Force in Pounds per sq/ft.

The first thing what is pounds per sq/ft? Do you mean pounds per square foot? If so, it would be pounds/ft² (psf or pounds /sq ft or lb_f/ft² where lb_f is pounds force as opposed to pounds mass). What you wrote means pounds per square per foot. Which is meaningless. At first I thought it might be typographical error, but noticed you used it twice the same way.

Secondly, you should that force is not measured in lb_f/ft². It's measured in force units which would be pounds (lb_f) or Newtons (or one of a number of other force units). The unit lb_f/ft² represents pressure.

I want to show some type of % of increase in pressure reacting to this ramp as it is lowered. If there is a better use of the word FORCE than that would be alright .

Sorry, but I am unable to express it any better than that....pressure on the ramp as it is lowered at a steeper angle is what i am after.

thank you.

"...pressure on the ramp as it is lowered at a steeper angle is what i am after." is perfectly fine.

That's what I thought you asking for. I just wanted to be clear.

Do you mean the barge speed is 4 mph relative to land, so 20 mph relative to the current?

As a rough guess I would say P = 0.5*ρ*V_n² where P = pressure in lbf/ft², ρ is water density in slug/ft³, and V_n is velocity normal to the ramp.

V_n = V*sin θ where θ = 20° or 30°. ρ = 62.5 lb/ft³ = 62.5/32 slug/ft³ .

On those assumptions I make it ~ 1 lbf/ft² at 20°, 2 lbf/ft² at 30°. No point trying to be too accurate, and (sin 30°/sin 20°)² ~ 2.

But how big is the barge? Are you sure it will stay level? Ramp 20' x 6' is pretty big and might pull the bow under. Also could be dodgy as it's a positive feedback situation - barge tilting forward increases the angle which increases the force.

Cheers.....Codey

The surface speed of 4 mph is how fast the barge would going if you were standing on land and watching it pass by with NO currrent at all.

Knots is more appropriate but I have to break it down to something easily understood.

* I am not sure where you get the 20 mph from*

2 mph is the speed of just the current.

Just for the record, this idea was brought up as part of a fish sampling project. I want to clearly demonstrate that this conept even at these low speeds would NOT work due to pressure on this ramp, as well as making the barge totally unstable when that ramp was lowered.

However, without some good numbers the case is just me saying , "it won't work".

So your calculations say

20° = 1 lb. per sq/ft

30° = 2 lbs. per sq/ft

Thank you.

Quick one before I read the rest of your post - 20 mph was a typo, I meant 2 mph.

If you can come up with some numbers to plug in, this might help give you a rough idea. Somebody else may be able to help with the varying angles.

http://www.chemicalprocessing.com/experts/answers/2010/093.html

Thank you for the formula site.

Sure. I'm not a formula guy, but those numbers in post 5 sound way low.

I'm with you in thinking that it won't work. I can imagine the ramp getting torn off, and the barge getting unstable.

Unless maybe if the ramp was porous, and you were able to scoop up the fish while the water flowed through.

A possible way to do this test would be to build a device utilizing a spring scale and a 1 sq ft piece of plywood on a hinge, attached to an actual boat. Then you could go out in the water, adjust your angle accordingly and the scale would tell you the real life, actual pounds per sq ft. No guessing involved.

Using that formula site, and remembering (or misremembering?) that 60 mph is 88 ft/sec, so the 4 mph the barge is moving with respect to the current is about 6 ft/sec, squared is 36, and assuming a k of 2, you're looking at about 72 lbs. per square foot of surface normal to the current flow.

Now you have to consider the impact of the 20 and 30 degree angles and develop a conversion between the areas at that angle and the equivalent area normal to the current flow. That should be a simple trigonometric function--can I remember which one.

The actual length of the ramp is the hypotenuse, and the length you want (for the area normal to the flow of current) is the length opposite the (20 or 30 degree) angle, so what you want is opposite divided by hypotenuse (the sin) times the length (or area) of that ramp (that's not stated very well, but anyway).

(BTW: The ramp is 6' x 20' so area is 120 sq.ft.)

So, sin 20 = 0.342 x 72 lbs/sq.ft. ~= 25 lbs./sq.ft x 120 = 3000 lbs.

sin 30 = 0.5 = 36 lbs./sq.ft. x 120 = 4320 lbs.

Now, you want to move against that force at about 6 ft/ sec, so:

with the 20 degree angle: you need 3000 lbs. x 6 ft./sec = 18000 ft-lbs/sec / 550 ft-lbs/sec per Hp, you need something like 32 Hp. to overcome the force against the ramp.

with the 30 degree angle: you need 4320 lbs. x 6 ft./sec = 25920 ft-lbs/sec / 550 ft-lbs/sec per Hp, you need something like 47 Hp. to overcome the force against the ramp.

If my calculations and math are correct. ;-)

But if the current is going 2 mph and the boat is going 4 mph, (in the same direction), wouldn't that drop the boat mph, (as far as the formula goes), to 2 mph?

Another thing. Is the water going to be able to flow up over the top of the ramp, or forced off to the sides? I think that may come into play also.

Maybe a mock up with a 1 sq ft piece of plywood, aluminum, etc. is the way to go?

Sorry for bad graphics. Once your speed is set, just raise or lower your scale to give you the force at any given angle.

I guess we have to depend on the OP for final clarification--the words could be a little ambiguous.

I think he (the OP) said in one post that, without the current, the barge would travel 4 mph with respect to land. Then I took it to mean that when you add in the current, it travels 6 mph with respect to the land.

But we only have to consider the speed with respect to the water (for determining the force against the ramp due to moving through the water), which, by my understanding is 4 mph (or about 6 ft/sec.).

Having read again OP's post #5, I think you're right, speed to use is 4 mph. There's no need to mention the speed of the current, the barge is doing 4 mph relative to the water whatever the current speed is.

So my figures in #14 should be x 4.

Cheers........Codey

Won't the force against the ramp be less if traveling with the current, and more if traveling against it, assuming that the speed of the boat remains constant on the surface?

I would think that current speed should be subtracted from boat speed if going with the current........and added to boat speed if going upstream.

It depends on how you interpret the OPs explanation of the speed of the boat. As I understand it, the speed of 4 mph is against the speed of the water. (Which means that, with respect to land, the speed of the boat is 6 mph if traveling with the current, and 2 mph if traveling against the current.)

As a further check, a certain amount of horsepower driving the barge will cause the barge to move at a certain speed with respect to the water. The speed of the current will add to or subtract from that speed with respect to land.

I think your conversion for the 20° is wrong. Basic figure should be x (sin 20°)² because of the V² in the formula.

With that and assuming 2 mph (about sft/sec) instead of 4 (kramarat #10) get 72 lbf/ft²/2²*(sin 20°)² ~ 2 lbf/ft². That's about 2*0.5*ρ*V², I guessed 0.5*ρ*V² (basic formula for velocity pressure) ~ 1 lbf/ft². As ρ = 1.9 slug/ft³, the site formula 2*V² ~ ρ*V².

For the power - the calculated force is normal to the ramp, not in the direction of barge travel. So another sin 20° factor is needed. Assuming 2 lbf/ft² normal, I make power 2 lbf/ft²*120 ft²*sin 20°*3 ft/sec ~ 250 ft.lbf/sec.

Cheers.........Codey

I don't think so, but the operative word is think--I'll probably have to cogitate on that overnight ;-)

Trying anyway:

The speed of the boat is 4 mph relative to the water. The speed of the ramp through the water is the same as the speed of the boat, it doesn't vary with the angle of the ramp. (These statements are true only to the extent that there is sufficient power (horsepower from the motor) to maintain 4 mph with the ramp in either position.)

Thus, I don't think you have to apply a sin correction factor to the speed (which is what V is).

I think the sin correction has to applied to the force, not the speed (though the result is the same). The calculated force is normal to the ramp. We need the horizontal component to multiply by the speed ie force x sin θ. Or you could look at it as the projected area = 120 ft²*sin θ.

That's my take on it, I could be wrong.

Cheers.........Codey

I agree, and that is what I did in my calculation (post #9).

I use the velocity of the ramp with respect to the water to be 2 mph (2.933 ft/sec). As the pressure acting on the ramp will be due to the relative velocities.

In Fluid Mechanics by Frank M. White, there is a reference to the drag coeff of a flat plate normal to flow. The value stated is 2.0. So that's what I'll use here.

Also from Frank M. White's Fluid Mechanic's book, F = ½ Cd ρ v² A and since the OP asks for pressure F/A = ½ Cd ρ v²

ρ = 62.428 lb_m/ft³ for fresh water (63.989 lb_m/ft³ for sea water) or converted to slugs/ft³ will be 1.9403 slugs/ft³ (1.9888 slugs/ft³ for sea water).

So that gives us

F/A = ½(2 )1.9403 * 2.9333² (slugs/(ft-sec²)) = 17.1127 slugs/(ft-sec²))

or

F/A = 17.11 lb_f/ft² given that 1 lb_f = 1 slug ft/s²

The angle will make a difference in the final force being applied as the effective area will be different for the 20 degree orientation vs. the 30 degree.

The effective area for 20 degrees is A₂₀ = 72*20*sin(20) = 292.5 ft²

The effective area for 30 degrees is A3₀ = 72*20*sin(30) = 720 ft²

F₂₀ = F/A * A₂₀ = 17.11 * 292.5 = 8,428.2 lb_f

F₃₀ = F/A * A₃₀ = 17.11 * 720 = 12,321.2 lb_f

I believe that Velocity MPH should be used instead of ft/sec. Also, I read the speed relative to the water as 4mph but that is a simple adjustment. Judging by work done designing some large highway signs, that Cd looks high, the signs get in the region of 1.3, although for safety reasons, a margin is added in other ways.

More importantly, there is a volume of stagnant, pressurized, water next to the ramp so that the ramp should receive the full fluid pressure on both pressure and suction sides.

@Netmaker, you will need to support the bottom of the ramp on both sides, with ropes from the bow or it will twist. If you clear up the questions and tell us what you want to calculate, the bending moment at the hinge? the pressure on the projected area or the actual area of the ramp etc.?

Re: I believe that Velocity MPH should be used instead of ft/sec

I think that depends on the formula that you are using and the units it requires. I based my calculation (in post #9) on the formula in the link given by kramarat in post #6.

That page is How do you calculate the force of a moving fluid against an obstruction?

And the formula given is:

"However, a simple but rough estimate that is commonly used for water is Force (lbs) = Area (ft2) x K x Velocity2 (ft/sec) where K is a drag coefficient. For a rectangular flat obstruction, K can be between 1.8 and 2.0."

I was referring to #18, I am more familiar with that formula, we use it to calculate wind loads in structural engineering. The one you have may include the effect of mass acceleration hidden within it but I don't see it.

My result:

V_n=4mph through the water, V_n²=16

Drag coefficient, use 2.0, (1.3 is common for, squarish, flat road signs with clearance all round, but we are long and thin and don't have clearance at the top, the barge is there.

p= 62.5/32 slug/ft³, say 2.0

P=0.5_xp_xC_{d x}V_n²= 0.5x2.0x2.0x16=32lbs/sqft.

Notes: If this is a net, several things change.

Since the drag coefficient is dimensionless, it does not matter what units are used for velocity, so long as the units are consistent and converted to what the OP asked for in the end which was lb_f/ft².

I used velocity in ft/sec and density in slugs/ft³ to make getting to lb_f/ft² relatively straight forward.

I won't argue over whether Cd should be 1.3 or two or one. I just used what I could find in a book.

Regarding which speed to use (4 mph or 2 mph)....The OP stated and then clarified that the 4 MPH was the speed of the boat as viewed from shore (on land) and that the current of the water is 2 mph. If the boat is moving at 4 mph relative to a fixed reference point (ground) and the current is also moving in the same direction at 2 mph with respect to the same reference, then the relative speed is 4 mph - 2 mph which equals 2 mph or 2.933 ft/sec.

The formula we use for wind, and I have been told by an ME colleague that it works for liquids as well, uses slugs,ft³ and mph with no other modifier. It is shown in Wiki, don't bother with the modifications, they deal with the wind speed increases with height and stuff;

Wind load calculations, Generic Formulas (0.00256 is the density of air in slugs/ft³)

Re: Regarding which speed to use (4 mph or 2 mph)....The OP stated and then clarified that the 4 MPH was the speed of the boat as viewed from shore (on land) and that the current of the water is 2 mph.

What the OP said in post #5 was: The surface speed of 4 mph is how fast the barge would going if you were standing on land and watching it pass by with NO currrent at all.

You wrote The effective area for 20 degrees is A₂₀ = 72*20*sin(20) = 292.5 ft^2.

Please try again. The 72 is inches, not feet!

Cheers.........Codey

Doh!!! Good catch. I didn't even notice the extra tick mark.

Actually the 292.5 is incorrect for even 72 feet.....I missed typed it, for 72 feet it would be 492.5 ft²

Using 72 inches (6 feet)

A₂₀ = 41.0 ft²

A₃₀ = 60 ft²

F₂₀ = 702.3 lb_f

and

F₃₀ = 1026.8 lb_f

Thanks

Just catching up to this.

1. At 4 mph, and this ramp at 20°, what can I say will be the pressure on the 72" x 20' ramp?

2. ...And again, at 30°.

Gentlemen, I do not need an exact scientific analogy. I am asking for a general answer in pounds per sq/ft exerted on this type of set up. My aim is to establish that this is NOT the way to go with this project.

I have a net plan that will adequately work for these researchers .I just wanted to establish a scientific ( mechanical) reason NOT to attempt this as it will cause serious grief and more than likely fail. The easiest thing I can think of is to show the tremendous pressure that will be exerted on this ramp system.

So... 20° =

30° =

thank you all for your time and effort in this.

So, per the calculation I used in post #9:

So... 20° = 3000 lbs. (and requiring 32 Hp. ("in the water") to move at 4 mph)

30° = 4320 lbs. (and requiring 47 Hp. ("in the water") to move at 4 mph)

It's all above my pay grade. Maybe this can help.

http://www.engineeringtoolbox.com/drag-coefficient-d_627.html

At 4 mph (2 mph relative to the current) the pressure will be the same for 20 degree or 30 degree ramp angle and would be approximately 17.1 lb_f/ft².

The forces acting on the ramp and transmitted to the boat will not be equal. F₂₀ = 702 lb_f and F₃₀ = 1027 lb_f.

My pay grade was surpassed when these guys asked me to calculate WHY this type system would fail. My response of " my gut feeling and 32 years on the water says it just will snap off and/or flip the barge in a cross current"....... was not sceintific enough.

Thank you all for the scientific answers.... even at taking averages, you've given me enough facts .

If the image posted in your Op is anywhere near accurately scaled, the gut feeling I have is the volume of water diverted up thru the ramp will likely swamp the boat!

....exactly what I tried to convey. ha ha

Wow! You even know doors if they're being opened under water. Very impressive.

Yeah, I've seen a lot of doors...

I sure learned a lot from this one:

The original complaint was, "Submarine too hot." So, we installed the screen door.

Now the end user is unhappy. The current complaint is "Open door, screen door lets in water, submarine fills with water."

Yeesh! It would have been nice to have that particular performance criteria BEFORE we recommended the screen door. I can't think of everything!

Pesky clients........they want it all!!!!!

We only live to make them happy........................not only that, but we're getting paid too much to do it.

How can we be so selfish?

I'm not grasping the entire concept. Is this suppose to be a braking mechanism? I can't see the practical use of the "ramp".

It's supposed to scoop up fish, for study..........and it won't work. That pretty much sums up the thread.

Ah, I thought they invented nets a couple of thousand years ago for that very purpose.

The fish they want to take are too big , too fast and scatter too quickly for this.

When they mentioned underwater ramp, a little red flag went up.

It might have worked in that James Bond movie where they scooped up submarines but in the real world its a No-Go!

I only wanted some scientific data to kill the idea so they would be more attuned to looking at one of our systems that incorporate netting to catch what they want.

The ramp will wrench itself apart at speeds of 4-6 knots , it will cause the vessel to become unstable in a cross current and there will be horrendous problems trying to steer this with the ramp deployed.

Thank everyone again.

I think I can assist you in killing this idea. Tell your people that as they deploy the ramp deeper it will create drag and slow the barge. In order to over come this the motor(s) will have to increase in speed and noise. Fish don't generally like noise, so you'll be scaring them off before you"re close enough to scoop em up. The power required when you get the ramp at steep angles will be quite high just to maintain the desired speed. Sell them a net!

Ummm, what fish (that's too fast) can't do 4-6 knots?

So exactly what Speed did they have in mind for this?

Do you have a 'net solution'? Or - Do you need ideas on ways of achieving whatever it is they need?

After all; the scoop 'principle' can, and does work, in other applications.

What you have above is just an 'embodiment sans 20 principles' (and common sense).

THAT was the slickest plane handling i ever saw. I 've seen the old prop jobs do that but not a turbo jet.

Note, they did not have a ramp hanging down either....I think it is a recessed door of some type that slides.

AS for the fish.... they are collecting YOY ( Young of the Year) . Most YOY of most species have a hard time maintaining even 4 knots for any length of time.

These however are the infamous Asian Carp, part of the great Asian Invasion heading into the Great Lakes . They move more like Blue fin tuna.

We are looking at a critter that can move quicker and with their Shrieken response, can scatter faster than most fish after the first fish bleeds out that pheromone in the water.

The ramp idea was part of a scoop 'em up system that someone with no fish sense dreamed up.

Basically all of my YOY designs ( benthic , pelagic and neustonic) operate at speeds of only 2-3 knots and catch about 80% of their target species.

Thank you for the video.

So it's a shallow water exercise?

Hitting something with the 'ramp' would be "interesting"

Good you have the net for it - ^{_{I can stop thinking 'how to make it work' - though 'air lift' did cross my mind.}}

BTW, they have been known to drop fish on fires

Did you notice in the first run the hills at the end "suddenly got higher" and the other runs started a fair way 'earlier'.