UKC

If I was fishing and I had caught a fish, I took in 1m of line, does the fish move more than/less than/exactly 1m towards me?

Assume the fish stays on the surface of the water.

The combined brains of our Maths and Physics teaching department failed to prove a solution last night, I got there just before I went to sleep. I thought it was a nice puzzle, fairly easy to figure out what the solution is, but tricky to prove, although possible with A level maths, with a very pleasing final step (at least how I did it).

*how about if it was on a treadmill?

In reply to ablackett:

Not explained very well at all I'm afraid and very little info to go on.

However I think your talking about displacement that can be solved using vectors.

M

In reply to MGC:

I'm afraid it's explained plenty well enough, I just want to know how much closer is my fish if I take in 1m of line.

In reply to ablackett:
Hi, are you considering the stretch with the line (monofilament line has a varied dynamic) and fishing rods have different actions I.e tip, medium and through?
I'm an angler more than a mathematician.

In reply to ablackett:

If using a fishing rod, pythagoras would kick in - so less then 1 metre - and would depend on the angle of the rod. In reality, you could take in zero line and the fish move considerable distance 'kiting' across the water.

Whats your answer?

In reply to ablackett:

You haven't given enough information (angle of rod, does it bend, does the line stretch), but I'm going to go for a "bit" closer.

In reply to mick taylor:

It's a mathematical fish, so it doesn't go kitting across the water, the line doesn't stretch, indeed I am using a fishing rod.

In reply to ablackett: How tall are you and how is your rod of unknown lengh inclined. I am assuming that you have caught the fish in still water rather than a current that is dragging, bending the line

In reply to Sir Chasm:

Exactly what I was getting at.

No angles , expects an sensible answer with no info other than 1 meter and a fish .

M

In reply to MGC:

"If I was fishing and I had caught a fish, I took in 1m of line, does the fish move more than/less than/exactly 1m towards me?"

Assuming that the situation is hypothetical, that is, you ignore stretch in the line and bending of the rod and restrict movement of the fish to that caused by the taking in of fishing line and assume that the surface of the water is a static plane, the question is still silly.

Either it is trivial: given some unknown parameters such as the length of the rod, angle of the rod, height of "me" above the surface of the water, it's a simple case of trigonometry: define the triangle given by the fish (approximated as a point), "me" and the static tip of the rod, reduce the length of the edge between the fish and the rod-tip by 1 metre and determine the length of the edge between "me" and the fish relative to the original length, according to the geometry of the triangle.

Alternatively, it's a trick question and you're waiting to reveal something - however obvious - that wasn't stated in order to prove how "dumb" everyone is.

In reply to ablackett:

The fish moves exactly 1m toward you.

In reply to Xharlie:
>

Unless he's either lying flat on the ground or holding the rod vertically above himself it won't be a triangle - rather a four sided shape. Anyway, the fish will move somewhat more than the 1 metre.

In reply to graeme jackson:

Agreed and you don't need A level maths, late nights or elegant solutions to demonstrate this

In reply to ablackett:

The fish doesn't move towards you when you take in line it moves towards the end of the rod (assuming the line is already taut and you aren't just removing slack without moving the fish at all).

Whether moving towards the end of the rod involves moving towards you depends on where the end of the rod is and where the fish starts from. If the fish started very close to you and the rod was held almost horizontally the fish could even move away from you.

In reply to ablackett:

I normally just pull the fish straight in and Mrs Num Num stuns it.

In reply to ablackett:

reckon my theory was right, but application wrong. Its more than 1 metre.

In reply to ablackett:

The diagram is like this - rod tip is X distance above water and doesn't move. Distance Y between the fish and the rod tip decreases by 1m to become distance Z (Z=Y-1). We assume you know the distance Y, therefore you can find out distance Z (Z=Y-1). Using pythagoras you can work out the original distance to the fish from a point directly below the rod tip using distances X and Y, and also the new distance using X and Z. Then you can work out how much closer the fish has got.

In reply to mick taylor:
If the fish stays on the surface and moves horizontally it will be less than 1m as the line is the hypotenuse of the triangle and the surface of the water the adjacent?

In reply to graeme jackson:

I did consider that but, for the purposes of the calculation, it is a triangle between the fish, "me" and the rod-tip. The rod and "me" don't need to create a straight line but, once the distance between "me" and the rod-tip is found (and the angle between that imaginary line and the plane of the water), all other geometry is irrelevant.

One must remember that pulling in line does not necessarily move the fish towards "me" - the answer depends on the geometry. For example, if the fish was actually closer to "me" than the rod-tip, pulling in line moves the fish away from me. However, if the fish stays on in plane of the water, it does not move 1 metre towards the rod-tip unless the rod-tip is also in the plane.

In reply to ablackett:

If your assumptions simplify the question as much as people here have done, the treadmill is irrelevant too. Had the combined brains of your maths and physics dept's had a few beers before the question was contemplated? My mates and I once spent about 10 hours trying to figure out how a light circuit with two switches worked, which is similarly simple if you know a bit about circuits.

In reply to ablackett:


It's not explained anywhere near well enough. It's not even explained well enough to work out which part of the fishing mechanics/geometry it is you're interested in, is it: The rod/line/man-fish triangle geometry, including or excluding deck height (or waders?) and water surface curvature? Where are the reference points in 'fish' and 'man', does the geometry include the catenary curve of the line and the rod flex or is the line light and inextensible?

If you answered your question other than to assert 'possibly' it's with more far than you've provided in the OP.

jk

In reply to Toerag:

Does this assume that the fish stops moving on the surface of the water after the line has been reeled in? The OP doesn't tell us that the fish applied any braking, therefore, going on what we have been told, and applying the physics of the world in which we live when applying energy and locomotion to a body on a horizontal plane, the fish will continue to move more than 1 metre. Unless of course the angle of the rod is such that the horizontal movement is restricted to below one metre. Too many unknowns to answer this accurately IMHO.

In reply to Xharlie:


Only if the rod doesn't bend under load - which it would.

In reply to ablackett:

Surely you need to know the distance from the fish to the rod tip initially? Or are you looking for a function, rather than a number?

In reply to ablackett:
More than due to the line on the spool amplifying the 1m?

In reply to jkarran:

Maybe OP is looking for a general solution. From a bit of scribbling, assuming fish is on surface, rod tip is static, and line has no stretch:

dx = (x_init*cos(theta_final)-L_final)/(cos(theta_final)

Could be wrong, I'm late for work so a bit rushed. Deffo could be converted to all initial terms fairly easily.

The man is irrelevant if the rod tip is static.

In reply to ablackett:

Perhaps I have done too many A level mechanics questions, but I assumed that people would make some assumptions such as inextensible string and a rigid rod. Also assume no side to side movement. I solved it first by making assumptions as to the length of rod, string, etc, then as a general solution with constants for initial fish distance, rod height and variable for string length.

I feel like I have just set this to a class and they have all declared it impossible, or decided they don't get it before giving it any thought.

In reply to ablackett:

The fish doesn't move closer to you, but you move closer to the fish

In reply to ablackett:


I think you should feel like you have assumed too many assumptions.


Edited to add: "Solved" it based on "making assumptions". Really?

In reply to ablackett:

Yes, everybody else is at fault.

In reply to Johnny_Grunwald:



Agree that OP made too many assumptions, without detailing them. If assumptions are required for a solution they should be spelled out in the question. However there are many engineering questions which are truly solved by starting with assumptions. They usually involve iterative calculations though.

Assuming the line breaks when you reel it in, the fish moves no closer to you.

In reply to ablackett:

Let's see your solution then.

In reply to ablackett:
If the angle between the line and the water does not equal 0 (i.e. the rod does not sit level with the surface of the water) then the fish moves less than 1m towards you.

I thought that was common sense but if not:

Triangle with hypotenuse as length of line reeled in (doesn't matter if this is before or after rod tip or both) and adjacent side is distance across water surface. Cosine of this angle is adjacent side divided by hypotenuse. Cos 0=1, Cos(>0) is less than 1 therefore hyp*cos(angle) is less than hyp length.

Dave Kerr - pretty much what you said

In reply to ablackett:

Am I being really stupid or is this not really really easy and just involves a manipulation of the cosign rule?

In reply to Sir Chasm:

Nobody is at fault, I'm surprised by how people have reacted. To solve the puzzle you have to make assumptions. It's up to you what assumptions you make, it's up to you if you have a go at it.

In reply to A Longleat Boulderer:
No. You're not being stupid.

However I am concerned at the quality of the entire maths and physics department at a school somewhere.

In reply to ablackett:

And what if you were on a treadmill?

In reply to Sir Chasm:

Okay, really not enough information here but assuming you want an idealised geometry problem we can ignore the finite size of the fish and the man and any flex, extension or curvature or momentum of anything involved. I can see two possible versions of this problem, both of which have been solved correctly above.
1)
If the fish is 'anchored' to the surface of the water and your rod is above the surface the fish will move toward spot on surface of the water directly under the rod tip. The amount the fish moves along the surface depends on the angles but it will always be more than the amount of line taken in. Assuming that the man is sat att eh water line, then the answer is 'more than 1m' as long as the fish remains further away than the end of his rod (otherwise it would move away from the man).

2)
If the fish is not anchored to the surface but simply floats with neutral buoyancy within the volume of water, then the fish will move 1m toward the top of the rod. Unless the man, the rod tip and the fish are perfectly aligned this will always result in a movement of less than 1m towards the man.

In reply to edordead:

Wrong . try drawing it on a piece of paper. mark a baseline - the distance between you and the fish. Draw in the rod and line making the line e.g. 7cm. then draw in a 6cm line from the tip of the rod and you'll find that when it intersects the baseline it's more than 1cm back towards you.

In reply to Oujmik:

from the OP..

"Assume the fish stays on the surface of the water."

In reply to ablackett:

Less than - you have caught the fish and it is not moving, as it sitting in your net. If you had hooked a fish you would be in the process of catching a fish.

Using your drawing, this is true steep angles but not at shallower ones. Drop the height of the rod tip and it's the opposite way around.

In reply to ablackett:

He says he has caught the fish. That means it's on the bank. Taking in 1m of line doesn't bring it any nearer.
If you mean hooked fish, that's different

In reply to ablackett:

If the tip of your rod was 4m perpendicularly above the water and 5m of line was between the tip of your rod and the fish. If you reeled in 1m of line, then the fish would move 3m towards you.

In reply to ablackett:
As any angler can tell you it isn't your fish until you land it. The fundamental premise of the question is invalid.

In reply to Xharlie:
Assuming the usual 'applied maths' rules of everything inelastic, water surface is flat, 'distance to you' is horizontal distance from fish to the vertical point under the tip of the rod, line perfectly straight, then consider the following

let h be the height of the rod above the surface of the water
let L be the length of line between tip of rod and fish
let r be the horizontal distance from fish to the vertical point under the tip of the rod

then, for any L and h, we can find r by Pythagoras

r = sqrt(L^2 - h^2)

So, differentiate r wrt L to find the rate of change of r with L (dr/dL). I'll leave that as an exercise for the student.

Or use excel calculate r for every L between max L and h (when the line would be vertical), and find the deltas between each step.

When L is large, delta r is just a little bit larger than deltaL (delta r tends to deltaL+ as L tends to infinity)
When L is small, delta r is large, tending to [erm, possibly infinity * deltaL] as L tends to h.

[lots of edits to fix errors]
[and I replied to the wrong person...]

In reply to ablackett:

Assuming the fish is a shark, you're gonna need a bigger boat.

In reply to ablackett:
<Sets vague question>
Then

You were vague ambiguous and imprecise in your question.
You bordered on rude in responding to those who pointed it out.

But, hey, don't worry the problem is everyone else.

If what you actually mean is that you are assuming a right angled triangle where the hyponenuse is shortened by 1 unit, the vertical is fixed and all lines are stright and inelastic, and asking what happens to the third side, then the solution is trivial (pythagoras) and I can't see the debate.

If there's a particular subtlety that you want considered (and several have been suggested), then you're going to have to be more specific.

In reply to JJL:

If the fish starts at infinity then everything else can be ignored and you get 2 answers:

1. the fish is 1m closer (zero stretch line).
2. the fish is unmoved (non zero stretch line).

I suppose there is a negative stretch line answer, but the fish would have to travel faster than light.

In reply to ablackett:

Look at you all, very rough answers, nothing about weather and sea conditions and curvature of the earth etc etc.
And for those doing courses not related to maths and science her are your alternative questions:

1. How do you think the fish feels.

2. Draw a picture of the fish.



Toby

In reply to Toby_W:

Not all of us subscribe to the spheroidal earth hypothesis.

In reply to ablackett:

So another conundrum for you teachers out there. If you set your class a maths problem and you make the following statements, will you be surprised if you get a very mixed bag of reactions (and solutions)?

“The solution is easy, but tricky to prove, although possible with A level maths.”

“It's explained plenty well enough”

“I assumed that people would make some assumptions”

“It's up to you what assumptions you make”

In reply to ablackett:

If someone poses a question in terms of real world objects like fishing rods and fish and the question is inadequately constrained the only correct answer is "I don't know". Fishing rods could easily bend by a considerable fraction of the 1m the line is being wound in. Live fish are not passive: they could swim in any direction. You can't just assume an initial location of the fish which makes the maths easy without any physical justification for assuming the fish to be there.

Making a bunch of assumptions that don't match the real life situation for no other reason than that they allow you to apply a mathematical formula is just wrong. 'I don't know' is far superior to producing a numerical or algebraic solution you have no reason to believe is correct and it shows more engagement with the physics of the problem.

In reply to ablackett:

A few thoughts.

Is the fish or fisherman on a treadmill? What have they done on grit? Is Kinder in yet, it was frosty last night?

In reply to ablackett:

My answer for what it is worth.

The fish will move less than 1m towards you*.
I cant provide an equation to prove this, just seems to be the case in my muddled head.

* assuming the rod is between 1 degree to 44degrees to the waters' surface.

In reply to tom_in_edinburgh:

Everything you say is true, but it doesn't help me figure out if my fish is more than/less than or exactly one meter closer to me, which was the initial question.

In reply to ablackett:

Is it an European or African fish?

In reply to ablackett:

Have you not figured out yet that the answer to your question depends ENTIRELY on your assumptions? In cases such as that the assumptions must be listed in the question or the question is invalid.

Assumption: at the same time the rod moves 1m along the hypotenuse towards the fish.
Answer: less than 1m.

Assumption: fish happens to be swimming towards the angler.
Answer: more than 1m.

What a stupid question.

In reply to Bob_the_Builder:

OK, cheers Bob, I made a couple of different assumptions and solved what was a very pleasant algebraic geometry problem, but i'm not in the least bit worried if you choose to approach it in a different way.

In reply to ablackett:
As rightly pointed out, when I gave my two solutions I forgot that you'd already said the fish remains on the surface. In that case, how about this...

Line length^2=Horizontal distance from rod^2 + Height of Rod above water^2

I shall denote these as L, X and Y respectively

So, rearranging

X=(L^2-Y^2)^0.5

You are interested in the rate of change of X wrt L, so we differentiate

dX/dL = L/((L^2-Y^2)^0.5)

You can prove that this is always greater than 1 (or less than -1) because the denominator can never be of greater magnitude than the numerator.

In reply to ablackett:

How about if I state it like this.

ABC is a triangle, with Angle ABC being 90 degrees,
the distance between A and C is x,
point D is on BC being a distance y from C and a distance (x-1) from A,
is y>1, y<1 or y=1?

I personally think it looses something if I state it like this. But it certainly doesn't bend, stretch, go to infinity or swim away.

In reply to ablackett:

OK, greater than. If you want a proof that the sum of any two sides of a triangle is greater than the length of the third side, then I'd be tempted to use the cosine rule if you count that as a proof?

In reply to Toby_W:


Having helped my niece with her maths assignment a few years ago, those approaches seemed to be required for the answer. Unless it was my niece wibbling instead of using precise mathematical language... I think she wrote a 5-page answer, whereas my 'model answer' was about half a page of succint maths...

In reply to Oujmik:

That's cracking, I solved it without calculus, but I do like what you have done there.

I think that I posted it on here in the hope that someone might enjoy it, think about it and allow me to think about the problem in a different way. Which appears to have happened now.

Thanks.

In reply to jonny taylor:

I knew you would do that. It's bloody obvious that the sum of the 2 sides is greater than the third so y + (x-1) > x, so y>1.

I'm suitably embarrassed that I hadn't spotted that. It came from a poster encouraging people to take Further Maths, my solution was probably worthy of a Further Maths standard, where your solution is probably worthy of an infant school but equally valid.

In reply to ablackett:

My fish is 7.5 hours closer than when the thread began

I shall be having it with chips and mushy peas

In reply to ablackett:


It might be, but I'm ashamed to say that it took me a good 5 minutes to actually prove that first statement using the cosine rule. There's probably a more elegant way, but I couldn't immediately come up with it.

In reply to ablackett:

Doesn't the exact answer depend on where the tip of the rod is compared to the surface of the water?

In reply to ablackett:
An easier (but equivalent if you rearrange with some trig rules) solution is just to look at the triangle of vectors

The fish has a speed along the surface of X' this is related to the speed of the line being taken in L' as

X'=L' / cos(theta)

Where theta is the angle formed by the line and the water's surface. Since the domain of the function cos is +1 to -1, X' will always be larger than X' except when theta=0 which requires an infinitely long fishing line.

In reply to ablackett:

The maximum distance the fish could possibly move horizontally toward you is 3 metres

In reply to Oujmik:

I'm sure you are right here, but i'm not following how

X'=L'/cos(theta)

In reply to ablackett:

Are you familiar with the idea of resolving vectors into two orthogonal components (normally the x and y of your chosen coordinate system)? It's exactly the same as the triangle of forces you might use to calculate load on an anchor.

In reply to ablackett:


The correct answer to the question is 'it depends' because the problem is inadequately constrained. That answer and reasoning behind it (e.g. dependence on initial location of the fish, slack in the line, flex in the fishing rod and motion of the fish) is a step towards a practical solution for a real world problem. Inventing constraints until a particular formula can be applied and generating a result based on the invented constraints doesn't solve anything in the real world.

In reply to ablackett:


This is not algebraic geometry. What is it you said that you teach?

In reply to edordead:



Bugger me that's twice in a week I've been right about something on here. Lucky I was sitting down.

In reply to tom_in_edinburgh:

It is interesting to consider if mathematics has every been used to solve a real world problem without assumptions and constraints. I can't think of any examples.

In reply to planetmarshall:

I understand it's not algebraic geometry, it's a geometry problem which I solved algebraically. I spend most of my time teaching Year 7's how to add fractions, so it's not too important what algebraic geometry is, forgive my mistake.

In reply to ablackett:


That's probably true but then the assumptions and constraints would be stated. Yours weren't. Which is why folks got pissed off.

Have you told us what your answer was?

In reply to ablackett:

Returned home from fishing all day, we caught plenty and they all came in. Thank god I didn't read this before I left this morning, it may have spoilt my day attempting to measure line every time I had a fish on.

In reply to ablackett:

This is a question of grammar not maths, the fish doesn't move because you caught iy before you went fishing.

In reply to DaveHK:

It appears there are several much simpler ways of looking at this, but here is what I came up with.

I set it up as a right angled triangle, ABC, with the right angle at B, height a, fishing line length z, fish distance (x+y), changing to fishing line length (z-1) and fish distance x, question is if y>1.

From pythagoras,
z^2 = a^2 + (x+y)^2
(z-1)^2 = a^2 + x^2

eliminating a^2 gives
0=y^2 + 2xy - 2z +1
completing the square gives
0=(y+x)^2 - x^2 - 2z+1
rearanging
sqrt(x^2 + 2z - 1) - x = y

Then noticing that
z-1 >x so z>x+1

so
sqrt(x^2+2x+2-1) - x < y
sqrt(x^2+2x+1)-x<y
(x+1)-x<y
1<y

so the fish moves more than 1m.

In reply to ablackett:

FFS! Just go to the chippy!!!!
By the time you lot stop f'in about, it would be cold!

In reply to ablackett:

Folks here are just obviously not eating enough fish.

The maximum the fish can move horizontally toward you is three metres. illustrated by the Pythagorean triplet 3, 4, 5. If the tip of the rod is four metres above the water, and there is five metres of line between the tip of the rod and the fish, then the fish is three metres out from being perpendicularly below the tip of the rod. Reel in one metre of line, then the fish will move horizontally to a position perpendicularly below the tip of the rod, i.e.. three metres. Change any one of these parameters the height of the rod tip, or the length of line, then the fish will always move toward you less than three metres.

In reply to birdie num num:



Only if you assume the fish is dead and the water isn't moving. If the fish can move without being dragged by the line it could move past the point directly under the tip of the rod and nearer to you so the case where it runs out of line directly under the tip of the rod doesn't give the maximum distance towards the fisherman.

In reply to tom_in_edinburgh:

You're inserting variables that were not in the original question. I would recommend sardines for tea next week.
And a daily cod liver oil tablet.

In reply to birdie num num:

Big fish, small fish, cardboard box.

In reply to John Simpson:

That could be the correct answer.

In reply to ablackett:

As I would say to the kids

Where is your diagram? !

In reply to birdie num num:


Dead fish don't bite. When a question starts "If I was fishing" then it is reasonable to consider the normal case of live fish.

In reply to captain paranoia:

A picture is worth a thousand words &#128515;

Toby

In reply to ablackett:

Mathematicians (and teachers apparently?) try to make contrived questions more interesting by assigning them a scenario which requires a bunch of ignoring reality to make sense. Engineers try to make real scenarios solvable by ignoring reality to make the questions make sense. Its a funny world.

Your triangle is an acceptable phrasing of the question. Thank you.

In reply to ablackett:

In all probability it will try swimming towards you to release the tension in the hook, so it will travel more than 1m towards you regardless of what you do with the line.

In reply to ablackett:
Then consider the scenario where the fish is on the surface of the river directly below the end of the rod and the rod Tip is exactly a metre above the water. The fish will move exactly 0 metres towards you, it will just go up vertically to the rod tip.

In reply to ablackett:
Now consider the rod tip a metre off the surface . There is 2 metres of line out and it is taut. The fish at this point is closer to you than the rod tip. The fish moves away from you as the line is drawn in, till it is exactly 1m below the tip of the rod. So the fish moves away from you.

In reply to ablackett:
Now consider the rod tip is on the surface, 1m of line are out, and the fish is in a direct line with the rod and yourself I.e. There are no angles. The line remains taut. The fish moves exactly 1m towards you, and one metre towards the rod tip.

In reply to ablackett:


Now consider the rod tip is on the surface, 2m of line are out, and the fish is in a direct line with the rod and yourself I.e. There are no angles. The line remains taut. The fish moves exactly 1m towards you, as you take 1m of line in. When you stop reeling the fish keeps travelling towards you, due to the conservation of momentum (Newtons Laws). So in this scenario the fish moves more than 1 metre towards you.

In reply to ablackett:
So given your conditions and your assumption the fish is dead and the line is taut (it it is alive or the line isn't taut all bets are off as to how far it moves towards you).

The fish can move less than 1m, exactly 1m, and more than 1m metre and still satisfy the conditions you gave. So the answer is all three.

In reply to ablackett:
Interesting to see that some people can see exactly what the implied assumptions must be to reduce it quickly to the geometrical problem intended, and other people can't/won't/don't want to, but still want to type an humorous response.

Edit: *unhumorous response. Autocorrect. Nothing against the humorous responses!

In reply to ablackett:

Does the fish move at all if no one from the internet was there to see it?

In reply to Jon Stewart:

No, we can consider different scenarios, which satisfy the given conditions, and give different outcomes. We do not need to make assumptions. His stated conditions do not automatically lead to a single outcome, and hence why his physics and maths departments disagreed on the answer. There is no single answer. This is why others on the thread have also stated that his question is ill formed. I have to agreed with them. If you were marking this in an exam, and the student gave any of the possible answers, and showed why, you'd have to give them full marks.

In reply to Orgsm:

Perhaps if you marking an exam. But the OP described it as a "puzzle" and that word gives you clear direction on how to approach it. Consider the real-world engineering? Not a puzzle. Find special cases that don't apply to assumed geometry implied by "fishing"? Not a puzzle.

The point is not to find an accurate answer, but to work out what the puzzle is and then solve it. Clearly quite a good puzzle as there were a good few wrong answers prior to Oujmik getting it right. I thought Num Num's 3-4-5 triangle answer was good too.

In reply to ablackett:

I watched the movie 'Big Fish' once, and I liked it.

In reply to Removed User:

what a shit thread

In reply to Seocan:

It ain't my fault.

The answer to original question is trivial, no real maths required.

In reply to ablackett:
No maths used , as I imagined myself in a field, I hooked an old boot 100 m from a rod (held at 45 degrees ) I reel in 1 m and measure ( in my mind) how much nearer my boot gets.

I reset at 50m and then 25 M , in each case I should get a different answer as the angle of the line ( not the rod) changes with distance.

So in my mind there is no point in doing any calcs, as I agree there is no correct answer as it was stated.

Or am I an idiot?

(Forgot to say , if I point the rod directly at the boot( zero angle) then no matter what the distance is ( beyond 1m) then the boot will move 1 m closer- in my mind at least)

In reply to marsbar:


Cheers, it seems to me that the thread has now floundered, time for a tune maybe

youtube.com/watch?v=AEKbFMvkLIc&