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#31




F= ma = m*dV/dt
So: F*dt = m*dV (this is the impulse momentum theorem... a fancy way to reiterate Newton's Second Law) F*dt is the definition of impulse (applied force multiplied by the duration of time that the force is applied) m*dV is the change in momentum (where by definition, momentum is mass * velocity) The mass of the disc does not change, and since you are already holding it up in equilibrium against its weight to keep it on a straight line, you can ignore it's weight (which as others have said, would be 0.175*9.81=1.72N). However, this mass is accelerating sideways in your throw. Again, the mass does not change (it's inertia does not change), but the desired acceleration dictates how much force will be needed to be applied. Going back to this equation: F*dt = m*dV You have a fixed dV (80 mph  0 mph) and you have a fixed mass (175g) (I'm ignoring normalizing the units for now because we're not actually plugging in the numbers) Thus, the less time you take to bring that disc up to speed (smaller dt) requires greater F. Mathematically, F and dt are inversely related by a factor of mdV But now look at it from another perspective of work done on the disc: The simplified physics definition of work is W=F*dx which means that the work done on an object is the product of the distance the object travels multiplied by the component of the force applied to the object that lines up with the direction of travel. I'm making the assumption that the disc is being pulled in a straight line. Therefore, the work is going to be the same force we alluded to before multiplied by the difference in distance between where you let go (final position) and the start of your reachback (initial position). Work done on an object also directly corresponds to the change in kinetic energy of the object (0.5 m*v*v) where in this case, the "v" is the 80 mph So W=F*dx by definition, but also W=dK which in this case because initial velocity was zero so initial K was zero, means W = final K = 0.5*m*v*v Then, F*dx = 0.5*m*v*v And solving for F: F = 0.5*m*v*v/dx which translates as follows: For a given mass (175g) and a desired final velocity (80mph), the force needed is dependent on how long the distance the force is applied. Long distance (long arms and thus long dx) means less force needed. Short distance means more force is needed to achieve the same result. Look at sprinters and a lot of football tailbacks. They can be short with really powerful legs because they need that burst of power because their race is over in a short distance. Or if you want to think automotive, they are geared down for power and low end acceleration. On the other hand, look at longer distance runners or wideouts and you'll see longer legs because they are geared higher for maintaining top speed once they get there (Usain Bolt is an anomaly). Quote:
1. In the example used above, we made the assumption that you're trying to get the disc up to 80 mph. This involves getting your hand up to 80mph which means getting your arm up to speed (something less than 80 mph) to get your hand going that fast. You are now doing work to a whole lot more than 175g, as your whole body is involved. I was isolating just the point of contact of your hand providing the external force to the disc 2. Glide. Putters (unless you're throwing a RDG Scale, Kastaplast Berg, Innova Rhyno or some other brick), your putter is designed to stay aloft. A quick snap of the wrist or fingertips will have a very low dt and at the same time will have a low dx, but as you are only working on the disc itself (not trying to throw your whole arm), the force required will be small. Sponsored Links

#32




Quote:
Yes, you could drag a disc 80mph and not redirect it, but that would be wildly inefficient.

#33




Quote:

#34




Quote:
The disc has to rotate around your last point of contact just like the tip of a bullwhip. I was trying to focus on the "heavy" sensation that the OP mentioned when ripping a disc, but you're absolutely correct, thanks for pointing it out!

#35




a parsec is a measurement of distance

#36




This is why I quit teaching bio and physics and started fighting wildfiresthe students didn't want that depth of focus. Well, that, and I would've been a forest ranger already, if not for Ronnie.

#37




I fail to see what the issue is with a parsec being a unit of measure of distance? He threw the disc fast enough to warp space.

#38




Right. Han and chewie got their reputation by taking a ballsy shortcut, not just by being fast.

#39




Oopsie
Last edited by etdefender19; 03102020 at 08:42 PM. 

#40




Quote:
A good example is my friend who has a good form but really tries to make his arm go fast and the disc slips. He's much more efficient and has even more power when he just focuses on a smooth throw. Last edited by ALTJ; 03112020 at 06:09 AM. 
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