Athletic track(triloki)

1. ATHLETIC TRACK MARKING PLAN

2. DEVELOPED BY TRILOKI PRASAD M.P.ED GURU GHASIDAS UNIVERSITY BILASPUR

3. TYPES OF TRACK SURFACE 1: SYNTHETIC TRACK 2: CINDER TRACK 3: GRASSY TRACK

4. DIRECTION OF TRACK NORTH WEST EAST SOUTH

5. POINT TO REMEMBER IN TRACK MARKING 1 : LANE :- Lane is the path through which the athletics run .the width of the lane is 1.22m to 1.25m 2 : LINE :- The markings on both sides of the lane is called line .the width of these lines should be 5cms each . 4 : TRACKS EVENTS :-The competitions which are conducted in the track 3 : TRACK :- Track includes all the lanes.

6. Methods of finding the total area required for track marking NO.OF LANES ON BOTH SIDES = 8*2 = 16 1:- LENGTH OF STRAIGHT 2:- WIDTH OF LANE 3: - NUMBER OF LANES ( IN STANDARD TRACK THERE SHOULD BE 6 TO 8 – 9 LANES ) 4 : - CURVE DISTANCE RADIUS ( C. D.R ) 5:- EXTRA SPACE 5 MTS

7. WIDTHS OF LANES = 1.22MTS ONE C.D.R = 36.50 MTS ONE STRAIGHT = 84.39 MTS TOTAL LENGTH = ( ONE STRAIGHT ) + (2 * C.D.R ) + ( WIDTH OF LANE * NO. OF LANES ) + ( 2 * EXRTA SPACE ) TOTAL AREA = TOTAL LENGTH * TOTAL WIDTH

8. ONE EXTRA SPACE = 5 MTS ( ON BOTH SIDES ) TOTAL LENGTH = 84.39 + (2 * 36.50 ) + ( 1.22 * 16 ) +( 2* 5) = 84.39 + 73 + 19.52 + 10 = 186.91 MTS TOTAL WIDTH = ( WIDTH OF LANE * NO. OF LANES ) + ( 2* C.D.R.) + ( 2* EXTRA SPACES ) = 1.22 * 16 ) + ( 2* 36.50 ) + 2* 5)

9. THE TOTAL AREA NEEDED FOR A TRACK = TOTAL LENGTH * TOTAL WIDTH = 186.91 * 102.52 = 19162.013 SQUARE MTS = 19.52 + 73 + 10 = 102.52 MTS

10. METHOD OF FINDING THE TOTAL AREA FOR AN ATHLETICS TRACK MINIMUM RADIUS = 35 MTS MAXIMUM RADIUS = 38 MTS MEAN RADIUS = 35+ 38 / 2 = 73 / 2 = 36.50

11. WHAT IS ACTUALLY USED FOR MARKING THE TRACK IS CALLED CURVE DISTANCE RADIUS OR MARKING DISTANCE RADIUS 2:- CURVE DISTANCE RADIUS ( C.D.R.) IT IS THE IMAGINARY LINE THROUGH WHICH THE ATHLETE IS SUPPOSED TO RUN 1:- RUNNING DISTANCE RADIUS ( R. D.R ) TYPES OF RADIUS IN ATHLETICS TRACK

12. THE METHOD TO FIND CURVE IF CURVE DISTANCE RADIUS ( C.D.R.) IS KNOWN C.D.R. = 36.50 MTS R.D.R = C.D.R + 0.30 MTS 36.80 MTS CIRCUMFERENCE OF CURVE = 2^r r = 36.80 MTS ^ = 3.14159 = 2* 3.14159 * 36.80 = 231.22 MTS

13. 1.12M 2.34M 3.56M 4.78M 6.00M R.D.R 36.80 R.D.R 37.92 R.D.R 39.14 R.D.R 40.36 R.D.R 41.58 R.D.R 42.80 R.D.R = C.D.R + 0.30 MTS C.D.R = R.D.R – 0.30 MTS R.D.R = 36.80 MTS

14. METHOD TO FIND OUT STRAIGHT WHEN CIRCUMFERENCE OF THE CURVES IS KNOWN TOTAL LENGTH OF THE TRACK = 400 MTS TOTAL LENGTH OF CURVES = 231.22 MTS ONE CURVE LENGTH = 231.22/ 2 = 115.61 MTS

15. ONE STRAIGHT LENGTH = 168.78/ 2 = 84.39 MTS TOTAL LENGTH OF THE STRAIGHTS = 400 – 231.22 = 168.78 MTS

16. METHOD TO FIND OUT CURVES WHEN LENGTH OF THE STRAIGHT IS KNOWN TOTAL LENGTH OF TRACK = 400 MTS TOTAL LENGTH OF ONE STRAIGHT = 84.39 MTS TOTAL LENGTH OF TWO STRAIGHT = 84.39 + 84.39 MTS = 168.78 MTS

17. TOTAL CIRCUMFERENCE OF THE CURVES IS = 231.22 MTS LENGTH OF ONE CURVE = 231.22 / 2 = 115.61 MTS TOTAL LENGTH OF CURVES = 400 – 168.78 = 231.22 MTS

18. METHOD TO FIND OUT RADIUS WHEN CIRCUMFERENCE OF THE CURVES KNOWN NUMBER OF CURVES IS TWO LENGTH OF ONE CURVE =115.61 MTS LENGTH OF TWO CURVES 115.61 + 115.61 = 231.22 MTS

19. R = 36.80 MTS = 36.7998 MTS R = 231.22/ 2* 3.14159 = 231.22/ 6.28318 ^ = 3.14159 CIRCUMFERENCE = 231.22 MTS R = CIRCUMFERENCE / 2^ FORMULA OF FINDING THE RADIUS

20. CIRCUMFERENCE OF CURVES = 231.22 MTS STRAIGHT LENGTH = 84.39 + 84.39 = 168.78 MTS RADIUS OF CURVES = 36.80 MTS

21. LANES MARKING 1.21-1.23mts 1.21-1.23mt s 1.21-1.23mts

22. PYTHEGORUS THEOREM OR BI – SECTOR 3 - 4 – 5 6 -8 – 10 9 – 12 – 15 BASE + PERPENDICULAR  = HYPOTENUSE  A  + B  = C   3 + 4  5 9 + 16 = 25 25

23. PYTHEGORUS THEOREM FORMULA P A O 5 MTS 3 MTS 5 MTS 4 MTS 4 MTS 3 MTS

24. BI- SECTOR METHOD

25. 84.39MTS 42.26MTS 36.50MTS 1.22 * 8 9.76MTS 36.50MTS 400 MTS STANDARD TRACK

26. Method of pegging the nails for accurate measurement

27. LANE 2 = [ W ( N – 1 ) – 0.10 ) 2^ W = 1.22 MTS N = 2 ^ = 3.14159 = [ 1.22 ( 2 – 1 ) – 0.10 ] 2 * 3.14159 = [ 1.22 *1 – 0.1O ] * 6.28318 = 1.12 * 6.28318 = 7.04 MTS LANE 1 = NO STAGGER

28. LANE 3 = [ W ( N – 1 ) – 0.10 ) 2^ W = 1.22 MTS N = 3 = [ 1.22 ( 3 – 1 ) – 0.10 ] 2 * 3.14159 = [ 1.22 *2 – 0.1O ] * 6.28318 = [ 2.44 – 0.10 ] * 6.28318 = 2.34 * 6.28318 = 14.70 MTS

29. LANE 4 = [ W ( N – 1 ) – 0.10 ) 2^ W = 1.22 MTS N = 4 = [ 1.22 ( 4 – 1 ) – 0.10 ] 2 * 3.14159 = [ 1.22 *3 – 0.1O ] * 6.28318 = [ 3.66 – 0.10 ] * 6.28318 = [ 3.66 – 0.10 ] * 6.28318 = 22.37 MTS

30. LANE 5 = [ W ( N -1 ) – 0.10 ] 2 ^ W = 1.22 MTS N = 5 = [ 1.22 * 4 – 0.10 ] * 6.28318 = [ 1.22 ( 5 – 1 ) – 0. 10 ] 2 * 3.14159 = 30 .03 MTS

31. = 37 .70 MTS = [ 1.22 * 5 – 0. 10 ] 2 * 3.14159 = [ 1.22 ( 6 – 1 ) – 0. 10 ] 2 * 3.14159 N = 6 W = 1.22 MTS LANE 6 = [ W ( N – 1) – 0. 1O ] 2 ^

32. LANE 7 = [ W ( N – 1 ) – 0.10 ] 2 ^ W = 1.22 MTS N = 7 = [ 1.22 ( 7 – 1 ) - 0 .10 ] 2 * 3.14159 = [ 1.22 * 6 – 0.10 ] * 6.28318 = 45.36 MTS

33. 53.03 MTS = 1.22 * 7 – 0.10 ] 2* 3.14159 = [ 1.22 ( 8 – 1 ) - 0 .10 ] 2 * 3.14159 N = 8 W =1.22 MTS LANE 8 = [ W ( N – 1 ) – 0.10 ] 2 ^

34. THE FORMULA FOR FINDING THE HALF STAGGER HALF STAGGER = FULL STAGGER / 2 LANE NO STAGGER 1 0.00MTS 2 3.52MTS 3 7.35MTS 4 11.18 MTS 5 15.01 MTS 6 18.85MTS 7 22.68MTS 8 26.51MTS

35. 400 M STANDARD TRACK 0.0mts 7.04mts 14.70mts 22.37mts 30.03mts 37.70.mts 45.36mts 53.03mts

36. THANK YOU TO BE CONTINUE......