George said:I've finally had time to think about your last figure. Treating the problem in terms of energy instead of forces certainly clarifies some things.
Yes, I need to also consider the dissipation explicitly, which adds the necessary entropic element to the whole process and the sink in the energy budget.
George said:How does the potential energy fit in to your three different throws? I can see that the nose-down throw reaches an equilibrium point sooner than the level throw or the nose-up throw. I would assume that means that the latter portion of the flight is more energy efficient for the nose-down throw. (Or rather, that the nose-down flight makes efficient use of the gravitational energy sooner.) However, does the nose-down throw have more or less potential energy when it reaches that point? In other words, if the nose-down throw follows a very flat trajectory, it might be very energy efficient at the end but not have as much height and therefore as much potential energy to work with.
Its a good question to see how the apex of the flight trades off in this manner relative to nose angle. I haven't worked out the details of this kind of thing yet, but when I can get a good estimate for the dissipation integrated over the flight path, then I can put more secure numbers and see how all of those things trade off. Note that the actual velocity and kinetic energy for the nose down flight going down a curve from a given point in the plot is significantly higher than for the nose up flight (i.e., nose down cruise speed, around the apex, is greater than for nose up).
Even more important is the dissipation and efficiency, as you suggest. An interesting result hinted at on the figure, is that the equilibrium glide angle of descent is smaller for more nose down. Glide is where dissipation equals work done by gravity, so the disc goes at a constant speed (I put a Sidewinder into a glide today at hole 27, DeLaveaga, "top of the world," as part of my field work). So the nose down is indeed more energy efficient. The reason for this is probably that the drag force goes as a power of angle of attack that seems to be of order higher than linear according to Potts' experiments, while the lift force has been measured to be approximately linear in the angle of attack (this is also true of all sorts of airfoil shapes). This means that increasing nose up makes the angle of attack higher, and the drag force's quadratic dependence out-competes the linear lift force dependence...thus nose down=less drag upon descent, and hence feeds upon potential energy at a lower rate.
George said:Also, how significant is the potential energy? Is it a large or small fraction of the kinetic energy? I can see the drop in kinetic energy as the disc ascends, but how much of that is work against gravity and how much is it work against drag forces? (I tried a back of the envelope calculation, guessed a flight speed of 20 m/s and an apex height of 5m, and came up with about a 1:4 potential to kinetic ratio, but I'm not sure how realistic my assumptions are.)
g*h=50, while v^2/2=200, so in this case 1:4. Probably about right, although these numbers can vary a bit. If you get the disc to apex at 10 m you will need more velocity to get it there...to know exactly how much becomes a matter of dissipation once again. It seems clear that the greatest energy cost is getting the disc to the apex...I think for big arms, this is where all the effort goes...you probably need a nose down just to keep the disc from rising up like mad, and then you aim to put it through some apex...once it is there it simply glides back down. So probably a good amount of kinetic energy is eaten up during this process, although the potential energy is still not varying a great deal.
Thinking about it, in principle this information might be obtainable from the present plot and formula via a subtle manipulation, since the rate of increase of potential energy is proportional to the square root of the vertical axis times the sine of the horizontal axis...easy breezy, no?
George said:How does the rotational kinetic energy relate to the drag energy loss and to the linear kinetic energy? In other words, as the angular velocity of the disc gets smaller, where does the rotational kinetic energy go? My guess would be that most of the loss is to drag and to (slight) sideways motion of the disc caused by a Magnus force. I would also guess that the disc only loses a small fraction of its rotational kinetic energy during the flight. (I assumed that the kinetic energy on your vertical axis was the linear part of kinetic energy.)
I agree with this assessment. I also don't think there is much Magnus force for typical throws, and neither does the angular velocity change much over the flight (which is saying pretty much the same thing).
George said:Overall, it seems to me that there are four relevant energy terms in the problem--linear kinetic, rotational kinetic, potential, and loss to drag. It would be interesting to see all four of those plotted as a function of time. Alternatively, if you could replot your figure with the potential plus kinetic energy on the vertical axis that would give different and helpful information as well.
Very true. The kinetic plus potential energy would have to decrease during flight, owing to continual drag loss, so I think the curves might follow a similar trajectory, but maybe not as much upward curvature of the curves around the apex to reflect the extra potential energy in the flight around the summit of the path.
George said:I don't mean to be critical--your work and your figure are both outstanding and very helpful and I'm quite impressed. They just raise lots of interesting questions! Thanks for posting it!
Thanks for the queries. This is indeed a fun way to look at things, and I'm glad you also enjoy it. I intend to follow through with this to the bitter end (where ever that is). (In my day-to-day work in geophysics I am also a big fan of energy balances, so this is in keeping with my usual form.)