G'day James,

For one, I reckon that you think point A (the fly in your example) follows a path down the line towards the front of the loop - it doesn't. Point A oscillates on a path that is perpendicular to the centre line, it's position will simply be sin(t) somewhere along that path, with the centre being zero - it will always move parallel to any other point on the line.

No problems at all with that. Point A will move horizontally to the left of the image until it starts moving to the right, as will any other point on the line.

Can you see that the particles at the extreme right (the front of your loop) and the extreme left, marked A (the fly) have zero velocity? The particle that is moving the fastest is the one on the centre line - so this one will 'stretch' away from the fly. Between the two extremes there is a spread of velocities that are described, themselves, by a sine wave. The particles will therefore not move uniformly off to the left into a fly cast, they would simply smear out with the centre particles moving away from the extremities.

Okay, let's play

dx/dt then. Your call. When you discretise this motion, velocities are out of phase with acceleration. The line at the rod tip and centre have maximum absolute velocities but minimum acceleration (equal to 0 ) while the points at A and in the loop have maximum absolute rates acceleration and minimum velocity. Left to its own devices at this point in time, point A will achieve the same velocity as the centreline velocity and not "smear out" as you predict.

Here's an old favourite. Sorry about this. Really, I am. I hated it last time and I hate it this time too, but it's required.

In the image shown, the rod tip is moving to the right at exactly the same speed as the line in the centre is moving to the left. The front of my loop and point A have zero velocity as shown. All agreed, no problems at all with your analysis, except the part about the part moving the fastest being the one on the centre line. The other point moving the same speed is the one at the rod tip, but to the right.

Let's assign the velocity of line at the centre point X m/s and the velocity of the rod tip -X m/s. The loop and A have velocities of 0 m/s.

However, we do not pull the rod tip back while the loop is moving forward: We keep it stationary. If we look at the picture with that as the frame of reference, so that the rod tip is NOT moving to the right and it has zero velocity, we see EVERYTHING else moving to the left with an apparent velocity of X m/s added to the earlier velocities.

Relative to the stationary rod tip, the loop moves to the left at a velocity of X m/s, the line at the centreline is moving left at 2X m/s and point A is moving at X m/s (more on this later.)

In other words, the loop is travelling at half the speed of the fly leg. The rod leg is also moving at a velocity of zero. Does this look familiar?

Secondly, you seem to think that all of your 'fly leg' is moving to the left and will continue to do so if allowed. Well it's not in a transverse wave. The simple calculus shows that the two ends are stationary and the centre of the fly leg has the highest velocity. So your fly cast is going to end up with a pretty weird shape. You can pick absolutely any moment in time and you will not find a single instance where a half wavelength (as in your picture) is all travelling in the same direction (again this is fundamental to transverse waves).

Relative to the rod tip, the half wavelength between the loop and A is travelling at (approximately) the velocity of 2X m/s to the left. And as you say, it's moving parallel to all the other points, horizontally. The half wavelength shown is also called the fly leg.

The paradox of Point A travelling at X m/s instead of 2X m/s as the rest of the fly leg is doing is the result of the imaged waves being drawn from a "standing wave array" in a fixed boundary state. In reality, the fly hanging off the leader has already reached the equilibrium promised by being "left to its own devices long enough". Its acceleration maximum from the discretised figure before has meant its velocity has already reached the velocity of the rest of the fly leg because it has a mass approaching zero.

The other factor involved here is the "bend" in the rod and fly legs. Shown here, they follow the Sine wave. In reality, they do not follow a sine wave, but something closer to Chevron wave as shown below*. All points on that straighter leg are travelling in horizontally and parallel direction.

The above are the results that you get from using the transverse wave equations, there's no debating this because that's the transverse wave theory that's been laid out and accepted.

Agreed.

Cheers,

Graeme

(* Yes, I know it should be "Amplitude X 2" in the image. A typo I failed to notice at the time of posting the first time.)