Why is a Jump Roll More Efficient than a Static One?

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.

16 comments on “Why is a Jump Roll More Efficient than a Static One?

  1. […] reason for that inefficiency has already been covered in this previous article. You can also relate it to the case when we rush the forward stroke of an overhead cast and start […]


  2. Bill Keister says:

    I guest got introduced to your blog via Walter’s MCI Study Group. Two fly lines on one rod great.

    Comment on the 7 cast.

    This really goes back to the point being made in the video of the overhead and the roll coast. The energy stored in the fly line at the end of the casting stroke is governed by E=1/2MV(squared) or as I think of it mass times one half of the velocity squared. Mass is the key. So the ratio to live line (storing energy) dead line (which dead load) the answer. This being so the placement of the anchor is the key. It is pure geometry. The further in front of the caster the anchor is placed the less live line there can be. As the anchor point comes closer to the caster the more line there is in the loop. If carried to its extreme, moving behind the caster, the more live line there is until it becomes an overhead cast.

    You also touched on load. I like many like bend instead of load to indicate applied force. We tend to think of our casting as applying force to the line. But if we look at F=MA we can also see a different view. In overhead casts the mass is effectively fixed within narrow limits. For rollcasts and spey casts the amount of mass is a variable which we attempt to maximize with the geometry of the cast. But in reality the amount of force is just as much a result acceleration. We feel this force as resistance or load (and here I am comfortable in calling it load). So many of our casting problems are the result on failure to accelerate the line. Or maybe more appropriately as much of the line as is possible.


    • Aitor says:

      Hi Bill,

      I think that we are saying the same thing with different words. Glad to see that we are on the same page.

      I would add that the key behind our inability to make a roll go as far as a jump roll (even on grass) is due to our limitations in power (understanding “power” in its pure physics meaning: F x V).


  3. Joe Jordan says:

    I just discovered your website and really enjoy the “think outside of the box approach” to understanding fly casting. Every endeavor seems to have its orthodoxy. As an amateur historian, I constantly find lies that are accepted facts only because they have been repeated so many times.

    You prove your points very well using the laws of physics. I have discovered that the only way for me to get a good roll cast is to increase the mass of my line way over the recommended weight. I guess this is why skagit is so popular.


    • Aitor says:

      Hi there!

      I agree, Joe. Also the design of a skagit line helps in two ways: with all the mass “concentrated” in a shorter lenght our acceleration is applied more efficiently to the whole line; moreover a short head is easier to control for anchoring and D-loop forming.

      Anyway spey is spey. Whatever the name you want to give to your particular choice of tackle the underlying physics is exactly the same.

      Thanks and keep visiting, there is more to come! 🙂


  4. Nice read mate! It also explains why positioning of the rod rotation at the very end of a max long stroke in a static roll cast supports the final turn over better than anything else. Simply the D-loop increases before rotation starts and thus we accelerate more line. Regards Bernd


  5. Oh and same for the jump roll cast. Many casters bring the rod forward early: go, go, kiss. That pretty easily keeps the D-loop from growing. If I instead position (pause) my rod in the point of return between the back cast and forward cast I get mor line mass accelerated and cast further. Cheers


  6. […] I had been reading in all those books. The roll cast was an easy one —authors said— in fact easier than an overhead cast because you get rid of the backcast part. However, when practicing it my results were awful, to say […]

    Liked by 1 person

  7. Ivan says:

    io avrei una domanda,quale è la definizione di rotolato ?perchè in molte sequenze io vedo spesso un lancio in avanti,probabilmente sono io che ho qualche problema di comprensione,io ho imparato che per ottenere un rotolato efficente devo battere la punta della canna verso l’acqua,in questo modo riesco a mettere in movimento più linea,in effetti se lo faccio funziona e in questo caso si vede come se la linea formasse una ruota ,naturalmente utilizzando una linea normale dt,che sia diversa la definizione in base al tipo di linea utizzato? Oppure è solo dato per scontato che il vecchio rotolato è superato?


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