second 1.666666667

=Newton Metre Per Second 1.308
= Joule Per Second 1.308
= Watts 1.308
= Full Size Equivalent 163.5000 Watts
The formula was provided by my engineer,There is no need for fluid mechanics or Froude Number ( though i can supply them too )

Once again, I am not calculating theoretical hull performance, I am measuring actual performance to verify the theory, and this is a straight scalar calculation.

 

[Baluncore]

There is no need for fluid mechanics or Froude Number ( though i can supply them too )
Once again, I am not calculating theoretical hull performance, I am measuring actual performance to verify the theory, and this is a straight scalar calculation.

If you measure the torque and RPM on the propeller shaft you can compute the power used to turn the propeller. Measuring also the thrust along the propeller shaft will give you thrust on the hull and propeller and hence propeller efficiency.

Once the boat is moving it will have a kinetic energy but that is not important, once it has accelerated to cruising speed all power will go into overcoming the drag of the hull and propeller inefficiency.

There is not simple relationship between power input and hull speed. It requires you to know things like fluid dynamics and the Froude number. If you measure shaft power or thrust against speed you will be able to plot the curve.

Basically , if an Outboard Motor creates a force of 1 Kilogram at 2 Kilometers per hour can we calculate what Power it is producing in say Watts.

No.
Just because the thrust gives a force equal to say w kg, does not allow you to compute the power.
A brick sitting on the ground exerts a force on the ground, but does not require energy input to remain there.

 

[jack action]

There is no need for fluid mechanics or Froude Number ( though i can supply them too )

Once again, I am not calculating theoretical hull performance, I am measuring actual performance to verify the theory, and this is a straight scalar calculation.

Your equation in (ref 1) is based on the Froude number. The equation assumes that the Froude numbers of your model and original are the same:

$$Fr_M = Fr_O$$
$$\frac{v_M}{\sqrt{gL_M}} = \frac{v_O}{\sqrt{gL_O}}$$
$$v_M = \frac{v_O}{\sqrt{L_O/L_M}}$$
$$v_M = \frac{v_O}{\sqrt{M}}$$

But if a design provided a lot of drag resistance (air or water) from fully submerged part of the boat or from something that protrude on top of the boat, the Reynolds and Euler Numbers might be relevant as well. It might not be relevant to your case, but I thought you should know:

Similarity of flows

In order for two flows to be similar they must have the same geometry, and have equal Reynolds numbers and Euler numbers. When comparing fluid behavior at corresponding points in a model and a full-scale flow, the following holds:

quantities marked with 'm' concern the flow around the model and the others the actual flow. This allows engineers to perform experiments with reduced models in water channels or wind tunnels, and correlate the data to the actual flows, saving on costs during experimentation and on lab time. Note that true dynamic similitude may require matching other dimensionless numbers as well, such as the Mach number used in compressible flows, or the Froude number that governs open-channel flows. Some flows involve more dimensionless parameters than can be practically satisfied with the available apparatus and fluids, so one is forced to decide which parameters are most important. For experimental flow modeling to be useful, it requires a fair amount of experience and judgement of the engineer.

 

[Dnomyar]

If you measure the torque and RPM on the propeller shaft you can compute the power used to turn the propeller. Measuring also the thrust along the propeller shaft will give you thrust on the hull and propeller and hence propeller efficiency.

yes indeed, but how would one do that ?

Once the boat is moving it will have a kinetic energy but that is not important, once it has accelerated to cruising speed all power will go into overcoming the drag of the hull and propeller inefficiency.

yes, i am understanding that.

There is not simple relationship between power input and hull speed.

Yes, hence the actual testing. All the mathematical models will not cover every factor.

No.
Just because the thrust gives a force equal to say w kg, does not allow you to compute the power.
A brick sitting on the ground exerts a force on the ground, but does not require energy input to remain there.

I can't follow that. The 'weight' I am detecting is the Force while the motor is Pushing against the hull.

If the propeller is not turning, there will be no pressure against the stern.

As the propeller speed increases, the Force measure as a precalibrated number of Grams will be registered.

As I picture it, this would be the equivalent of tying a string to the bow of the boat, and putting the other end on a pulley with say a 1 Kilo weight hanging on it.

The top speed (velocity) attained with this fixed weight would indicate how much Work ( or Power required ) was needed to achieve this velocity, ( Engine efficiency adjusted of course )

 

[Dnomyar]

Your equation in (ref 1) is based on the Froude number. The equation assumes that the Froude numbers of your model and original are the same:

The Froude Number is calculated by
Speed to be scaled to Fn
FL=v/((g*LWL)^0,5)
FL= Froude Number by length
v= speed in m/s
LWL= Length on waterline
g= free fall acceleration 9,812m/s2


It has nothing to do with the Scale Formulae that i can see

But if a design provided a lot of drag resistance (air or water) from fully submerged part of the boat or from something that protrude on top of the boat, the Reynolds and Euler Numbers might be relevant as well. It might not be relevant to your case, but I thought you should know:

That would be true if the Froude Number was part of the equation

Where is the Froude component in this ?

Scale speed is calculated :
vM= vO/((M)^0.5)
vO= speed of original
vM= speed of model
M= scale

 

[Baluncore]

If you hold the boat hull still in a channel that has an adjustable water flow, you will be able to measure the drag on the hull at different speeds. You will need a spring balance in the line that holds the boat. The measured drag will give the thrust needed from the propeller to maintain that hull at that speed.

The propeller is a complication in that it is highly water speed dependent. Adjust the power to the propeller until the spring balance reads zero and you will know the drag = thrust and power requirement for your hull to maintain that speed.

Once you have an equilibrium reached between propeller thrust and drag you no longer have a dynamic system, you have a static system. You have static forces but no energy expended in a static system.

 

[Dnomyar]

Yes, this is the popular tank testing.

Measuring the drag is with a spring balance, or a Load Cell with the water current providing the flow requires an expensive tank setup. Putting the measuring device on the boat means I can run it in any flat water pond or dam.

Also, measuring the 'Weight' while the boat is running , simulates the prop effect more closely, I would expect.

 

[jack action]

The Froude Number is calculated by
Speed to be scaled to Fn
FL=v/((g*LWL)^0,5)
FL= Froude Number by length
v= speed in m/s
LWL= Length on waterline
g= free fall acceleration 9,812m/s2


It has nothing to do with the Scale Formulae that i can see



That would be true if the Froude Number was part of the equation

Where is the Froude component in this ?

Scale speed is calculated :
vM= vO/((M)^0.5)
vO= speed of original
vM= speed of model
M= scale

Say:

FLM= Froude Number of model by length
vM= speed of model
LWLM= Length on waterline of model

FLO= Froude Number of original by length
vO= speed of original
LWLO= Length on waterline of original

g= free fall acceleration 9,812m/s2
M= scale

Then we can say:

M = LWLO/LWLM

Since g/g = 1, then M = (g*LWLO)/(g*LWLM) is also true.

So, putting this into your scale formula:

vM = vO/((g*LWLO)/(g*LWLM))^0.5)

Which can be rewritten:

vM/((g*LWLM)^0.5) = vO/((g*LWLO)^0.5)

Or:

FLM = FLO

Meaning that your scale formula assumes that the Froude number of the model is equal to the Froude number of the original. It is the condition assumed for similarity of flows. That's what a dimensionless analysis is.

 

[Dnomyar]

Say:

... That's what a dimensionless analysis is.

I have no idea what you are talking about, sorry.

The formula I use is a straight conversion of model velocity to an equivalent full size Velocity.

As far as I know, its no more involved than scaling the Weights down to 1:5 scale

ModelWeight = FullSizeWeight /(5*5*5)

or scaling the weights up to full size weight

FullSizeWeight = ModelWeight * (5*5*5)

 

[Dnomyar]

Scale Model Formulae

Further to the Scale Spped Calculation discussion, I found further clarification at
http://www.charlesriverrc.org/articles/design/ibtherkelsen_scalespeed.htm

What scalemodelers should know
In order not to bore readers more than necessary, the outcome is presented first, the details and comments later for those who might want a closer look. In everything that follows, the subscript 1 stands for some property of the full size aeroplane, while subscript 2 is reserved for the model.

Rule 1 z2 = z1/r Wingloading, z, resulting from simple downscaling.
Rule 2 v2 = v1·sqrt(1/r) Velocity, v, resulting from simple downscaling.
Rule 2a v2 = v1/r True scale velocity.
Rule 3 w2 = w1/r4 Mass, w, required to fly at true scale speed.
Rule 4 svz = z1/r2 Scale velocity wing loading, svz, resulting from Rule 3

Rule 2 seems to be the one that I was given
======
1:5th Scale, Full Size Velocity = 22kph )
ModelVelocity = FullSizeVelocity x sqrt(1/r)
= FullSizeVelocity x sqrt(1/5)
= FullSizeVelocity x 0.447213595
= 22 x 0.447213595
=9.84

Rule 2a gives a very close similar answer

 

[jack action]

And from the website, the explanation on how rule 2 was found:

Rule 2
Of special interest for aeroplanes is the Lift Equation: L = CL·q·S
L = w·g = Newton ( N )
w = aeroplane total mass ( kg )
g = acceleration due to gravity = 9.81 m/s2 ( 32.2 ft/s2 )
CL = Lift Coefficient ( a dimensionless value to indicate the lift capacity of a certain shape )
q = d/2·v2
d = density of air = 1.225 kg/m3 ( 0.0765 lb/ft3 or, to be correct, 0.002377 slugs/ft3 )
v = velocity ( m/s )
S = surface area of wing ( m2 )
S = c·b
c = mean wing chord ( m )
b = wingspan ( m )
The lift equation in full: w·g = CL·d/2·v2·S
or, upon rearrangement: 2·w·g = CL·d·v2·S
Rearranging for velocity squared: v2 = 2·w·g / ( CL·d·S)
Comparing model velocity to full size velocity:
v22/v12 = (2·w2·g·CL·d·S1) / (CL·d·S2·2·w1·g) = (w2·S1) / (S2·w1) = z2/z1
Substituting z2 by Rule 1 leads to:
v22/v12 = z1/(r·z1) = 1/r
v22 = v12 ·1/r
v2 = v1·sqrt(1/r)
Velocity for model ( by simple downscaling ) = full size velocity · squareroot of downscaling

Of likely interest may also be to compare the original velocity, v2, with the resultant velocity, v3, in case the model mass ( and hence wingloading ) is changed:
v3 = v2·sqrt(w3/w2) or the equivalent: v3 = v2·sqrt(z3/z2)

Since it is an airplane, they use the lift equation which contains the dimensionless number called the lift coefficient. Because you have a boat, you have to use the Froude number. It is a coincidence that both give the same result. Do you understand what is written in the quote titled Similarity of flows in post #19? Read it carefully and follow the links, this is important because it states that there are other dimensionless numbers that can be of importance in certain situations.

 

[Dnomyar]

This is getting tedious.

There is no LIFT equation in the Scale velocity calculation.

They are not connected.

They do not affect each other, nor are they relalted at all.

They are as different as can be !

 

[jack action]

I don't know what to tell you anymore. I'm starting to think you are not reading the posts thoroughly and/or you don't have the capacity to understand mathematical equations.

In my previous post, I took a direct quote from the website where you found Rule 2 (http://www.charlesriverrc.org/articles/design/ibtherkelsen_scalespeed.htm) explaining how they got their scale velocity formula; And the author says clearly and directly that it comes from the Lift equation. Read my quote again, I even put the words Lift equation in bold.

In post #19, I gave you a quote coming from Wikipedia concerning the scaling factor when considering similarity of flows:

In order for two flows to be similar they must have the same geometry, and have equal Reynolds numbers and Euler numbers.

[...]

This allows engineers (<-- this is you) to perform experiments with reduced models in water channels (<-- this is what you want to do) or wind tunnels, and correlate the data to the actual flows (<-- this is the scaling effect), saving on costs during experimentation and on lab time. Note that true dynamic similitude may require matching other dimensionless numbers as well, such as [...] the Froude number that governs open-channel flows (<-- this is your situation). Some flows involve more dimensionless parameters than can be practically satisfied with the available apparatus and fluids, so one is forced to decide which parameters are most important. For experimental flow modeling to be useful, it requires a fair amount of experience and judgement of the engineer. (<-- again, this is you)

Not only you have to consider the Froude number when scaling, but the Reynolds and Euler numbers might be of importance as well.

I can't tell you more clearly than that: For experimental flow modeling to be useful, it requires a fair amount of experience and judgement of the engineer.