Zevachim 64b - throw to the makom hadeshen

Zevachim 64a, it says that sometimes a kohen would have to make a difficult throw, more than 30 amos, from the southwest corner of the mizbeach to the makom hadeshen. Rashi says that the insides and feathers of the bird were hard to throw well, so that was tough. He figures out the >30 amos, basically a diagonal across a square, 22 amos west to east from the SW corner of the mizbeach to half an amah east of the kevesh - and 22 amos north to south from that corner plus a little, to ten amos north from the very end of the kevesh: 22x22.
Rashi doesn't mention that you'd need to clear the kevesh as well. That isn't that easy, given that the kevesh was around 3 amos high at that point (10 amos along out of 32, from a total height of 9 amos), and you need to land only 1/2 amah to the side of the kevesh. That's a pretty steep angle, and I'm wondering how high you would need to throw it to clear the kevesh and still land on the makom hadeshen. (Unless they drilled a tunnel through the kevesh shaped just right for the throw!)
Since the throw is a parabola, ignoring air resistence [maybe a problem according to that Rashi above] what one would need to do is fit a parabola to three points (x,y). You can google a number of sources for it, including an online calculator.
It's easiest to measure directly x along the diagonal between the makom hadeshen and the cohen, and y is up from the ground. The points are:
a) the makom hadeshen; call that (0,0),
b) the point on the kevesh it just misses (0.5*sqrt(2), 3), [you can calculate "3" more accurately]
c) and the cohen's arm (22*sqrt(2), 9+3) [3=height of cohen's hand?].
Once you have the parabola, you just need the height of the vertex. It'd also be interesting to know the angle at which he has to throw it up.
Has anyone seen this before, or do I need to try and work it out?
Thanks!

Okay, if the cohen throws from 12 amos up (top of mizbeach + 3), I get
y=(-0.1238)x^2 + 4.238x, maximum height 36.27.
So it would need to go 24 amos up, in addition to the 31 amos sideways. The initial slope (for y' = (-.2476)x + 4.238) is -3.4655, which is an angle of 73.9 degrees.

If the cohen throws from 9 amos up (just top of mizbeach, or according to Tosefos who implies that there may have been a ramp to the SW soveiv = 6 amos up, + 3 amos height), I get
y = (-0.1270)x^2 + 4.241x, maximum height 35.41. It would need to go 26 amos up, in addition to the 31 amos sideways. The initial slope (for y' = (-.2540)x + 4.241) is -3.662, which is an angle of 74.7 degrees.

Either way, a very difficult throw indeed!

On further reflection, the presence of air resistance might make the problem easier. The calculations are more complicated and you can find treatments online. But just to simplify, you can picture that the thrown object loses horizontal velocity, and eventually falls nearly vertically.
If so, the best solution may be to throw the bird parts nearly horizontally, but just fast enough that they/it slows down when finishing its time over the kevesh. Then it curves down sharply and drops neatly onto the makom hadeshen. The air resistance actually helps.
A difficult throw indeed! It needs to be just the right speed.

I've summarized this discussion, along with an additional point, at https://drive.google.com/open?id=1yQzi-FTvUcWhJVsWPi6-803wYghoHG1D