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  #31  
Old 03-10-2020, 10:48 AM
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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:
Originally Posted by ALT-J View Post
The concept of acceleration is kind of hard. In disc golf its probably easiest to see the effect in putts. The best and most consistent and repeatable putts always feel like that the time span for the actual throw is very small and compact. Yet the disc flies 60-70 feet with a very small movement. Its like going from 5 mph to probably 40 mph in a fraction of a second that makes the disc fly. Thats very confusing and counter intuitive for me for some reason.
This last thing you said about the compact putt isn't any different (remember, the disc is only 175 grams) but there are two things that are overlooked:
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.
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  #32  
Old 03-10-2020, 11:47 AM
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HyzerUniBomber HyzerUniBomber is offline
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Quote:
Originally Posted by Blobfish View Post
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
Not quite. Your hand will be going slower than 80mph to have the disc eject @ 80mph. That's the whole point of redirecting the disc (hand on the outside), loading the wrist and using a hit. That is what accelerates the disc. I can barely move my hand and redirect the disc and get it to go much faster than my hand speed.

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

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  #33  
Old 03-10-2020, 12:04 PM
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Quote:
Originally Posted by BogeyNoMore View Post
Is it OK if I just throw my disc?


Kidding aside, while I really do think it helps to understand the basic physics of disc flight, the physics pretty quickly gets more complex than your basic two semesters of Physics 101 & 102 addresses.... which is why most people won't be able to follow the complete physics of disc flight.

However, you can get a great deal of practical knowledge by accumulating a wealth of observational info. That helps build a sort of internal database, which ultimately leads to a feel for your discs, and the game.

I believe most people's best throws are about "feel" rather than about their comprehension of disc physics.
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  #34  
Old 03-10-2020, 12:11 PM
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Blobfish Blobfish is offline
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Quote:
Originally Posted by HyzerUniBomber View Post
Not quite. Your hand will be going slower than 80mph to have the disc eject @ 80mph. That's the whole point of redirecting the disc (hand on the outside), loading the wrist and using a hit. That is what accelerates the disc. I can barely move my hand and redirect the disc and get it to go much faster than my hand speed.

Yes, you could drag a disc 80mph and not redirect it, but that would be wildly inefficient.
Yes, you're correct there. I made a lot of assumptions to simplify the problem. It's like a bendy shaft on a golf club that effectively whips the head. I tried to address the snap in a way in the second point, but yeah, good point about the hand speed.
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!

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  #35  
Old 03-10-2020, 04:24 PM
Disc Golf Doctor Disc Golf Doctor is offline
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Quote:
Originally Posted by etdefender19 View Post
I threw my Falcon (by Millenium Discs) through the entire Kessel course in 12 parsecs
a parsec is a measurement of distance
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  #36  
Old 03-10-2020, 04:55 PM
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Quote:
Originally Posted by Blobfish View Post
Inertia.


Way too long answer follows:

Mass is the way we quantify inertia (for all intents and purposes, inertia and mass are synonymous).

A qualitative description of inertia is an object's resistance to a change in its state of motion. Newton's....
This is why I quit teaching bio and physics and started fighting wildfires---the students didn't want that depth of focus. Well, that, and I would've been a forest ranger already, if not for Ronnie.
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  #37  
Old 03-10-2020, 04:57 PM
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Quote:
Originally Posted by Disc Golf Doctor View Post
a parsec is a measurement of distance
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.

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  #38  
Old 03-10-2020, 04:58 PM
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Quote:
Originally Posted by Disc Golf Doctor View Post
a parsec is a measurement of distance
Right. Han and chewie got their reputation by taking a ballsy shortcut, not just by being fast.

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  #39  
Old 03-10-2020, 08:40 PM
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etdefender19 etdefender19 is offline
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Oopsie

Last edited by etdefender19; 03-10-2020 at 08:42 PM.
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  #40  
Old 03-11-2020, 06:07 AM
ALT-J ALT-J is offline
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Quote:
Originally Posted by HyzerUniBomber View Post
Not quite. Your hand will be going slower than 80mph to have the disc eject @ 80mph. That's the whole point of redirecting the disc (hand on the outside), loading the wrist and using a hit. That is what accelerates the disc. I can barely move my hand and redirect the disc and get it to go much faster than my hand speed.

Yes, you could drag a disc 80mph and not redirect it, but that would be wildly inefficient.
So where does the term arm speed come from? Clearly its not the thing you want to look for and so many people including me are trying too hard instead of a controlled movement that gets the job done better power and controlwise. Perhaps theres a ratio between lower arm speed and disc ejection speed that could enlighten what we need to do and look for. If we know the length of a lower arm and its speed, is it possible to know how much faster it will eject in ideal conditions?

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 ALT-J; 03-11-2020 at 06:09 AM.
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