Consecutive captures on the same square

Since we are talking about a standard game of chess (although with both players co-operating), we know that there are four pieces that cannot possibly make a capture in the series:

• the two bishops on the wrong coloured squares
• one of the kings (the other can be the last to capture)
• the first piece that gets captured (cannot be any of the above).

Furthermore, getting the maximum number of pawns involved is tricky. It seems best to group the pawns in quadruplets (e.g. all four pawns on A and B files), and then sacrifice 1 pawn to promote the other 3. We can also feed the useless bishops from above to the lane-switching process, each sacrificed bishop allows 1 pawn pair to promote.

Since we can leave two pairs of pawns unpromoted and attacking the same square, and the useless bishops enable 2 more pawn pairs to promote, we will only have to sacrifice 2 pawns to promote the rest of them.

Therefore, the maximum number of consecutive captures on the same square is

To confirm that this is possible, let's construct an actual game.

To get started, let's make the unavoidable sacrifices. The plan is to do the captures at f5.

1.  b4   a5 
2.  bxa5 c5 
3.  d4   cxd4
4.  g4   g6
5.  Bf4  f5
6.  Bg3  f4
7.  h4   fxg3
8.  h5   Bg7
9.  h6   e6 
10. hxg7 g2 
11. e4

In this position we can confirm that now every piece can eventually reach f5 without any further captures, so it's just a matter of doing the tedious arrangement work: (this could definitely be improved, in efficiency and coherence both)

11. -     Bb7 
12. Rh5   g2
13. Rg5   h5 
14. c4    h4 
15. c5    h3 
16. Nf3   h2
17. Nc3   d3 
18. c6    h1=N
19. Nh4   Ng3 
20. Kd2   g1=N
21. Ke3   b4
22. f4    b3
23. f5    b2 
24. Kf4   Nf3
25. Bh3   d2 
26. Qc2   Qf6
27. Nd5   Rh5
28. Ne3   d5
29. c7    d4 
30. c8=R+ Kd7
31. Qc5   b1=R
32. Qe5   Rb5
33. Rc5   Ra6 
34. a4    Rd6 
35. a6    Rd5 
36. a7    Nh6
37. g8=B  d1=B 
38. Bh7   Bc2 
39. a8=N  Bc8 
40. Nc7   Nc6
41. Ne8   Ne7
42. Ng7   d3 
43. a5    Nd4 
44. a6    d2
45. a7    d1=Q 
46. a8=N  Qd3 
47. Nc7   Kd8 
48. Nce8  Bd7
49. Nd6   Qf8 
50. Ra5

Now all the ducks are properly lined up:

and it's finally time for the fireworks:

50. -     Qxf5+
51. Qxf5  gxf5 
52. exf5  Qxf5+
53. gxf5  Bxf5 
54. B3xf5 exf5 
55. Rxf5  Rdxf5+
56. Nhxf5 Ngxf5
57. Rxf5  Bxf5 
58. Ngxf5 Rbxf5+ 
59. Rxf5  Nhxf5
60. Ndxf5 Nexf5
61. Bxf5  Nxf5 
62. Nxf5  Rxf5+
63. Kxf5

As a final note, it's worth noticing that doing black's last capture with a rook is important: trying to do it with a bishop or a knight comes with a nasty surprise: the white King won't be able to complete the final capture, because