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!

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!