Quiz - casting mechanics

Walter & Group...

I cannot possibly send all of the answer sheets ..... however, rest assured that I take a lot of time to read each of them.

Here, I picked out some interesting answers which use differrentwords and /or concepts than the ones already sent.

Gordy

 

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Some answers from Lyth Hartz :

4.)  Can the fly rod bend without acceleration ?
No, not when casting. Acceleration is needed to bend/load the rod.   The rod can bend leaning against a tree but that is just bent, not loaded.

I should have added, "when casting".   However, as I think about it, when we pull that rod from the tree, it will straighten, thus unloading.    G.

5.)  What do you think the CASTING ANALYZER actually measures ?
I am not totally sure but I think it measures speed, acceleration and arch. 

Bruce Richards and Noel Perkins designed it.  They used the term, "angular acceleration".  I am in the process of giving some thought to exactly what angular acceleration is in view of the fact that acceleration is a vector quantity (force applied in a straight line.)    G.

6.)  There are two things which must match in order to achieve an (almost) straight line path of the rod tip.  What are they ?
The arch and the bend of the rod combine to form the straight line path when combined correctly. 

If by the term "arch" you mean the CASTING ARC, then I agree.   G.

15.)  Suppose you made a cast but didn't stop the rod.  Would the rod fail to unload ?
Yes, the rod will unload because at some point the rod will stop.  But, it will unload too slow to form a nice tight loop towards the target.

16.)  Your student asks you, "Why do I have to make a stop at the end of my cast ?".  What do you tell him ?
To make the rod unload briskly in order to make a nice tight loop that travels towards your target.

Your answers to #s 15.) and 16.) bring these questions :

1. If you do not stop the rod will it unload to the same degree as it would if you stopped it ?

2. Does the rate of deceleration of the rod during the stop affect the rate of rod unloading ?

I suspect that the answer to 1. is that the rod will unload to the same degree.   As for 2.) I don't know if the rate of deceleration of the rod during the stop affects the rate of rod unloading, nor could I find any study which addressed this.

Gordy

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

From Frank Harford:

15.)  Suppose you made a cast but didn't stop the rod.  Would the rod fail to unload ?

 

       I don't think you can make a cast without stopping the rod . However , if you could continue constant acceleration, as an abstract exercise, the rod would not unload.

 

Paul Arden has pointed out in prior messages that when trout fishing many of his casts are made without stopping the rod .... yet they are effective fishing casts.

 

Having said that, I do agree that if you could continue with constant acceleration, that the rod would not unbend or unload.    G.

 

16.)  Your student asks you, "Why do I have to make a stop at the end of my cast ?".  What do you tell him ?

      It enables the caster to unload a lot of stored energy at once and send the line flying.

 

Brings us back to the as yet unanswered questions which I posed, above in comment to Lyth Hartz's answers. You do, however, present a word picture which would be great for teaching students, I think.  G.

 

19.)  Which do you think is most important for an efficient long distance cast ... the direct ("swing") effort of the caster or the unloading and release of energy from a bent fly rod ("spring") ?

 

       I believe that it is about 80% swing and 20% spring

 

Perhaps you are correct.  I don't know of any studies which quantified in order to come up with percentages.    G.

 

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

From Bob Stouffer :

6.)  There are two things which must match in order to achieve an (almost) straight line path of the rod tip.  What are they ?  Acceleration and Arc both balanced in congruence with the length of line outside the rod tip and rod bending.

I'd call that correct since it is acceleration which determines the amount of rod bend.  Still fits with the concept that the CASTING ARC and ROD BEND must match.    G.

12.)  If you answered "yes" to question 11.), which do you think is more important ?  Stiffness

While both stiffness and mass distribution are important to the turnover of the leader, MASS is by far the most important.   G.

19.)  Which do you think is most important for an efficient long distance cast ... the direct ("swing") effort of the caster or the unloading and release of energy from a bent fly rod ("spring") ?  I think that the two concepts are inseparable.

I must agree, for an efficient cast both are necessary.  However, I asked which is the most important.  Granted that you can't make a long distance cast without each, in general the effort of the caster (swing) is more important than the energy realeased by the unbending of the rod.(swing).   Of course, one could argue that the rod would not have been bent in the first place without rotation produced by the caster.

One crude experiment :  If you rig up a broomstick with guides and tip top, you can make a poor but effective short cast.  If you hold the handle of a fly rod in a vise, then pull back on the tip with the fly line, the layout is usually a mess pretty close to the rod.  So there the broomstick casts wins out even though it can't be bent and therefore cannot be loaded.     G.

 

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

From Aitor Coteron re the effect of line mass and diameter .   I placed his well informed answers to the quiz in two attachments.   G.:

Hi Gordy,

I reply below.

From Aitor Coteron in answer to Al Buhr's statements :

Gordy,

I don't really understand what Al does mean with his statement.

IMHO the difference between two lines in its ability for turning over a long leader lies in their respective capacity of overcoming air drag with the minimum loss in their momentum.

An example taken form the daily life will illustrate my idea about this issue:

Two cars of the same make and model begin a 700 km trip with their tanks full. One driver likes to speed more than the other so he reaches his destination earlier. This car which has gone faster will have less fuel left in his tank at the end of the trip.

Between two lines of the same mass and cast with the same initial speed, that whose mass profile makes it travel faster (that is, that with the longer and thinner front taper) will reach the connection with the leader with its "tank" more depleted, so this faster line will have less capacity to turn over a long leader.

The problem lies in the fact that air drag increases proportionally to the square of the velocity, so the faster the line goes the higher the loss in its momentum.

In vacuum two lines of the same mass travelling at he same initial speed will turn over the leader the same, whatever its mass profile.

I could be wrong but this is how I see this issue until proved wrong.

Regards,

Aitor

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`

Aitor...

Wind resistance does make a major difference.  We must consider, too, that a line with greater mass will also be traveling with greater momentum.  The ideal line for casting in high winds would have minimal diameter and highest mass, as I see it.  G.

Well, IMO air drag (wind resistance, if you prefer) is not the major difference but the only one. Of course in that regard the line with the minimal diameter and the highest mass is the more efficient, that's the reason which explains why a sinking line can be cast with less effort from the caster.

The problem arises when we try to compare two floating lines with the same density and the same mass but different mass profiles, that is, different front tapers.

You are, I think, correct with you statement that the faster the line goes the higher the loss in momentum... However, if you have much more momentum to begin with you have a lot more to lose before collapse.  G.

If we consider two floting lines of the same density and the same mass that we cast with the same initial speed, their respective momentums ae the same; p = mv, and m and v is the same in both cases. However how that mass is distributed along the length of the line has a significant impact in the ability of the line to overcome air drag.

For instance, taking those two floating lines, with the same density and the same mass, which we cast with the same speed the differences in the taper can determine which one is more suitable for casting into the wind.

For most presentations, we'd like to have complete dissipation of energy at the very conclusion of the leader turnover.  If we have lots of energy left, we could look at that as wasted (unless we were trying for a forceful splash down as with a bass bug.)  G.

Or we need to cast into the wind, or turn over 6 m long leaders.

The question is to understand why different line tapers have a different capacity to overcome air drag. And to understand that we must have an idea of two key concepts: momentum conservation and the relation between the increase in velocity and the increase in air drag.

I might be wrong but I this is how I see it so far.

Regards,

Aitor

 

Attachment: Gordy's quiz.doc
Description: Binary data

Attachment: 7.doc
Description: Binary data