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Hauling, Rod-loading.
On Nov 10, 1:35*pm, "
wrote: Whether the mathematical model is entirely accurate, I am no longer sure. *I will have to check it again. I believe the magnitude of the air resistance to be relatively inconsequential to the casting mechanics (though on a windy day, it's effect is undeniable afterward), so with your permission, I'll leave it out for simplicity. That makes your equation to be: Frt = Fi * Fa * Flt It's my contention, however, the the force terms should be grouped, and that the unbalanced force causing the line acceleration is actually (Frt - Flt). This allows your model to fit into the F=ma equation as: (Frt - Flt) = Fi * Fa I don't think that's a great deal different than you were describing it verbally; but it *is* a great deal more accurate as a physics equation. The units issue I mentioned is manifest in this series of equations from your post: 30g * 1m/s² * 0.3 = Flt / 0.01kgm/s² 30g * 1ms² = 0,03kgm/s² * 0,3 = 0.09 kgm/s² Note that between the first and second equations, you simply discarded the kg.m/s² units in the demoninator, using only the same units that go with the Flt force in the numerator. Had you retained those units as you should have, the numerator and denominator units would cancel and you'd have ended up with no units at all on the right side, but g.m/s² on the left. |
Hauling, Rod-loading.
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