Vanity: Need to Calculate Drag on Sq Plate in 6 mph current

Email | 05-04-2006 | R J Cook

[Robert A Cook PE]

My dad (long time civil structural engineer) sent the following question this evening to me:

Tomorrow morning, I need the drag of one square foot of plate in 6 mph water. Not edge effects, downstream turbulence, corner losses, etc. Just how much force will I have to provide to hold a square foot of plate still into a 6 mph current? No, not in channel. In open water.

I can't find it ANYWHERE! Durn! Got a structure in a pickle in water, and I really need that drag force to figure piling forces. My Olde Books are EMPTY on the subject.

Anybody got a suggestion of how much it will take? Please?

Ramon in a mess, Durn!

[Robert A Cook PE]

Passing along this email from my dad for your comment and assistance, while I'm Googling for the calc's separately.

His email is available, freemail me if you want to discuss directly with him.

I'll try to get him to login on new his account in FreeRepublic as well.

My other questions: If it's a square plate being modelled (then his pier would have to be a rectangle or square itself incross-section. That resistance would not be the same as a flat (very thin) plate since a rectangle has drag along both sides, as well as face, corner, and exit losses.

If it's a flat square )thin) plate in the current, then it won't be able to carry a load: only create resistance like a brake.

[Robert A Cook PE]

You guys have covered naval physics/design in the past, any recommendations?

[hole_n_one]

You wanna field this one?

[RedBloodedAmerican]

Scrstching head trying to recall; dont you do the drag coeeficient
My guess is zero.

[RedBloodedAmerican]

bah typos
should be:
Scratching head trying to recall; dont you do the drag coefficient

And my guess of zero is wrong but its been about 25 yrs since I messed with this type of equation and plugging in numbers thats what I got, but obviously something I did was wrong :)

[AlaninSA]

R = 0,5*Ro*c*Aw*v^2

[Robert A Cook PE]

Can't be zero: You have for example, a 1 ft by 1 ft column in flowing water, perhaps 12 feet deep.

That's got to be calculable for water resistance on a resistance per foot length basis.

[359Henrie]

I know you need 6lb drag setting to bring in a 7lb rainbow.

[AlaninSA]

There's also material, thickness and shape to consider. Is it a perfect square, how thick is it and what's the material? Density plays a role in drag/resistance.

[Robert A Cook PE]

R = 0,5*Ro*c*Aw*v^2

Ro is what units?

C is coefficient of flow for a square cross-section, right?

[AlaninSA]

Right - and Ro can be in any measurement you care to use - it's portable as long as you're consistent in your equation.

[RedBloodedAmerican]

yea zero doesnt make sense. I must have remembered the formula wrong, or maybe the wrong formula! Would it be drag coefficient?

[spunkets]

Is the flow perp to the face and supported only by the neglected supports? Open channel here means the cross sectional area of the pipe is not significantly effected by the plate right?

[Robert A Cook PE]

There's also material, thickness and shape to consider. Is it a perfect square, how thick is it and what's the material? Density plays a role in drag/resistance.

The pier of course, in this case, would be concrete as the resisting solid, with a cross-section as a square (I'm assuming) facing the flow at alpha = 90 degrees; and the fluid would be water at atmospheric pressure/temperature (1 gm/cc metric, 62.4 pounds per cubic foot English) at 6 mph.

[Cyber Liberty]

I'm gobsmacked. I am watching, though. That formula has promise...the v^2 makes intuitive sense.

[Robert A Cook PE]

As I understand it, the width of the river's net cross-section is not significantly affected by the width of the pier: Thus, assume it is a 100 foot wide river obstructed by a single pier at midpoint, perpendicular to the flow that is a simple 2 foot by 2 foot square shape.

The river is assumed NOT to "dam up" or block the flowing water or change average height upstream of the pier, but will simply try to push the pier downstream.

[Flightdeck]

"Thus, assume it is a 100 foot wide river obstructed by a single pier at midpoint, perpendicular to the flow that is a simple 2 foot by 2 foot square shape."

Well, I had an answer all ready for you for a flat plate, but then I checked before I posted and saw that. I could get back to from work tomorrow, though.

[spunkets]

Coeff drag for a square is 2, so the eq is: drag = area * density * v²

Assuming 2ft plate in middle depth of "deep" 100ft wide river?

[Tuscaloosa Goldfinch]

I just PRAY there's a DUer lurking, having his preconceived notions blown to smithereens. I read this thread just out of curiosity -- now I wish I hadn't. It didn't do anything for my self-esteem realizing that I couldn't understand half of what was written.

Can anyone help my daughter with her algebra? You'd be much better at it than I am LOL.

[martin_fierro]

The answer is 3.14.