Breaking strain recommendation for dams?

[Ko7Ad]

How much does a 1kg fish weight in water?

I have to laugh at myself for my previous reply. Cause the Buoyancy Force which will be the "weight" of a fish in water is actually the same as is is out of water.

F(buoy) = (displaced fluid density)x(gravity acceleration)x(displaced volume)
= (1000kg/m³)x(9.8m/s²)x(0.001m²)
=9.8N
Thus the same weight as above water

This is asuming a fish of 1kg has a volume of 1lt, which is most likely not the case.

[smallstreams.co.za]

This is asuming a fish of 1kg has a volume of 1lt, which is most likely not the case.

And I would asume it would move too. So 1kg of fish does not equate to 1kg of "pulling power". Not sure if this was mentioned by Albert de Jager in his post as above

[Ko7Ad]

And I would asume it would move too. So 1kg of fish does not equate to 1kg of "pulling power". Not sure if this was mentioned by Albert de Jager in his post as above

Well I am sure we could work this out! :tongue:

How fast would a 1kg trout swim? We could possibly use the rmp of a reel and the spool diameter to get an accurate estimation of its speed and work that back to what force would be required to move at that speed?

Anyone here that took physics at University? We may need some assistance here...

I just have to correct my previous post: I found that raw fish has a density of approx 946.79kg/m3, this will give its "weight" in water based on the Buoyancy formula of 10.35N!!!!

[markdej]

According to this research paper: http://m.jeb.biologists.org/content/63/2/451

Trout have a median acceleration of 13.4m/s and a maximum acceleration of 42m/s (hell thats fast.)
So as F=M*A the force required to stop a 1kg trout at median acceleration is 13.4n and the force required to stop a 1kg fish at maximum acceleration is 42n. Assuming 5x stroft has a breaking strain of 2.4kg = 24N then a 1kg fish could potentially pop 5x with absolute ease.

[smallstreams.co.za]

According to this research paper: http://m.jeb.biologists.org/content/63/2/451

Trout have a median acceleration of 13.4m/s and a maximum acceleration of 42m/s (hell thats fast.)
So as F=M*A the force required to stop a 1kg trout at median acceleration is 13.4n and the force required to stop a 1kg fish at maximum acceleration is 42n. Assuming 5x stroft has a breaking strain of 2.4kg = 24N then a 1kg fish could potentially pop 5x with absolute ease.

Have you taken water viscosity into account? Hehehe

[Andre]

Weight and mass are two different things. Mass in kg's, is the amount of matter a body is made up of. Weight measured in pounds (or ounces) is the force with which gravity acts on it. so to answer the question, we might need to think about a few things. If there are tow fish, one of 1kg, and one of 10kg, if they are floating, they would both weigh the same, in other words, gravit is having an equal effect on both. When they are out of the water....different story.
Like a hot air baloon....when in flight, what does it weigh?......answer.....nothing, its weightless. But it still has the same amount of mass regardless if it is airborne or not.

How much does a 1kg fish weight in water?

[Ko7Ad]

Weight and mass are two different things. Mass in kg's, is the amount of matter a body is made up of. Weight measured in pounds (or ounces) is the force with which gravity acts on it. so to answer the question, we might need to think about a few things. If there are tow fish, one of 1kg, and one of 10kg, if they are floating, they would both weigh the same, in other words, gravit is having an equal effect on both. When they are out of the water....different story.
Like a hot air baloon....when in flight, what does it weigh?......answer.....nothing, its weightless. But it still has the same amount of mass regardless if it is airborne or not.

If the fish has the illusion of weightless in water is does not apply in our fishing scenario. Cause the weightlessness will only apply on the vertical axis, the weight of the fish still determines the force required to move it (reel in or fight the fish) on the horizontal axis.

[Andre]

Ya I know. its about equal and opposite forces acting on the object......for it to be suspended in water, the upward force of the water is equal to the downward force of gravity. Im aware that the fish is still responding to gravity...Just throwing another curved ball into the mix. Hopefully everyone is as confused as I am,.

If the fish has the illusion of weightless in water is does not apply in our fishing scenario. Cause the weightlessness will only apply on the vertical axis, the weight of the fish still determines the force required to move it (reel in or fight the fish) on the horizontal axis.

[Glans]

Let's add another angle to this complicated story
We will have to take the length and width off tail and body the fish compared to the nose( where you hooked it) in consideration as well. If it moves at a certain speed the power used to get at that speed was due to a side movement from tail placing a force on the nose in the opposite direction. So the leverage force on nose forced by the tail will be more on to the tippet than the moving force going forward.
Who wants to add that force to the equation. It looks like it is pure luck that we land any fish from all these considerate calculations.

[Scythe]

Attachment 21471

Now that is a beautiful fish.

What is a recommended breaking strain to use for dams? Im considering using 0,15mm (2,4kg) Stroft GTM

For my money, use the highest rated breaking strain with the thinnest diameter you can find

Naturally, you do this while taking into consideration :

What you are presenting (viz streamer v dry fly)
Where you are presenting (viz structure, snags)
How you are presenting (Rod, line, leader construction)
Feeding behaviour (apppropriate to the type of fly, dry fly = sip ... streamer = m@er hom)
What you know to be in the stillwater (at stockie joints v wild empoundments where you know there are bigger fish)

I believe in keeping it simple ...

(but then again I'm not that keen on doing sums while fishing and I don't have any desire to break my or anybody else's rods ...)