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Leo Moore

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Reply with quote  #1 
I've spent a lot of time learning how to cast but now I'm wanting to understand the physics of why a longer line needs to be heavier for the same rod. The Rio "A" recommendation for a T&T 1308-4 is listed as follows.

Skagit 525
Scandi 460
Short Head 520
Power Spey 570

I know the Skagit is fat and short comparied to the Scandi, I sort of understand that. But why do the longer lines need to be heavier? How come X amount of grains doesn't load the rod?

Thanks for any help.

Leo

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Gene Larson

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Reply with quote  #2 
Skagit 525 23 525 22.82608696
Scandi 460 33 460 12.43243243
Short Head 520 47520 11.81818182
Power Spey 570 61570    9.344262295

The key is grains per foot of head.  The last number is the head weight/length of head.  The total weight is needed to load the rod and with a longer line, the lighter per foot weight needs to be carried by the length of line.  
Bruce Kruk

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Reply with quote  #3 
Its heavier because it is longer, simple as that.
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Bruce Kruk
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June Kim

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Reply with quote  #4 
Interesting question. I'm also curious about the exact answer. I think Leo is asking why the total weights differ.

To cite Simon G.'s explanation, longer spey lines have "loading weight" at the back. That's probably why the longer lines have to be heavier in total.

I leave the link below: 

Understanding Spey Lines by Simon G.

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Leo Moore

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Reply with quote  #5 
What I'm asking is, why 460 grains stretched out thinner and longer won't load the rod or 520 stretched out thinner and longer won't load the rod? I understand it's so but I can't explain why.
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Bruce Kruk

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Reply with quote  #6 
The weight is in the back of the line and the front is more tapered in longer lines, that front taper adds extra weight to the over all weight of the line but the part that is loading the rod (the rear and or mid section depending on line) does not overload the rod.
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Bruce Kruk
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June Kim

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Reply with quote  #7 
Leo, I understood your question now. It seems more complex than thought. I think Gene and Bruce's answers explain it almost though. I'd like to hear fuller explanation too. 
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Leo Moore

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Reply with quote  #8 
I have another question. When making a cast is there more line on the water with a longer head than with a shorter head? I suspect there is, so the longer head would need more weight to keep it anchored or am I off base?
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Paul Metcalf

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Reply with quote  #9 
F = ma

next...
Dennis Kulhanek

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Reply with quote  #10 

F/w=a/g  Next. Leo Try to understand that that the only uppper part of D loop is the force loading the rod and apply the power on the forward cast. Any line faceing away from your anchor! Is opposing force[weight,drag] will hinder your forward cast.


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Gene Larson

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Reply with quote  #11 
Way Yin in "Spey to Z" makes remarkable casts with a 100' head.  He is using a very long rod rated for a 10 wt line.  The heavy line is needed to fully load the rod thus enabling him to cast the long line, shoot some running line, and amaze us.  The key elements for me have been identified--long taper on the front of the line, heavy weight at the back, turn over of the line assured by the heavy line rolling over the lighter tip.  The physics formulas explain it scientifically, but the best explanation is found by watching Way cast and accepting that with time and practice we may be able to do half as well.  
Paul Metcalf

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Reply with quote  #12 
F = d(mv)/dt

One day I'll have this casting thing figured out and I can make a useful contribution to the discussion.
Poppy Cummins

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Reply with quote  #13 
Quote:
but the best explanation is found by watching Way cast and accepting that with time and practice we may be able to do half as well.



There is a powerful message in the above quote. Awesomeness @ spey casting doesn't really come with equipment or physics formula, but with spending A LOT of time practicing.

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Poppy=Red Shed Spey Rod Pimp http://www.redshedflyshop.com FRSCA-Founding Member How you get the line out and fishing is personal preference so as long as it works and is easy no one should care but the caster. MSB
Dennis Kulhanek

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Reply with quote  #14 
F=d(mv)dt  Thank's for calculus notation of momentum. But I think that no one is interested in that. Gene and Popy posts are right on and only practice will make better casters!
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Paul Metcalf

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Reply with quote  #15 
I'm always intrigued with the "ignorance is bliss" approach to explaining things.  I thought the OP asked for a physics lesson?  All of my physics classes utilized equations[confused]

Here's an easy one for everyone - keep tension in your line at all times, slack is a killer or energy (oops, sorry) waster.  Better?  Equations are so much more friendly[smile]
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