Upward Force from Form Drag

[Merlin]

Hi James

Thanks to bring us back on earth. If we compare a vertical cast and an horizontal cast (idealized), the only thing which is changing is the position of the aerodynamic force on the loop, for the rest (fly leg, rod leg), nothing really changes.

Let's take a few figures: the weight of the loop is about 0.01 N, and for about 30 m/s fly leg speed the tension at the top and the bottom of the loop are about 0.2 N, 20 times more. Now instead of having an aerodynamic force opposing the weight (and about 0.01 N), this force is in the horizontal plane and gently pushes the loop on the side.

It is unlikely that the weight of the loop will drastically change the rate of fall of the legs and that the now lateral aerodynamic force will drive the fly leg on a side. They are too small and we are not going to see evidence of that (the loop is too small by comparison to legs).

So even if in vertical plane the loop weight can be compensated by aerodynamic forces (the subject of calculations), the capability of the aerodynamic force to deviate the legs is questionnable. What we can suspect is that there can be a sufficient launch angle for the fly leg, even small, that gives this impression of a climbing fly leg. The exaggerated launch angles demonstrated by Graeme in his videos are consistent with this statement.

Merlin

[gordonjudd]

What we can suspect is that there can be a sufficient launch angle for the fly leg, even small, that gives this impression of a climbing fly leg.

Merlin,
I agree.

That was the observation that made back in in post #4:

I agree the propagation of the loops in your video seemed to be dominated by their launch angle and don't have much of an inclined section of line on the bottom of the loop.

Maybe the direction of rho_l*v_loop.^2 force on the rod leg dominates the direction the fly leg is being pulled

Thanks for putting some relative numbers on how big the rho_l*v_loop.^2 acceleration force is compared to the drag forces.
Gordy

[gordonjudd]

So, we must also be in agreement that the fly-leg then performs work on the rod-leg, i.e. lifting its mass?

James,
I see it as the changing the momentum of the fly leg is produced by a force (tension in the line) at the top of the loop. You get the same rho_l*v_loop.^2 tension value if you compute the change in the momentum of the line going around the loop due to the change in the direction the mass moves going around the loop. Remember the chain fountain analysis?

That tension at the bottom of the loop (there is an equal amount of tension at the rod tip) then is sufficient to pull out the slack in the line so that the rod leg tends to take on a catenary shape as the loop propagates.

I am a bit fuzzy on whether or not an internal force can produce work.

Well, I think the issue must be that all those references that claim that internal forces don't do work must refer to RIGID BODIES or PARTICLES. If that's the case, they're right. But in general, for deformable bodies, that's not true and your textbook is right.

Since our fly line is deformable I guess that internal tension can change the shape of the line and thus change its potential energy.

but surely it's obvious that the greater the fly-leg momentum then the greater the rate of change?

The momentum momentum change is equal to rho_l*v_loop.^2. Therefore you could have two cast with different amounts of carry (and thus the longer line would have more momentum) launched with the same velocity that would have the same momentum change as long as the lines involved had the same linear mass density. So I don't think your "surely" actually is so sure.

Gordy

[gordonjudd]

The tension in the rod leg is caused by the fly leg pulling it. Newton's 3rd Law.

Graeme,
How does the fly leg pull on the rod leg?

I think the pull is at the top and bottom of the loop and it is produced by Newton's second law, i.e. a momentum change.

It is the force from the momentum change of the mass going around the loop that produces line tension. That tension then pulls on the fly leg and the rod leg. I don't think the fly leg pulls on anything. Just the reverse. The fly leg itself is being pulled.

Gordy

[James9118]

launched with the same velocity that would have the same momentum change as long as the lines involved had the same linear mass density. So I don't think your "surely" actually is so sure.

Gordy

Gordy,

If you've read anything of what I've written you'll know I talking about increasing the launch speed. It's almost like your being deliberately obtuse to avoid discussing the main issues.

Daniel,

Here's a picture of me in a distance casting comp - I suspect my launch angle is fairly typical for a distance cast.

[Graeme H]

Hi Graeme

It is DE.

Merlin

Hi Merlin,

If I were to make a video of that cast with the marked line, do you think any marker on the line between D and E would have a velocity approximately equal to half the fly leg velocity? Or do you think it would have a velocity of approximately zero?

If the video shows it to be approximately zero, what does that do to your calculations? Would it make you reconsider the importance of "loop lift"?

Cheers,
Graeme

[Graeme H]

The tension in the rod leg is caused by the fly leg pulling it. Newton's 3rd Law.

Graeme,
How does the fly leg pull on the rod leg?

I think the pull is at the top and bottom of the loop and it is produced by Newton's second law, i.e. a momentum change.

Hi Gordy,

The fly leg has (generally) forward momentum that we imparted by accelerating it with the rod. The only thing preventing it continuing on that trajectory is the rod leg. The rod leg stops the fly leg elements going forward with tension, since it's attached to the rod. The fly leg maintains that tension at the loop. For every action, there is an equal and opposite reaction.

It is the force from the momentum change of the mass going around the loop that produces line tension. That tension then pulls on the fly leg and the rod leg. I don't think the fly leg pulls on anything. Just the reverse. The fly leg itself is being pulled.

You've got this bass-ackwards. Momentum change at the loop is the result of forces being applied, not the cause of of the change.

What is "pulling the fly leg"? If it's the loop, what is pushing the loop? It has to push against something before it can pull anything along. If the loop is pushing against the rod leg, why is there any tension in the rod leg?

That tension then pulls on the fly leg and the rod leg.

So you're saying tension within the loop itself pulls both the rod leg and the fly leg? How is that consistent with any of Newton's laws?

So if the loop is pulling the fly leg and the rod leg, can you explain why my line continues to crash into my car well after the loop has been destroyed in this video?



The rod leg falls because the fly leg is no longer imparting tension at the loop. Its tension has been destroyed.

The fly leg continues to crash into the car without a loop because its momentum is maintained until something stops it (here, it's the car rather than the rod leg.)

Cheers,
Graeme

[Merlin]

Hi Graeme

Here is an illustration of the travel of a point in the inclined section of the loop:

aero4.JPG (16.34 KiB) Viewed 3096 times

This is a simplified illustration, DE is the incline section of line and is 1 m long (Gordy’s data), and so the height of the “loop” is 0.19m.

At time t the triangle is in E, while the dot is 1m behind. The triangle travels downwards and slightly forward at Vfly * sin 5.5. Let’s take figures and assume that Vfly = 30 m/s. Then the speed of the triangle is 2.9 m/s in the red arrow direction. It takes about 66 milliseconds for the triangle to get to the end of the inclined section. During these 66 ms point E has moved forward and the dot is now at this point E at the top of the loop. The triangle has moved along the inclined section at 15 m/s = 1m / 0.066).

You can see that the dot has moved about 2 meters forward (30 m/s *0.066 = 2 m), whilst point E which is the top of the loop moved by 1 m (15 m/s * 0.066 = 1 m). All this is consistent with the graphic I posted before. The loop is travelling at Vfly/2 and a point on the inclined section moves down along it at 15 m/s.

If you shoot a video, you will see a marker on the inclined section going down nearly vertically (the red arrow).

It's just simple high school level physics.

Agree

Merlin

[Graeme H]

If you shoot a video, you will see a marker on the inclined section going down nearly vertically (the red arrow).

Hi Merlin,

A complex way of demonstrating it, but yes, I'm happy with the diagram and the end conclusion quoted here. I see what you're saying now. (There are some mixed frames of reference which had me perplexed from while.)

The triangle moves down quite slowly compared to the fly leg's velocity. Using your figures, the triangle also moves forward 2cm in that time (0.19*Sin5.5), which surprises me. I'll shoot the video and see how it works in real life.

Regardless, there can be no lift generated in that inclined section, can there? That would be analogous to claiming a ball dropped at 3m/s experiences lift and never hits the ground.

It also has no application in a non-sexy loop, does it?

Cheers,
Graeme

[Merlin]

The « lift » value depends on the fly leg speed, and it can oppose the « loop » weight sligthly above 30 m/s. Even if we do not know the actual speed during the video, It is doubtful that something less than 0.01 N upwards will change drastically the picture. The launch angle has more effect IMHO.

Merlin