PROBLEM: How Far Will THIS Move?

Found this.....good luck

http://www.ocean-sci.net/7/733/2011/os-7-733-2011.pdf

you come up with some pretty crazy questions, I'm gonna give my calculator Sunday off on this one

Well Pilgrim, if it too hot to handle..... HA! HA!

This is my busy time of the year for weird questions from my customers. I only have until the supplies show up this week to get some of this figured out and then there will be NO time for anything except work.

This one kind of snuck past me...its a generalized adaption of a sounder device attached to a deep water net. I am not sure of the distance that the two different cable venues will travel. Somebody will figure it out...its why I stay on here...everybody is a GENIUS!

Hey its Mardis Gras week end...no parades for this old fart, but its a good excuse to fry up some fish and hush puppies!

Just fried fish and hush puppies....its 78F right now and dat' ain't Boudin weather!

Thank you again.

It's an interesting question, and a hard one IMHO. If it's any help, I've estimated the horizontal force on the ball from the water. I've got various formulas for settling velocities of spheres in different conditions, determined by Reynolds No. Assumed steel density 7900 kg/m³ to work out sphere dia, and water 1030 kg/m³ for seawater. Then tried varying sphere density to give settling velocity your 1.34 m/s. Works out to net downforce 13 newton, so taken this as the horizontal force in your situation. Net downforce of steel sphere 290 newton.

Greatly simplifying the problem by ignoring the dia and mass of the cable, if θ = angle of cable from vertical, θ = atan(13/290) = 2.6° and horizontal distance = 900*sin θ = 40 ft.

Your Calculations show the STEEL cable will move 40 feet , correct?

thank you.

Not exactly, I assumed the cable has no mass or thickness (but still takes the force, obviously). I calculated the ball is then 40 ft behind the cable connection point at the surface.

How deep's the water? Will the ball drag the bottom? I assume yes.

Completely different kettle of fish if the ball is suspended and not dragging.

This gets really interesting when you consider the 450# of steel cable pulling down on the ball. I don't think you can ignore that weight, and the drag it will produce on the bottom. You don't have a 75# ball with it, it will act like it's much heavier.

I'd GUESS that the HDPE line will be mostly straight, since the drag on the ball will overcome any sag.

It's the steel that is bothersome.

And, is it on the bottom, or no?

BTW, netmaker, should post his short video that I saw. It shows his nets at work.

once he has a net on that it will be an underwater sail subject to current, so depending on the surface area of the net and debris or fish.........

• No net.
• The gadget rides independently ahead of the net by itself.
• Net drag is already calculated.
• The 75# weight is free swimming...never touches the bottom... the depth is well over 2500 feet of water

If the current is only 3 mph and there is no boat speed to add, I don't see either case moving the ball very much.

OK. My concern was for all of that 1/2" line x 900' x 12" with a 3 mph steady current pushing on it. I sure don't know. ha ha !

thank you. We'll find out in about 6 weeks what is what.

Well the speed that the ball and cable are moving through the water will determine the depth....In this case sitting still in a ~3mph current, won't have much effect...but can the current be considered uniform from surface to depth? ...probably not....In any case this is guesswork at best and I think 40' trailing would have little effect on depth, maybe around ~1'....

Love that drawing. With skills like that, why buy CAD?

Current will NOT be uniform, but I had to make a guestimate at an average based on highs and lows. All I need is an average.

hopefully nothing hits a line.....

.......Oh I have heard stories. Both EAST and WEST coast....

After reading all the previous replies, it sounds to me like it's time for a scale model experiment.

....that what I was trying to avoid. However, unless someone sees something really off-center, I believe we can live with a 1-40' swing at 900' depth.

I've had another look at this. Checked my calculated drag force 13 N against velocity pressure 0.5*ρ*V². That x projected area = 29 N, giving a "shape factor" about 0.5, which seems reasonable.

Looking at the steel cable, for any say 1m length of it, the drag force is the same at any point along the cable, and so is the net downward force, so I conclude that if there were no sphere to support, just the cable hanging free, it would hang in a straight line (obviously not vertical). Applying the same 0.5 shape factor to calculate the drag on the cable, estimated angle is ~ 45°. As the angle increases the drag force per m decreases, varying as cos θ, but I haven't gone into that much detail.

Looking at sphere + cable, the drag on the cable increases the angle. To get an estimate, made an arbitrary assumption that the angle is equivalent to adding net downforce and drag from the bottom 20% of the cable to the figures for the sphere, all acting at the sphere location. That gives an angle ~ 30° and horizontal distance 447 ft. You may or may not think that's reasonable!

One more thing - if you want to minimise the angle, you could use a thinner cable. Taking worst case, the sphere in air, ½ " gives only 2.6 MPa. Dia 0.1" gives 65 MPa, which is still low, and gives angle 10° and distance 66 ft (on same assumptions as above).

Revisioned....

I have to agree with the direction of curvature of NickName's graphs. Near the boat the drag forces (horizontal) of the entire tether will be acting, while near the ball only the drag forces of the remaining tether and the ball will be acting.

At 0.46lb/ft, the steel tether weighs 414 lbs., ≈5.5 * the weight of the ball. Since the ball is such a small fraction of the weight, the tether is going to be rather close to Codemaster's straight line. At 0.07lb/ft, the plastic tether weighs 63 lbs., ≈0.84 * the weight of the ball, so the ball is going to have much more influence, and there will be much more curvature. The tether will be farther from vertical near the boat, and more nearly vertical near the ball.

I'll leave the drag calculations to those with more knowledge and/or better software...

Note that NickName's two graphs can be misleading, since they have different vertical scales. At first glance, the two graphs appear to show the plastic tether as more nearly vertical, while the opposite is true.

If I read this read this right, the steel tether will have less movement than the HDPE?

If by 'movement' you mean 'deviation from a straight line', that is correct.

"....That gives an angle ~ 30° and horizontal distance 447 ft. You may or may not think that's reasonable!"
You are saying 447' movement beyond point "X" and point "Y" ? Or am i reading this wrong?

I was only working on the steel cable case, and the 447' was the horizontal distance from X to the steel sphere. I assumed steel and HDPE were 2 options for the tethers, rather than both being used. In any case, you did not not give a distance between X and Y.

Be good to hear what the trials show. Do you have some way of measuring the distance? As it's several 100' down it doesn't sound easy!

X and Y were just for reference sake. We will use either one or the other but not both.

Are you saying that I should expect the sphere to move 447' horizontally away from y?

No, from X. In other words, 447' from the point where its own cable is anchored at the surface. Just looking at one cable/sphere at a time. In any case it's very much an estimate, in an attempt to move the discussion along.

Ref your #30, I agree the steel tether will have less movement than the HDPE. For a given cable dia, both have the same drag, which increases the movement, but the steel has higher downforce, which decreases it.

Stupid me... I meant "X" also.

• OK, so we have a wide spread of answers here from 1' through 447'.
• I understand this is a bit tricky and has some variables that could seriously affect the answer either way.
• If I had to make a call this week, judging by all of this, I should stick with the steel cable regardless, as it will have the least movement...whatever that amount is... Correct?

No problem.

Yes, stick with the steel cable if you want to minimise movement, but this is (I think) first time you said you want to minimise it, rather than just trying to find out what it is.

I still think you can reduce the movement by a thinner cable, but I dropped a cod in my calcs for #23. For 0.1" dia should be ~ 15° and 225'. For cable alone thicker cable has less movement, as drag varies as dia, but downforce as dia². But for cable + sphere, thinner cable means cable weight is smaller fraction of total.

...so you are suggesting 250 feet movement?

Re-revision...

Qualitative only; I'm not going to try to finger it out! But a question: Is 900' the unloaded length? Is so, the HDPE will hang deeper than the steel (due to differential stretch,) and that will probably let it move even more.

The ball on the HDPE rope will swing farther. Reasoning: Both cable and rope are the same diameter, thus the same drag; however, the rope will have less restoring force from its weight. Thus the HDPE tethered ball will move farther because of greater swing. Also the HDPE will stretch more under the increased tension load (vectors and all that stuff), which will let it move even more.

Now, when you change to 3/8" rope, there will be less drag, but also less weight to produce a restoring force. I don't know what will happen numerically, but I'll assume the effects cancel. However, it will stretch more than the 1/2" which will allow it to move a bit farther.

900' with the 75lb. weight attached is measured on the boat, dry.

Right now I am just trying to gather some facts before I run it all by the researchers. They have their own ideas and we will soon be at odds.

I don't know the answer, but here is how I would attack the problem. Write a computer program to figure the forces at small increments in the vertical direction (say 1 foot) starting at the bottom. The weight is 75 lb. The drag is calculated for flow rate squared times cross section 1/2 " x 1 foot times coefficient of drag. Figure the vector direction, and this is the angle at the bottom, so you can calculate the amount of cable length in this vertical foot. Continue on to the next vertical foot, using the weight vector directed along the direction of the cable and the drag vector horizontal. Calculate the slant cable length from the angle. Sum up the cable payout length until you sum to 900 feet. Also sum up the horizontal component because this is what you want.

Once you write the simulation (say in MATLAB), you can vary the constants and see the effect of different weight per foot of line or diameter.

This also sounds very logical. I have no idea how to write a program like though. But your suggestion is to START calculating from the weight up, NOT from the boat down? correct?

Yes, start from the bottom up. Otherwise, you don't know the force pulling down from the weight and the weight of the cable below. Take the location of the weight as your origin. Take increments in the upward direction. Each little piece of cable will have balancing forces on it: its own weight pulling down (known), drag (from the current) pulling sideways and upward (also known), tension from the piece below (which you calculated in the previous step), and tension pulling up at an angle with cancels out the other forces (the unknown, which you are calculating at this step).

Force on your cable is a vector directed in the direction of the cable. If you examine a small piece of cable the tension pulling upward is slightly larger than the tension pulling downward. If you separate out the vertical and horizontal components of the upper and lower tension vectors, the difference in vertical components will be the weight of the cable section minus the lift on the cable caused by the current, and the difference in the horizontal components will be the horizontal (drag) force from the water current on the cable segment. Its a matter of summing up from the bottom, keeping track of the sums of the diagonal (cable length) of each segment and the horizontal component (for your desired answer).

I hope this explains it a little better.

You can use Excel for this kind of program, one row for each iteration (distance increment above weight) and one column for each variable.

Hi

I wrote a program back in the 1990's to calculate just this. It assumes the cable takes the shape of a catenary. Problem is ...... I need to find it!

Give me a day or so..

The standard catenary is the shape of an infinitely flexible cable with no forces other than gravity acting on it, except t the ends. Here we have both gravity and drag, so it won't be a standard catenary. It just might be a catenary rotated 90...

It is a catenary but with variable loads in both directions since as well the weight as the drag force are NOT constant in a given rope section.

It is also possible to integrate along the rope considering -as mentioned in my other comment- that the rope can support only traction forces. this is equivalent to say that the resulting force of vertical and horizontal loads is tangent to the curve in every point of it. For all graphs the integration is for the given rope length (rounded to 275 m)

I am sorry for the error it was not the right picture here are all at same scale.

Above pictures were computed with a Cd=1

Above pictures were computed with Cd=1 but the drag was computed considering an average angle between rope and drag force

Above pictures were computed with different Cd but no angular correction. The differences made me ask for a better more reliable source for cable drag Cd values.

Any way the trends are physically correct and the model responds to the parameters variations the logical way.

I am very glad to have generated such a discussion it shows once more that it is good to do some thing instead of only giving counsels.

Nick,

You are still saying that there is a very good probability that the distance could be outwards to 250' with your calculations?

Thank you taking the time to do these graphs.

Nick is using metric units (which does simplify calculations, so that's 250 meters, not 250'.

However, I think there is something wrong yet; I find it hard to believe that the heavy cable extends farther back from the boat than the lighter cable.

Hopefully, soon I'll be able to do some calculations of my own...

Where have you seen this ? I see it not: the heavier the cable the more it goes down and thus less length behind the boat!

I am glad some body else will make the effort to make a model and bring a solution, in fact it is what was asked for by Netmaker.

I repeat to make it clear I made simplifications since the problem is quite difficult to solve straight. I also am not sure about the drag coefficient which as you see in the lowest 2 pictures has an influence. there is an other aspect I neglected again for simplification. The angle the rope makes with the flow direction has 2 consequences: a) the drag is not normal to the rope so that it could lead to a portance force I did not estimate which will increase even more the angle since it is opposite to the weight.

I only made a correction (simplified again) considering the rope as full straight with constant angle and reduced the specific drag (N/m) proportional to the cosine of the angle.

As for the computations if one looks it is very simple and it takes half an hour to do it with an XLS sheet.

I am sorry but being in Europe I have a better feeling for metric units as you in the States have it for the other type so that every time I make the conversion in order to avoid errors. Do not forget that an accident was due to such mixed units usage.

I haven't seen it...

The total vector forces on the cables must, as you have pointed out, act tangential to the cable at any point. If there were no weight, the cable would extend out horizontally. The drag (horizontal) forces are roughly the same on both cables (though greater on the more nearly vertical one). Vertical forces pull it below horizontal (from the boat's perspective). More vertical forces MUST create a greater deviation from horizontal, not necessarily in a linear fashion.

"portance force"? I'm not familiar with that term.

I'll try to do some calculations this afternoon (currently 1PM) in Excel. Although I'm in and from the US, I do prefer metric. When I retired from teaching physics and started working in aerospace (1996), I was aghast to discover that they still use inches... You are probably referring to the failed Mars expedition, but I'm sure there have been other accidents from using mixed units.

OK, I took the time...

I have low confidence in the accuracy of my drag values, but high confidence in everything else.

I started with a free-body diagram of the ball. With a weight of 334N, and a ball drag of 6.7N, the cable must pull on the ball with an angle of 1.15° from vertical. I then made free-body diagrams for each 30ft segment of cable. The upward force on the top of each segment must be equal to the downward force on the bottom of that segment (the weight of the ball for the first segment) plus the weight of 30ft of cable. The horizontal force on the left side of each segment must be equal to the horizontal force on the right side of that segment (the drag of the ball for the first segment) plus the drag force of that segment. I assumed that the drag on each segment would be equal to the drag on a vertical rod of the same diameter and height as the height of that segment. I do know that is only a rough approximation.

I used the bottom angle to calculate a first-order approximation of the height of that segment, calculated the forces using that height, then used the forces to calculate a second-order approximation to the angle of that segment, then used that angle to calculate a second-order approximation to the forces and a second-order approximation to the angle and the ∆X and ∆Y.

This is then repeated 29 times to obtain the curve. This is all done in an Excel spreadsheet, which I'd be glad to share, especially if someone will take the time to check my logic and calculations. Here's the result:

Revision-004

Don't you have Day Job? Does your boss know you are doing Cartoon CAD instead of work?

Which ever one of these proves to be T-H-E correct one, I want to know if I have your permission to copy and send this to the customer?

Where is my Giant Squid in all of this? Anybody can find an octopus.

He's down in the pit.....we need a longer line...lol No happily retired....I'm bidding on Ebay, so I'm just killing time....Anybody can use any of my images....

Neither do I I'm retired!

I tried again using a different approach. Took moments about X, ACW due to dynamic forces on the sphere and cable, and CW due to the net weight in water of the sphere and cable (negative in the case of HDPE). Assumed the cable is straight, as I can't think of an easy way to calculate curvature. But for a given angle of the cable at X, the curvature decreases the distance from X to the sphere, while increasing the ACW moment because the length of cable furthest from X is more vertical, thus tending to increase the distance, so the 2 effects should partially cancel.

Results are

½" steel - angle 34°, horizontal distance from X 200'.

½" HDPE - angle 59°, distance 774'.

3/8" HDPE - angle 54°, distance 728'

Distance figure for ½" HDPE is close to dkwarner's, but we don't agree (yet) about dia of the sphere.

Thank you too for the work you put into this.

If you could simplify this for me, please.Using the boat as "0", what are the Depths at D1 and D2?Using the boat again as "0", what are the Horizon Distances of H1 and H2?
Basically, H1 and H2 will tell me how far behind this sphere is and D1 and D2 will let me figure an estimated depth .

Those values are on the graphs, but I know CR4 does not allow high definition images. Here is the bottom right of that graph, enlarged. My X's are your H's, and if you remove the negative signs, my Y's are your D's. The boat end is 0,0. PM me your email and I'll send you the Excel Document and a higher resolution copy of the graphs. The original graphs are to scale on both X and Y; I can't guarantee that after posting here, although they look pretty good.

Correct me if I am wrong.

Using my graph, D1 will be 527 feet? and H2 will be 771 feet?

No. X is horizontal (H1), Y is vertical (depth, D1). I'm recalculating, so I won't repeat any numbers at the moment.

I just looked again at your illustration. I had assumed (commonly a bad thing) that the numbers 1 & 2 referred to Ball locations, so ball location 1 would be H1, D1 and ball location 2 would be H2, D2. Now I see that on your illustration, one location is H1, D2, and the other location is H2, D1.

In any case, my figures were labelled using the standard mathematical symbols of X for horizontal and Y for vertical.

A few more comments:

I assumed that each 30 foot section of the cable was a straight line. I used a drag coefficient of 0.5 for everything. The ball is 5.8" or 14.7cm in diameter, based on a density of 0.285 lb/in^3 or 7.9 g/cm^3.

I don't know how to guarantee that the scales in Excel are equal in both directions, so the graphs were produced in CAD. For each cable, I drew 30 rectangles, all having a common bottom-right corner and each having the calculated width and height (X and Y values) for the corresponding segment. On a different layer of the drawing, I then drew a polygon starting at the bottom right and passing through the upper-left corners of all the rectangles. I then moved the polygons so their upper-left end was at the boat, and added the boat, balls, and text.

You can see that the general shapes agree with all of Nick's graphs, even though Nick still didn't get the horizontal and vertical axes to the same scale.

I used different Cd values but results are converging if values will be the same

my last graphs show how it does change the form. You confirmed my computations

Thanks

Nick

PS.: How big is the ball in reality ?

Nick,

Customer has not shown us the weight, only given us an estimated weight and that it is spherical.

Can't agree with your ball dia. I make it 7.95" or 20.09 cm

I'll double-check it. I'm working on the 3/8" HDPE as well.

Of course you are right! Somehow I ate the π when solving V=4/3πr^3 for r, so I was off by the cube root of π. Must've been hungry!

I'm redoing it all...

I'm in agreement with the 7.95", but make it 20.2 cm (keeping 3 digits precision).

Yes you're right. 20.185 cm by Mathcad, which is the figure I used in calcs. Another typo!

BTW who marked my #87 off-topic? Seems well on to me. There have been quite a few comments on this, but few of us have come up with actual figures!

That sometimes happens when the NSA starts data mining the thread and confuses the software into thinking the remarks are OT.

Final....

That is a lot further than I ever thought with just a 3+ meter /sec current.

It IS what it IS though and you can't change physics.

Another OOPS. I thought it was 1.34 m/sec, which is close to 3 mph.

It is, from the OP:

A current of 1.34 meters/ second is pushing on both

YEP..... a 3 mph current ( average) or 3.5 nautical knots

You multiplied instead of dividing by the conversion factor (or vice-versa).

3.0 mph is 2.6 knots.

Not to get too picky but 1.34m/sec is 2.6 knots.

There's an echo in here.

.......that's why I leave the math up to you all.... I went the wrong way.... ha ha!

Everyone I know, including myself, needs their math checked by someone else. I just realized that I hadn't taken buoyancy into account, and I hadn't thought that it would be sea water. I'm doing that now.

OK, this has to be it for a while! I have corrected my ball diameter and its drag (thanks to Cody), corrected another small numeric error, and compensated for buoyancy and the density of sea water. All calculations use the acceleration of gravity for New Orleans: 9.792 m/sec^2. I did NOT take into account the lift on the cable due to its angle in the water, so all of these depths are at least slightly greater than the probable real value. This of course means the horizontal values are at least a bit short. I was quite surprised how close the 3/8" rope came to the 1/2" rope.

The editor would not let me post a high resolution JPG image, so here's a lower res. one: I think I've got the text large enough to be readable.

Good. This will be accurate enough for my purposes. thank you and everyone else who worked on this.

I can make adjustments as required as long as I have a reasonable starting point.

You're welcome!

I meant to point out that the steel cable will be pulling down on the boat with nearly a ton of force (1900lb), and pulling back with a force of 1360lbs, for a total force of 2340 lbs. It's amazing for only 3mph!

For the 1/2" HDPE, those values are 268 down, 835 back, and 877 lbs total.

If anyone sees another flaw, do let me know!

Dick

Thank you. This is more than enough information to make rigging decisions . Yes, water force is amazing. ......its why I invest so much time in net design using the high tech nettings and lines.

However, for my engineering calculations I have learned to come here........ as long as the folks on CR4 don't mind my abusively strange requests for information.

Thank you all.

It does seem excessive....but accumulative effect from 900' of line, weight reduction due to water displacement, water density...I guess the only way to resolve this is a real world test(I'm betting on the 250' scenario)...Maybe we should take everyone's guess and average that....lol

About HOW FAR UP from the weight would you guess THIS curvature area is?Just approximate the distance up, a good guess... within say, 10-20 meters + or -?????

I think the assumption of constant uniform force of current is incorrect....I think it is more feasible that the current will vary at different depths and this will skew the model...Although these stylized models look great, they don't always represent reality....that curve, I would guesstimate at 150' from the ball of doom....

IF the current is constant at all depths down to the ball, The curvature is greatest at the attachment to the ball; the distance from the ball to the curve is zero.

On the other hand, Solar Eagle is absolutely correct that in reality such uniformity is unlikely. In that case, there is no precise way of predicting the location of the ball without knowing the motion of the water at all depths.

Yes quite so, this illustration shows a stronger current from surface to about 600' depth....

The one thing I have learned over the past three decades is that "everything" looks good on paper. This scenario is a model only. There will be cross currents but they will change every day, twice a day in some cases. This will be impossible to track. However, if I have a number, even an average number, I will be able to guestimate where everything is.

Thank you all. I never dreamed this was such a task, but obviously it is. I would have thought maybe, 10-20 meters at most from the stern of the boat. Go figure , eh? right?

This sure looks like homework. Have you enrolled in some night classes Netmaker?

We've heard from Netmaker before, and I'm sure this is not homework--except maybe the part of your occupation that you have to do at home.

..... homework was never as hard as the Real world! But not as much fun either. I enjoy what I do and the stranger the request the more intrigued I am in the job. Its why I stopped making batting cages and soccer nets...too dang boring!

I'd drive an Engineering Professor to madness with some of the crazy @$$ scenarios I have to deal with on a monthly basis! But CR4 always pulls me through. ha ha ha

Here we go with another obvious homework question. This is the third time this week that some young student has posed this exact question.

Ha Ha Ha! Thank you all for the compliment.

Ever see a student THIS ugly???

Once again, CR4 comes through on "homewerk" "homework".