Physics of a diver used in trolling?

[Stephen Tashi]

What is a simplified physical model for a "slide diver" used in trolling for fish?

As I understand a slide diver, it is a small disc like device attatched to the end of a fishing line. The fishing line is attached to something on a boat and the boat pulls line through the water at a constant velocity. The purpose of the slide diver is to cause the line to float along at a constant depth and, in some uses, to make the line float off the the side of the boat instead of trailing directly behind it.

(This post is prompted by a post in the Mathematics section by harlyhar which asks how to fit an equation to some empirical data for a slide diver. https://www.physicsforums.com/showthread.php?p=3479961#post3479961 I doubt the empircal data is sufficient to supply all the variables involved, but it would help to know the symbolic form of the physics formulas.)

The plane of the disc is not necessarily perpendicular to the line. One thought is that the force exerted by the disc is due to drag and that the drag is a function of the rate of flow of water normal to the surface of the disc. In equilibrium, I think the diver remainis at contant position in a frame of refernce that moves with the boat. In this model, the net force down on the disc (due to gravity and any bouancy) plus the drag force would be balance by the tension on the line. Is that a plausible picture of what's going on?

 

[BruceW]

That would be my first guess. But this is the first time I've heard of such a device, so I'm not an expert.

You said the diver keeps at constant depth. So it must be made in a certain way. The gravity plus bouyancy is a constant force upward that doesn't depend on depth or speed through water. The drag will be proportional to speed squared through water (since I'm assuming it moves through the water with relatively great speed). The angle of the plane of the diver will also affect the drag, (the perpendicular area is proportional to the drag). The depth will affect the forces on the diver only because of the angle at which the tension in the line acts.

So somehow these contributions all combine so that the diver stays at a constant depth. If we say that $\theta$ is the angle the line makes from horizontal and $\phi$ is the angle the plane of the diver makes from horizontal, then we have:

The total drag force is proportional to $v^2 sin(\phi)$ (i.e. greatest when perpendicular to the water movement). I will call the constant of proportionality D (which will depend on the density of water and drag coefficient). So the horizontal component is $D v^2 sin^2(\phi)$ and the vertical component is $D v^2 sin(\phi) cos(\phi)$.
The total force due to the tension in the line takes whatever value is required to keep the diver traveling at the same horizontal speed as the boat. The horizontal component is: $Tcos(\theta)$ and the vertical component is: $Tsin(\theta)$ (where T is the total tension).
Also, we require that the vertical forces sum to zero, since we assume the diver is in equilibrium.
So we have:
$$bouyancy - gravity + D v^2 sin(\phi) cos(\phi) + Tcos(\theta) = 0$$
(where gravity is the positive value of the gravitaional force). And for the horizotnal forces:
$$D v^2 sin^2(\phi) = Tsin(\theta)$$
Combining the two equations we get either:
$$T(\frac{sin(\theta)}{tan(\phi)} +cos(\theta)) = gravity - bouyancy$$
Or, if you prefer:
$$Dv^2 (sin(\phi)cos(\phi) + \frac{sin^2(\phi)}{tan(\theta)} ) = gravity - bouyancy$$

Either of these equations give the conditions required for the diver to stay at a fixed point relative to the boat.
Note that: $D=\frac{1}{2} \rho C_d$ (where $C_d$ is the drag coefficient and $\rho$ is the density of the fluid). And gravity is the mass of the diver times g. And the bouyancy is the weight of the fluid displaced, i.e. the volume of the diver times the density of the fluid times g.

So in other words, the speed of the boat restricts the angles of the diver and the line.

But it is probably more likely that the shape of the diver causes the drag to be greater at certain parts of the diver, which controls the angle of its plane.
So all the calculations I have done are not the full picture, I think. So it is not enough to just consider the forces on the diver as a whole. I reckon you need to consider forces on different parts of the diver (since the tension will act at a point and the drag will be different at different parts of the diver).

I'm not sure how I would go about calculating that, maybe someone else on this forum already knows the answer?..