Guest Dan Report post Posted November 15, 2008 Found this on another board. Believe it or not, the author is serious. "Some considerations on casting and rod loading. A simple calculation for casting in air is; Frt = Fi * Fa * Ff * Flt Where Frt = the force on the rod tip in kg.m/s², Fi = inertia (mass) in grams, Ff = the coefficient of fluid kinetic friction µk ( air resistance), Fa = the acceleration of the line in ms², and Flt = line tension in kg.m/s² As may be seen, the equation depends on line tension being greater than zero to produce a positive value. The higher the line tension, the greater the value. If the value is zero or less, then Frt=0 This equates to " No tension= "no force on the rod tip" To find line tension, rod loading, acceleration,friction, etc. one may simply substitute the equations, you can calculate all the variables, and also prove that line tension is a major factor. Frt = Fi * Fa * Ff* Flt Flt = Frt / fi * Fa * Ff Fi * Fa * Ff = Flt / Frt So, as the rod loading at any given point is known, ( it is simply the curve of the rod, can be measured statically for any weight). We will assume a rod loading of 0.01 kgm/s² (10 grams ). The line mass can be weighed. Assume 30g here. As the static line tension is exactly equal to the static rod loading this must also be 0.01kgm/s² but only when the rod and line are static! What is left when the rod/line is moving must be the acceleration. Tension is required to accelerate the line. However, the actual acceleration of the line, and friction, are extra variables we don?t know yet. Plugged in to the first equation, we get; 0.01kgm/s? = 30g * Fa * Ff * Flt Second equation; Flt = 0,01 kgm/s? / 30g * Fa * Ff Third equation; 30g + Fa * Ff = Flt / 0.01kgm/s² We still need to know the coefficient of friction and the acceleration. Unfortunately, as this coefficient is not a fundamental force, it can not be derived from first principles, and must be observed empirically. In this case we will simply assume it to be 0.3. We don't know the acceleration either, but we will also simply assume a value here, of 1ms² That gives; 30g * 1ms² * 0.3 = Flt / 0.01kgm/s² 30g * 1ms² = 0,03kgm/s² * 0,3 = 0.09 kgm/s? Therefore, Flt= 0,09kgm/s² / 0,01kgm/s² The units cancel, and Flt = 9kgm/s² Add the values to all equations; Frt = Fi * Fa * Ff * Flt 0.01kgm/s² = 0,03kgm/s² * 1ms² * 0.3 * 9kgm/s² Second equation; Flt = Frt / fi * Fa * Ff 9kgm/s² = 0.01kgm/s² / 0,03kgm/s² * 1ms² * 0.3 Third equation; Fi * Fa * Ff = Flt / Frt 0,03 kgm/s² * 1 ms² * 0.3 = 9kgm/s² / 0.01kgm/s² This proves all equations. Plugging in the values you have for any particular conditions will tell you the line tension, the rod loading, the acceleration, and the friction. If you graph the information, you can read it off directly. You can also see how changing the mass changes the tension and acceleration, how friction affects the model, and a lot of other things. You can also plug in the force for a haul, and see how it affects the setup. Lots of things are possible. These equations are rudimentary, but cover all major factors. I am still working on equations for the conversion of line tension to line momentum. The equation shown is also primarily designed to show what happens on the forward stroke. From when the rod begins loading. One may of course adjust it, and add other factors if desired. There are a couple of points worthy of note. The fluid friction varies according to the amount of line outside the rod tip, as of course does the mass, and its velocity. Once line has rolled out and is shot or released, the tension on the line itself is governed by the momentum of the line pulling on the backing. This retains some tension on the line. As long as the line stays straight, as a result of this tension, it will fly further. Once it starts to "crinkle" it collapses. In order for the line to turn over completely, there must be sufficient tension for it to do so. The equation shown is just one of a series which I am trying to use to set up a casting simulator, first as a mathematical model, and then including programmed graphical elements. The target is a dynamic model of casting, into which one may plug in any rod or line, and also show the optimal length and weight for shooting heads etc etc. Hopefully it will also show the effects of differences in rod tapers and action. One of the main things of note here, is that it is rod and line tension which keeps the line swinging back and forth when false casting with a fixed line. The force applied to the butt only adds sufficient force to account for "losses" to fluid friction. Also, one does not "throw" or "cast" the line, one rolls it out. When the line is rolling out, it is the tension on the bottom leg of the line loop which causes this. In order to convert the rod and line tension to line momentum, when shooting line, the point at which tension is released, and how this is done, is of major importance. In order to "force" turnover for instance, more tension is required. This can be done by "overpowering" the cast, or by using a "check haul", pulling back on the line before it has unrolled will cause tension to increase, and the line turns over faster. Pulling back on a line which is already unrolled will of course merely brake it. This also demonstrates how hauling works, it does not accelerate the line, or load the rod much, it increases system tension, mainly line tension, which is converted to momentum. This theory, and the related equations, are my original work, if you use it, please credit where you got it from. TL MC" Mike Connor-ROFF Quote Share this post Link to post Share on other sites
Guest Nick C Report post Posted November 15, 2008 Well... I can't wait to try this method tomorrow This person sure knows their physics... maybe he should get his phD in fly casting. All kiding aside, The effort and research that this person has done is definatley appreciated. Nick C Quote Share this post Link to post Share on other sites
Bear 77 Report post Posted November 15, 2008 What? You making my head hurt. I think I will stick to my bass rods and some nice chunks of plastic....see structure-throw bait-feel bite-set hook-fight fish-repeat as necessary. Bear Quote Share this post Link to post Share on other sites
Guest Nick C Report post Posted November 15, 2008 Exactly how fishing should be be Bear and thats it. Nick C Quote Share this post Link to post Share on other sites
Castnblast 86 Report post Posted November 15, 2008 What? You making my head hurt. I think I will stick to my bass rods and some nice chunks of plastic....see structure-throw bait-feel bite-set hook-fight fish-repeat as necessary. Bear good one Bear, that's funny s**t right there Quote Share this post Link to post Share on other sites
Guest Nick C Report post Posted November 15, 2008 I agree Nick C Quote Share this post Link to post Share on other sites
Guest fish_on Report post Posted November 16, 2008 Very interesting facts, thanks for sharing that Dan Quote Share this post Link to post Share on other sites