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Chapter:   Werfertreffen - Event Calendar - Sonwend-Werfertreffen Pottenstein 2002 - Supplement (.) Rules and Protocols

Draft - not yet complete

Since we had no rules yet, we quickly made some.

Anno has written down the protocol for the contest.
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The above sketch means the following. It are only two disciplines: Knife and Axe. In each discipline several distances are offered. Knife: 3 m, 5 m and 7 m, Axe 3 m and 4.5 m. Each thrower has 3 rounds with three throws each, means 9 throws for one discipline. One throw counts maximal 2 base-points. Nine throws with 2 base-points makes maximal 18 base-points. The base-points are multiplicated with the distance. So for 3 m are maximal 18 * 3 = 54 end-points. For 5 m maximum is 18 * 5 = 90. For 7 m distance you have maximal 18 * 7 = 126 end-points.

Jury notes (part 1 of 3).
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Jury notes (part 2 of 3).
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Jury notes (part 3 of 3).
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The notes above mean the following.

First the knives ...

 

Contest Protocol Knife
Distance	Franz	Anno	Tarzan	Mümmel
Runde 1
3 m	1 + 1 + 2 4 * 3 = 12	0 + 1 + 2 3 * 3 = 9	0 + 1 + 2 3 * 3 = 9
3 m		0 + 1 + 2 3 * 3 = 9
5 m	0 + 1 + 2 3 * 5 = 15	0 + 0 + 1 1 * 5 = 5	0 + 1 + 1 2 * 5 = 10
5 m			0 + 0 + 1 1 * 5 = 5
7 m	0 + 1 + 1 2 * 7 = 14
Summe Runde 1	4 + 3 + 2 = 9 12 + 15 + 14 = 41	3 + 3 + 1 = 7 9 + 9 + 5 = 23	3 + 2 + 1 = 6 9 + 10 + 5 = 24
Runde 2
3 m	1 + 1 + 2 4 * 3 = 12	1 + 2 + 2 5 * 3 = 15	1 + 1 + 2 4 * 3 = 12
3 m		0 + 1 + 1 2 * 3 = 6
5 m	1 + 1 + 2 4 * 5 = 20	0 + 0 + 0 0 * 5 = 0	0 + 1 + 2 0 * 5 = 0
7 m	0 + 1 + 2 3 * 7 = 21		0 + 1 + 2 3 * 7 = 21
Summe Runde 2	4 + 4 + 3 = 11 12 + 20 + 21 = 53	5 + 2 + 0 = 7 15 + 6 + 0 = 21	4 + 0 + 3 = 7 12 + 0 + 21 = 48
Runde 3
3 m	1 + 2 + 2 5 * 3 = 15	0 + 1 + 1 2 * 3 = 6	1 + 2 + 2 5 * 3 = 15
5 m	1 + 2 + 2 5 * 5 = 25	1 + 1 + 2 4 * 5 = 20	0 + 1 + 1 2 * 5 = 10
7 m	0 + 1 + 2 3 * 7 = 21	0 + 0 + 0 0 * 7 = 0	0 + 0 + 0 0 * 7 = 0
Summe Runde 3	5 + 5 + 3 = 13 15 + 25 + 21 = 61	2 + 4 + 0 = 6 6 + 20 + 0 = 26	5 + 2 + 0 = 7 15 + 10 + 0 = 25
Gesamt Messer	9 + 11 + 13 = 33 41 + 53 + 61 = 155	7 + 7 + 6 = 20 23 + 21 + 26 = 70	6 + 7 + 7 = 20 24 + 48 + 25 = 97

... then the axes. Here all throwers threw the same distances in each round, each throw once 4 m and once 7 m. Therefore the stick-points are directly comparable. Interrestingly, with the sticking-points and with the distance-points, the same winners order is achieved. Fine -- so that multiplication method may not be too unrealistic perhaps.

 

Contest Protocol Axe
Distance	Franz	Anno	Tarzan	Mümmel
Axt Runde 1
4 m	0 + 0 + 0 0 * 4 = 0	1 + 1 + 2 4 * 4 = 16	0 + 0 + 1 1 * 4 = 4	1 + 1 + 2 4 * 4 = 16
7 m	0 + 0 + 0 0 * 7 = 0	1 + 1 + 2 4 * 7 = 28	0 + 0 + 1 1 * 7 = 7	0 + 0 + 0 0 * 7 = 0
Summe Axt Runde 1	0 + 0 = 0 0 + 10 = 0	4 + 4 = 8 16 + 28 = 44	1 + 1 = 2 4 + 7 = 11	4 + 0 = 4 16 + 0 = 16
Axt Runde 2
4 m	1 + 1 + 2 5 * 4 = 20	1 + 2 + 2 2 * 4 = 8	1 + 1 + 2 5 * 4 = 20	0 + 1 + 2 3 * 4 = 12
7 m	0 + 0 + 1 1 * 7 = 7	0 + 0 + 0 0 * 7 = 0	0 + 1 + 1 2 * 7 = 14	0 + 0 + 0 0 * 7 = 0
Summe Axt Runde 2	5 + 1 = 6 20 + 7 = 27	2 + 0 = 2 8 + 0 = 8	5 + 2 = 7 20 + 14 = 34	3 + 0 = 3 12 + 0 = 12
Axt Runde 3
4 m	0 + 1 + 1 2 * 4 = 8	0 + 1 + 2 3 * 4 = 12	1 + 2 + 2 5 * 4 = 20	0 + 1 + 2 3 * 4 = 12
7 m	1 + 2 + 2 0 * 7 = 0	1 + 1 + 2 3 * 7 = 21	0 + 1 + 1 3 * 7 = 21	0 + 1 + 2 0 * 7 = 0
Summe Axt Runde 3	2 + 0 = 2 8 + 0 = 8	3 + 3 = 6 12 + 21 = 33	5 + 3 = 8 20 + 21 = 41	3 + 0 = 3 12 + 0 = 12
Gesamt Axt	0 + 6 + 2 = 8 0 + 27 + 8 = 35	8 + 2 + 6 = 16 44 + 8 + 33 = 85	2 + 7 + 8 = 17 11 + 34 + 41 = 86	4 + 3 + 3 = 10 16 + 12 + 12 = 40

 

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© 2000 - 2003 by Norbert C. Maier   # 20020630.1701