Lift force acting on a fly line

[Torsten]

Paul from the "Two heads cast at the same time" thread:

On a separate issue, I'm now not convinced that there is an upwards force keeping the loop aerielised. If you look at the 170 backcast video on a log you can clearly see the point of the loop following a parabola. And that's exactly the sort of loop shape that should be lifting. I think that's probably wrong, which annoys me more than anything because we had an argument with Burgess and Tim Rajeff vs Bruce Richards and Noel Perkins very many years ago on the Board before the last Board, which was when and how those PDF documents surfaced. I don't mind Tim being right of course...

The title of Noel Perkins/Gatti Bono's paper can be misleading, lift means not that the fly line is actually lifting but rather the lift force due form/skin drag, which may slow down the downfall for sections of the fly line. (Not that the line is actually lifting, or only for cases where this force would exceed the gravity)

We can have a closer look at the question by examining a section of a fly line:

Lift Force.gif (4.54 KiB) Viewed 5439 times

Let's say we have an section of a fly line in a air flow with the velocity U with the inclination angle alpha. Due to the form drag an upward component of this force - the lift force FL - is acting on the section of the fly line.

According to [1] we can determine FN by the following equation (angle alpha > 20°):

FN = 1/2 * CN * rho_f * D * (U * sin(alpha))²

where

FN: Normal force per unit length (N/m)
CN: Normal force coefficient; ~ 1 for a cylinder
rho_f: Density of the fluid (kg/m³), air ~ 1.225 kg/m³
D: diameter of the cylinder
U: Incoming flow velocity (m/s)
alpha: Inclination angle of the cylinder

I'll leave it to you as an exercise to compute the lift force and compare it to the gravity.

If you can spot a section of a fly line in a video, with an inclination and you know the velocity, you can compute the lift force. It's here also a question of significance and this can be determined.

--

[1] 2014, Divaret, Forces exerted on a cylinder in near-axial flow
https://www.researchgate.net/profile/Ah ... C%82ow.PDF

[Graeme H]

If you can spot a section of a fly line in a video, with an inclination and you know the velocity, you can compute the lift force. It's here also a question of significance and this can be determined.

Well, this part is pretty easy. I'll post the actual velocities tonight when I get home.

In this case, U will be around 0.5m/s for the bottom of the loop. That's about 1.8 km/h on a calm day. As noted in the other thread, there's not much scope for lift from the bottom of the loop with a velocity that would hard pressed to outrun a toddler ...

We can also do this for the fly leg, where it's often inclined down. However, in that case the line is usually travelling parallel to its orientation and experiencing skin drag rather than form drag.

(This image came from a study of the dolphin nose you see there.)

Cheers,
Graeme

[gordonjudd]

In this case, U will be around 0.5m/s for the bottom of the loop.

Graeme,
I don't know how you came up with .5m/s for the relative wind velocity on the inclined section of the loop, but it is easy to see that it equals v_loop for a tethered cast.

Put yourself in a moving frame so the shape of the loop appears to be fixed in a frame that is attached to the front of the loop that is traveling at a constant velocity as done below.
https://vimeo.com/342277007

I know you might think that Newton was wrong about getting the same results for observations made in any inertial frame, but that is the way physics works. I think you even agreed with James statement that:

You know that changing the frame of reference in no way changes the physics or any derived or predicted results right?
and you responded that:
Yep, I know changing the frame of reference has no impact on the physics when it's done correctly.

Maybe you think there is something "incorrect" in that video.

Now in that moving frame it appears that the air (and the of the world) is rushing by your fixed line shape at relative velocity of v_loop. Simple no. That is why people like to use a moving frame to analyze what is going on with loop propagation.

We can also do this for the fly drag on fly leg

and because of that tilt and the fact that the coefficient for form drag is about 200 times larger than the coefficient for skin drag the form drag can be larger than the skin drag for the kind of tilt angles we see in casting.
Gordy

[gordonjudd]

According to [1] we can determine FN by the following equation (angle alpha > 20°)

Torsten,
Why does the angle have to be greater than 20° for the sin(alpha) term to hold? Noel did not use that constraint in his analysis.
Gordy

[Graeme H]

In this case, U will be around 0.5m/s for the bottom of the loop.

Graeme,
I don't know how you came up with .5m/s for the relative wind velocity on the inclined section of the loop, but it is easy to see that it equals v_loop for a tethered cast.

Try -v_loop for the rod leg.

Loop nose is moving into the wind at v_loop.

The line in the rod leg is moving away from the loop nose at -2v_loop

-2v_loop + v_loop = -v_loop.

Oh, and how did I come up with 0.5 m/s? I measured it. You know, like actually looking at real life data ....

[gordonjudd]

Try -v_loop for the rod leg.

Graeme,
Thorsten clearly stated that:

U: Incoming flow velocity (m/s)

If the incoming velocity was -v_loop (coming from the right of the inclined section) that would produced a downward force on the line. Angle your arm down out the window of a moving car and you will experience the difference the tilt angle makes.

Oh, and how did I come up with 0.5 m/s? I measured it. You know, like actually looking at real life data....

Then if Newton was right about inertial frames, there is something wrong with one of our measurements.

Gordy

[Graeme H]

Have a very close look at your video Gordy.

On a calm day and with a moving FoR, the air is traveling at the same speed as the wall in the background. The line in the rod leg is also traveling at the same speed as the wall. The markers on the line confirm that finding.

So what is the velocity of the line in the rod leg relative to the air? The markers move at the same speed as the air which is the same speed as the wall.

Air speed = wall speed = rod leg speed.

Are you trying to tell us that the rod leg markers are travelling at the same speed at the loop nose? Real life says no.

[Graeme H]

If the incoming velocity was -v_loop (coming from the right of the inclined section) that would produced a downward force on the line. Angle your arm down out the window of a moving car and you will experience the difference the tilt angle makes.

In your moving frame of reference, v_loop = 0.

In that same moving FoR, -v_loop = 0.

U = 0 (or nearly so in real life.)

[gordonjudd]

Are you trying to tell us that the rod leg markers are traveling at the same speed at the loop nose?

Graeme,
Your are right that point particles in the rod leg are moving at the same speed at the air and thus there is no skin drag on the moving rod leg in a moving frame. That is reasonable, since there is no skin drag in an earth frame either since the velocity of the air and the rod leg are both zero in an earth frame. Score one for Newton.

It is the shape of the loop (not the moving particles in the line) that has a relative wind velocity of v_loop. If you were to mark the positions of the inclined section of line (not the marker points but the same points relative to the front of the loop) and then compute the velocity of that inclined section of line you would also get a relative wind velocity of v_loop. That is the way it works out in the moving frame where that inclined section does not seem to be moving but it is exposed to an incoming wind velocity of v_loop.

At any rate the apparent velocity of the inclined section of line is v_loop not .5 m/s. I guess it is important that we are measuring the same thing in both frames.

Maybe it is time for Torsten to have a go at this. It appears that you and I cannot agree on anything.

Gordy

[Graeme H]

So this only works for the bottom of the loop?

The top of the loop (which is inclined down) does not push the loop down but the bottom of the loop does push it up?

Cheers,
Graeme