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Marker rod/fishing rod, clip at same distance?


Miltz308
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Still waiting for cm to explain how he suddenly jumped from 2ft to 8ft ..... as I'm sure most of us are especially Androoooo.

So come on cm ! Nige is right in saying you have completely contradicted yourself on this thread .

 

Also agree with Nige this was a good thread last night & what a forum is for 8)

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CM has dropped me a message with some details on how he got there. I need to work it through and see what I get.

 

I've said, my theory with pythagoras was similar but as with all theories if they can't be replicated they are wrong.

 

Personally I think it probably is due to a differential on drag between air and water. Which would be really quite difficult to calculate and so a rule of thumb may be better.

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This 'rule of thumb' you mention, Androoooo - how long is it? - does it have a knuckle? Is there a black nail on the end? Is the nail clipped? More importantly - is it your thumb or that belonging to somebody else. I ask the latter question due to it possibly changing to an index finger by the end of the thread if it doesn't belong to you. :wink:  :mrgreen:

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This 'rule of thumb' you mention, Androoooo - how long is it? - does it have a knuckle? Is there a black nail on the end? Is the nail clipped? More importantly - is it your thumb or that belonging to somebody else. I ask the latter question due to it possibly changing to an index finger by the end of the thread if it doesn't belong to you. :wink::mrgreen:

Turnip... Nige's experience seems to confirm that 3 to 1 is near enough, especially for a fishing situation.
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I wasn't pretty close, I was spot on! You can theorise all day long but getting out there and doing it is the only way to see if that theory works. Which with you saying 2ft swing back in 20ft of water, doesn't!

I even went out in the boat to see the distance side on, which makes a massive difference in perspective than looking at it from the bank!

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Casting distance and fall back on a tight line.

The maths in this can get quite confusing but it isn't with a few diagrams to see clearly (I hope).

I'll make a few assumptions as there are many outside influences, but the idea here is to get an idea of the affect fall back has on where a bait lands in relation to where it enters the water (i.e. compared to where sprockembs [spod rocket spomb etc..?] are cast).

Assumptions:
How long your line is when it hits the water.
How deep the water is.

I'm going to use metres as I understand it better and it's easier to follow than yards, feet and inches since there is only unit to follow, But I will indicate approximate feet again rounded.

We are casting from near enough (so's not to get feet wet!) water level.

The depth of water is 6 metres (20ft rounded up)

Height of the cast is worked out based on roughly where I think the butt of the rod is when I cast out and I'm rounding it to keep things simple - 1.4 metres, then add the length of the rod 3.6 metres (12ft) gives us 5 metres from the ground at water level to the tip of the rod.

You have clipped up at 30 metres (98ft).

We now know the height [A] and length [C] and can therefore work out the distance from casting to landing in the water.

6Y5IuIR.png

Hard part alert, since we have a right angled triangle we can use Pythagoras

formula c2 = a2 + b2
transforms to b2 = c2 – a2

A=5 B=? C=30

(30x30) – (5x5) = 900 – 25 = 875 rooted = 29.58m i.e. 0.4 metres shorter

Phew, that was easy!

After the lead has sunk, it will be closer to the bank since it was held on a tight line (assumption), for this I am going to assume there are no undercurrents or dodgy gravel bars interrupting a tight line.

YC80TRB.png

Now we're in trouble, we no longer have a triangle. Not really, we can draw a new line to the tip of the rod [line A]

v8TWBjN.png

Trouble again, we don't have a right angle triangle to use Pythagoras on. However there are a few solutions to this, the one I have chosen is below:

A6C6iXa.png

We know C (length of line 30 metres) and we can work out A by adding tip of rod height to depth of water = 5 + 6 = 11

A = 11 C = 30

(30 x 30) = 900 – (11 x 11) = 779 metres rooted = 27.91 metres

This gives us a difference of 29.58 – 27.91 = 1.67 metres (5 1/2 feet)

So let's play with a few more extreme numbers

Length of line 100m
Height still 5m
Depth 10m

Length cast = (100x100) - (5x5) = 9975 rooted = 99.87m
Fall back = (100x100) – (15x15) = 9775 rooted = 98.87m

Difference 99.87 – 98.87 = 1m

Removing most of the calculation
Length of line 12m
Height still 5m
Depth 5m

Length of cast = 144 – 25 rooted 10.9
Fall back = 144 – 100 rooted 6.63
Difference = 4.28

I've done a table showing a few results using a spreadsheet (LibreOffice calc).

nrj25pw.png

I hope that this is all clear and above all correct :oops:

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Please check the above, and if it's way off then feel free to delete it.

 

Having spent some time trying to think about this with what Androoo had mentioned in a post earlier with other pertinent comments from CM Nige, Phil and Nick (sorry I've lost track of who else but you know who you are :oops:).

 

I think that this can put it to bed - it needs two calculations (Androoos two triangles) or gps trackers.

Edited by ianain
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Kit

 

Can some kind moderator go through my post and change the relevant 2 to superscript 2 as in C2

 

i.e. this bit [should be able to copy and paste]

 

formula c2 = a2 + b2
transforms to b2 = c2 – a2

 

Oh and delete this incriminating evidence :oops:

 

And this: This gives us a difference of 29.58 – 27.91 = 1.67 metres (5 1/2 feet)

 

to this

 

And this: This gives us a difference of 29.58 – 27.91 = 1.67 metres (5 1/2 feet)

 

Ta

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Ianain, I think it's wrong based on Nige's experience.

 

I think it would be right if all of this happened in a vacuum. It doesn't.

 

I think to work it out, you'd need to look at fluid dynamics to account for the different drag between water and air. It would get incredibly complex.

 

Nige's experience suggests the 3 to 1 rule is near enough and I'd happy with that in a fishing situation. I wouldn't be as stupid as to imagine 1. Hitting the exact spot everyone. 2. That surface currents wouldn't move my spod before it emptied. 3. That everything would descend directly horizontally.

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Remember Nige was asking for the distance between two marker floats, they create a 90 degree drop, Pythagorous kicks in, no scalenes, i factored in line stretch and other factors, i left it for Androooo to look at, i came up as the same as Nige but pure calculation based on averages, too many variables to be exact, the falling weight will contract back with the line stretch degrading at an exponential rate rather than a pure radian, factor in the energy of the cast and the weight of the lead and unles every cast is the same and all materials have the same properties, it is impossible to put an exact value on the distance.

it wasn't yesterday though cm you said 2ft drop back in 20ft of water ?

It would appear I wasn't as far out of my depth as you thought :wink:

 

20150607_135839_zps3yn5uqor.jpg

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I think it would be right if all of this happened in a vacuum. It doesn't.

 

I think to work it out, you'd need to look at fluid dynamics to account for the different drag between water and air. It would get incredibly complex.

 

Yes I was thinking more of leaving environment out of it and keeping it simple. Will look back through the thread as I thought this was around what Nige uses.

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Nige's experience suggests the 3 to 1 rule is near enough and I'd happy with that in a fishing situation. I wouldn't be as stupid as to imagine 1. Hitting the exact spot everyone. 2. That surface currents wouldn't move my spod before it emptied. 3. That everything would descend directly horizontally.

 

Yep, my calcs give 0.8m (2.6ft).

 

To my mind though 3 - 1 can't work at all distance and depths :confused:

 

At least I tried :hrumph:

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On the bank, actually fishing, 1 in 3 seems to work fine. It also seems to be near to the theory that some have stated on here. Phil was branded as being out of his depth for suggesting the 1 in 3 guide though. Only for it to be pretty much what the outcome had been with all the complex maths! Lol

Closer than the 2ft in 20ft of water that was being suggested last night! Lol

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Yep, my calcs give 0.8m (2.6ft).

 

To my mind though 3 - 1 can't work at all distance and depths :confused:

 

At least I tried :hrumph:

It can't, but it's a good starting point with all the other variables involved!

The only true way of getting an exact amount of swing back is measuring it n each specific situation. That's where my two marker float technique comes in!

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On the bank, actually fishing, 1 in 3 seems to work fine. It also seems to be near to the theory that some have stated on here. Phil was branded as being out of his depth for suggesting the 1 in 3 guide though. Only for it to be pretty much what the outcome had been with all the complex maths! Lol

Closer than the 2ft in 20ft of water that was being suggested last night! Lol

 

Often the case Nige, rules of thumb are good things to have. I just can't get my head around it :oops: but what is, is what is :)

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Yep, my calcs give 0.8m (2.6ft).

 

To my mind though 3 - 1 can't work at all distance and depths :confused:

 

At least I tried :hrumph:

Ian I think the pythagors element accounts for some of the shortfall. I then think that there is a difference created by drag. The line that remains out of water falls faster than the line and lead in water, essentially pulling the lead nearer to the bank.

 

It's all well and good having a theory but if the evidence presented proves you wrong then the theory is proved as incorrect.

 

Something Nige has done.

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It can't, but it's a good starting point with all the other variables involved!

The only true way of getting an exact amount of swing back is measuring it n each specific situation. That's where my two marker float technique comes in!

 

Ah thanks Nige, I can at least scrape that out of my poor head :lol:

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