Claremont Posted October 22, 2008 Report Share Posted October 22, 2008 I was telling my son how finals and championships were run back in the days when we used to have 16 riders each having 5 races in a 20 race final or championship. Then it occurred to me that I really had no idea how this was done...how the draw for each heat was done so that each rider was ensured of racing every one of his opponents once over his 5 races. I even tried working it out for myself, but it seems to be a mathmatical thing and thats why I'm still so broke...just bloody awful at maths! lol Is there some sort of mathmatical formula that was used and does anyone know what it is or how it works? I know there must be some sort of trick to it because I know it can be manipulated for entertainment value. I saw this years ago while watching the Australian Championship at Claremont in about 1970. By the time we got to heat 20, each of the 4 riders involved had a change of lifting the title. And thats a huge coincidence! lol Quote Link to comment Share on other sites More sharing options...
Grachan Posted October 22, 2008 Report Share Posted October 22, 2008 For what it's worth it's the same formula as they use in the GP but without the semi-final and final. Don't have it to hand though! Quote Link to comment Share on other sites More sharing options...
norbold Posted October 23, 2008 Report Share Posted October 23, 2008 (edited) 1-2-3-4 5-7-6-8 10-11-9-12 15-14-16-13 13-1-5-9 14-10-2-6 11-15-7-3 4-8-12-16 6-16-1-11 12-5--15-2 8-9-3-14 13-4-10-7 7-12-14-1 2-13-8-11 16-3-10-5 9-6-4-15 1-8-15-10 9-2-7-16 3-12-13-6 5-14-11-4 Edited October 23, 2008 by norbold Quote Link to comment Share on other sites More sharing options...
Jeff. Posted October 23, 2008 Report Share Posted October 23, 2008 1-2-3-4 5-7-6-8 10-11-9-12 15-14-16-13 13-1-5-9 14-10-2-6 11-15-7-3 4-8-12-16 6-16-1-11 12-5--15-2 8-9-3-14 13-4-10-7 7-12-14-1 2-13-8-11 16-3-10-5 9-6-4-15 1-8-15-10 9-2-7-16 3-12-13-6 5-14-11-4 And it is entire possible, though unlikely, for any one of those last four races to be contested by four unbeaten riders Quote Link to comment Share on other sites More sharing options...
BOBBATH Posted October 24, 2008 Report Share Posted October 24, 2008 correct-I think it was a great system. Quote Link to comment Share on other sites More sharing options...
arnieg Posted October 25, 2008 Report Share Posted October 25, 2008 I was telling my son how finals and championships were run back in the days when we used to have 16 riders each having 5 races in a 20 race final or championship. Then it occurred to me that I really had no idea how this was done...how the draw for each heat was done so that each rider was ensured of racing every one of his opponents once over his 5 races. I even tried working it out for myself, but it seems to be a mathmatical thing and thats why I'm still so broke...just bloody awful at maths! lol Is there some sort of mathmatical formula that was used and does anyone know what it is or how it works? I know there must be some sort of trick to it because I know it can be manipulated for entertainment value. I saw this years ago while watching the Australian Championship at Claremont in about 1970. By the time we got to heat 20, each of the 4 riders involved had a change of lifting the title. And thats a huge coincidence! lol A quick bit of degree level number theory. It is what is known as a Steiner system. A steiner system is where 'n' numbers can be put in groups of size 'a' such that they share a common group with any other number 'x' and only 'x' times. Known as an (n,a,x) Steiner system. The 20 heat formula is a (16,4,1) Steiner system. In an individual competition 16 riders meet in groups of 4 (a heat) and meet each other only once. The World Cup race-off formula (when they had 5 teams competing) was also a truncated (i.e. they didn't race the heats where 5 riders from the same team met each other!) Steiner system. In this case it is a (25,5,1) Steiner system. There are relatively few sets of numbers for which this works [(9,3,1) is another handy one] and there are various mathematical proofs that demonstrate the circumstances in which a Steiner system will work. 30 years ago I got a question about this on my Oxford entrance exam and once I realised they were asking me about the formula for a speedway meeting I found it a remarkably easy question. It probably made the difference between me getting in and not getting in to Oxford Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.