Fishing lure question

  • Thread starter LTLhawk
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Can anyone help me with a formula to calculate the depth of a fishing lure behind a boat traveling at a constant velocity. Known: boat velocity, lure weight, distance behind the boat. Assumption: fishing rod tip is at water level.

Obviously I want to know how far down my lure is. Thanks for the help
 

BruceW

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Since the lure travels behind the boat at constant velocity, the forces acting on it must cancel in each direction. This gives the angle of the line. Using this and the total length of the line gives the depth of the lure.
[tex] depth = L \times \frac{weight \ difference}{tension} [/tex]
Where L is the total length of the line, tension is the tension in the line, and weight difference is the weight of the lure minus the weight of an amount water that has the same volume as the lure.

But I think an easier way would be to measure the angle of the line going down into the water.
When the angle is defined such that it is zero when it lies on the water surface, then:
[tex] L\sin(\theta) = depth \ of \ lure [/tex]
 

SpectraCat

Science Advisor
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Since the lure travels behind the boat at constant velocity, the forces acting on it must cancel in each direction. This gives the angle of the line. Using this and the total length of the line gives the depth of the lure.
[tex] depth = L \times \frac{weight \ difference}{tension} [/tex]
Where L is the total length of the line, tension is the tension in the line, and weight difference is the weight of the lure minus the weight of an amount water that has the same volume as the lure.

But I think an easier way would be to measure the angle of the line going down into the water.
When the angle is defined such that it is zero when it lies on the water surface, then:
[tex] L\sin(\theta) = depth \ of \ lure [/tex]

No .. I don't think that is correct. The buoyant force can probably be neglected in this example ... the upward force on the lure comes from the component of the line tension normal to the water surface. That must balance the gravitational forces, while the parallel component of the tension would balance the viscous drag (friction) force on the lure. So, it seems like there must be some additional information about the shape of the lure and it's cross-sectional area, or some other way of calculating the friction forces on the lure.

Anyway, the depth of the lure will be given by the distance behind the boat time the tangent of the angle between the line and the water's surface. That tangent will be equal to the ratio of the gravitational force to the drag force.
 

BruceW

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My equation is correct, I have accounted for the component of line tension normal to the water surface in the correct way.
Whether the buoyant force can be neglected or not depends on the density of the lure. If its density is much greater than the density of water, then the buoyant force can be neglected.

The weight and drag of the line also contribute to this problem, although I haven't accounted for them in my equation.
 
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This is good stuff... thanks guys.
 

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