That is a rather usual question and also a very interesting one. It has been asked to me again recently via the internet by a couple of fellow casting instructors. Let’s go for it.
Let’s take a look first at the Roll vs. Overhead video used in a previous article:
I will describe the scenario:
Just one rod rigged with two lines: Royal Wulff #7 and Rio Tournament #6.
The TT has the ideal taper for roll casting; the Rio is designed for long casts overhead.
The Rio is unrolled behind the caster; the TT set in a roll cast configuration with its leader anchored by means of a screwdriver stuck in the ground (is there a more solid anchor than planet Earth itself? If there is let me know).
In this way the very same casting stroke applies force to both lines. What the video shows, however, is that the overhead line reaches its target whereas the rolled one falls short. It is an interesting experiment that any caster can try by himself (just in case somebody doubts of the result).
Anyway the result is that, if you use a stroke with the amount of energy needed for the overhead line just to straighten, the roll cast line falls short of the target. Always. Why?
I have said that the very same casting stroke applies force to both lines simultaneously, but is it the same amount of force for both? No, it isn’t.
What our stroke is doing is applying the same acceleration to the rod, and the rod to both lines, but the length of line actually accelerated in one case is much longer than in the other: we accelerate the whole length of the overhead line whereas only a short piece of the roll line is subjected to acceleration.
Do you remember the basic formula of Force?
F = m.a.
For a refresher this is a nice and easy source Force
Acceleration is the same for both lines but mass isn’t. So the force we apply to the roll cast line is much less than that we apply to the overhead line. For the same, identical, casting stroke getting the same distance with much less force sounds impossible, right?
If you prefer we can address the problem from the standpoint of Energy.
By applying force to the lines over a given distance we are doing Work on them. Do you remember the formula for Work?
W = F.d
For a refresher this is a nice and easy source Work
The work done on an object amounts to the energy transferred to it; that is, more Work = more Energy transferred.
But we have already discovered that the roll line has been subjected to less force than the overhead one; less force amounts to less work done on the roll line and, consequently, less energy in it.
How do you expect to get the same distance with less energy?
So what we have is that the casting stroke that works for an overhead cast isn’t good for a roll cast. By the same token the stroke good for a jump roll isn’t enough for a static roll.
Some may say that, after all, that rod load shown in the video is just enough for propelling one line but not both. Another of the blindfolds that the casting paradigm based in load puts over our eyes.
Yes, only one line is propelled all the way forward but, how is it that it is always the one with the longer piece of “live line”?
So, as a corollary, IMO we have to think in terms of how long is the piece of line I am accelerating directly to the target, and not of the mythical “load” and “anchor loading” or the impossibility of a “highly energized V loop”.
The key to understand this issue lies in the fact that the rod in the video isn’t applying the same force to both lines, just the same acceleration. It seems that it isn’t easy to grasp so an additional (more graphic) example to clarify this is in order:
Let’s say I have a video showing three model railway coaches.
Two are connected together and laying on a straight rail. Parallel to them there is another rail with the third coach laying on it.
In the front part of both convoys we have a string, one string for each convoy.
At a given time the strings get taught and the two convoys start moving with exactly the same acceleration.
Since the track has a ruler alongside, by means of Tracker or any other application we are able of easily calculate the magnitude of that acceleration.
We also know the respective masses of each convoy: isn’t a surprise that mass A has a value X and mass B has a value of X/2.
By solving the F = m.a equation we get that the force applied to mass A is double the force applied to mass B. Right?
Well, after that first video I show you a second one with a general view of the scene. Now we can see that there is a guy pulling on both convoys at once by holding both strings in one hand.
So now, out of a sudden, we discover that he is applying the same force to both convoys? Not al all, the force exerted on each of them is different, one being exactly half the value of the other.