Need help with Fluid Mechanics equation

[Doug]

I am trying to design a "fin" for my water rudders for my experimental
floats. I need the equation for the amount of forces exerted by water
going 0-50mph on a flat surface at right angles to the water flow. It
would be dependent on the speed of the water, the depth of the fin and
the angle of the fin to the water. I am hoping one of you tech types
in homebuilt might know how to find this equation. The reason I want
the equation, and not just the answer, is I want to be able to fool
around with location of this fin to find the best spot. I have a
degree in Electrical Eng, and had a course in statics, but I never
took Fluid Dynamics. Hopefully the equation is not too complicated.

|
----->water flow---> |<-----Need force F
|
side view of fin

Doug

[Corrie]

What do you mean by "best location?" Best in terms of ...?

I'm not a hydrodynamicist, but I'm pretty sure that the standard
equation for dynamic pressure would work: density x (velocity^2/2).
http://wright.nasa.gov/airplane/bern.html

This assumes an incompressible fluid, and water certainly is. The
depth should not have an effect, since the fin isn't going to be feet
deep in the water. The pressure difference at the surface and at one
foot is probably negligible.

Just a swag to check the assumptions: Say we have a velocity of 15
m/sec, or 1500 cm/s. Assume water density is 1 g/cc (that's for pure
H2O; "real" water is slightly higher by a small percentage).

1g/cm^3 x 1500^2cm^2/s^2/2 = 1.125x10^6 g/cm s^2.

Converting g/cm s^2 to kg/m s^2 (divide by 1000, multiply by 100) we
get 1.125 x10^5 Pascals of pressure. (A Pascal is 1 kg/m s^2, or 1
newton per square meter)
http://whatis.techtarget.com/definition/0,,sid9_gci541172,00.html

To convert Pascals to to psi, divide by 145.03
(http://www.springfixlinkages.com/tech_conversion_factors.htm )

775.7 pounds per square inch.

Seems like a lot! But remember, this is water at 15 meters per second
- 33 mph, faster than most ski boats. Hitting the water at that speed
is like hitting a brick wall.

At a taxi speed of one m/s, the dynamic pressure is 1g/cm^3 x
100^2cm^2/s^2/2 = 5,000 pascals, or 34.5 psi. Think about paddling a
canoe at a good clip, then sticking the tip of the paddle in the
water, flat-on, and trying to hold it. 35 psi sounds about right.

Anyway, that's the way I figure it. Feel free to check the math and
correct my asusmptions. Like I said, I'm not a hydrodynamic engineer.

Corrie

(Doug) wrote in message >...
> I am trying to design a "fin" for my water rudders for my experimental
> floats. I need the equation for the amount of forces exerted by water
> going 0-50mph on a flat surface at right angles to the water flow. It
> would be dependent on the speed of the water, the depth of the fin and
> the angle of the fin to the water. I am hoping one of you tech types
> in homebuilt might know how to find this equation. The reason I want
> the equation, and not just the answer, is I want to be able to fool
> around with location of this fin to find the best spot. I have a
> degree in Electrical Eng, and had a course in statics, but I never
> took Fluid Dynamics. Hopefully the equation is not too complicated.
>
> |
> ----->water flow---> |<-----Need force F
> |
> side view of fin
>
> Doug
>

[Doug]

Thanks Corrie, I think that is what I need to get started. As for the
best location, well, I've never done this before so I don't know. But
there are some options. One problem is as the rudder comes up, the fin
changes angle to the flow, but I think I have a handle on that one.
The other is, we don't want the fin to raise the rudder any more than
"all the way up". You would have to see the rudder and all the
brackets to see what I mean. Anyway, thanks for giving me some ball
park figure. Now I need a tensionmeter, something I can hook to the
rudder and get some exact "pounds of pull" readings. Anyone know where
I can get a good cheap tensiometer? that reads 0-20 lbs?

(Corrie) wrote in message >...
> What do you mean by "best location?" Best in terms of ...?
>
> I'm not a hydrodynamicist, but I'm pretty sure that the standard
> equation for dynamic pressure would work: density x (velocity^2/2).
> http://wright.nasa.gov/airplane/bern.html
>
> This assumes an incompressible fluid, and water certainly is. The
> depth should not have an effect, since the fin isn't going to be feet
> deep in the water. The pressure difference at the surface and at one
> foot is probably negligible.
>
> Just a swag to check the assumptions: Say we have a velocity of 15
> m/sec, or 1500 cm/s. Assume water density is 1 g/cc (that's for pure
> H2O; "real" water is slightly higher by a small percentage).
>
> 1g/cm^3 x 1500^2cm^2/s^2/2 = 1.125x10^6 g/cm s^2.
>
> Converting g/cm s^2 to kg/m s^2 (divide by 1000, multiply by 100) we
> get 1.125 x10^5 Pascals of pressure. (A Pascal is 1 kg/m s^2, or 1
> newton per square meter)
> http://whatis.techtarget.com/definition/0,,sid9_gci541172,00.html
>
> To convert Pascals to to psi, divide by 145.03
> (http://www.springfixlinkages.com/tech_conversion_factors.htm )
>
> 775.7 pounds per square inch.
>
> Seems like a lot! But remember, this is water at 15 meters per second
> - 33 mph, faster than most ski boats. Hitting the water at that speed
> is like hitting a brick wall.
>
> At a taxi speed of one m/s, the dynamic pressure is 1g/cm^3 x
> 100^2cm^2/s^2/2 = 5,000 pascals, or 34.5 psi. Think about paddling a
> canoe at a good clip, then sticking the tip of the paddle in the
> water, flat-on, and trying to hold it. 35 psi sounds about right.
>
> Anyway, that's the way I figure it. Feel free to check the math and
> correct my asusmptions. Like I said, I'm not a hydrodynamic engineer.
>
> Corrie
>
> (Doug) wrote in message >...
> > I am trying to design a "fin" for my water rudders for my experimental
> > floats. I need the equation for the amount of forces exerted by water
> > going 0-50mph on a flat surface at right angles to the water flow. It
> > would be dependent on the speed of the water, the depth of the fin and
> > the angle of the fin to the water. I am hoping one of you tech types
> > in homebuilt might know how to find this equation. The reason I want
> > the equation, and not just the answer, is I want to be able to fool
> > around with location of this fin to find the best spot. I have a
> > degree in Electrical Eng, and had a course in statics, but I never
> > took Fluid Dynamics. Hopefully the equation is not too complicated.
> >
> > |
> > ----->water flow---> |<-----Need force F
> > |
> > side view of fin
> >
> > Doug
> >

[Corrie]

Well, a teniometer is generally used to read tension in a cable. If a
fin changes its angle to the flow, it's going to act like a rudder.
You're right, I think I'd need to see the system - or at least some
drawings or photos - to have a clear idea of what you're looking for.

(Doug) wrote in message >...
> Thanks Corrie, I think that is what I need to get started. As for the
> best location, well, I've never done this before so I don't know. But
> there are some options. One problem is as the rudder comes up, the fin
> changes angle to the flow, but I think I have a handle on that one.
> The other is, we don't want the fin to raise the rudder any more than
> "all the way up". You would have to see the rudder and all the
> brackets to see what I mean. Anyway, thanks for giving me some ball
> park figure. Now I need a tensionmeter, something I can hook to the
> rudder and get some exact "pounds of pull" readings. Anyone know where
> I can get a good cheap tensiometer? that reads 0-20 lbs?
>
> (Corrie) wrote in message >...
> > What do you mean by "best location?" Best in terms of ...?
> >
> > I'm not a hydrodynamicist, but I'm pretty sure that the standard
> > equation for dynamic pressure would work: density x (velocity^2/2).
> > http://wright.nasa.gov/airplane/bern.html

[Blueskies]

..
"Doug" > wrote in message om...
> rudder and get some exact "pounds of pull" readings. Anyone know where
> I can get a good cheap tensiometer? that reads 0-20 lbs?


Digital fish scale?


--
Dan D.