MARKING PLAN

TRILOKI PRASAD

M.P.ED

GURU GHASIDAS UNIVERSITY

BILASPUR

1: SYNTHETIC TRACK

2: CINDER TRACK

3: GRASSY TRACK

DIRECTION OF TRACK

NORTH

WESTEAST

SOUTH

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.

Methods of finding the total

area required for track marking

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

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

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)

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

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

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

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

1.12M

2.34M

3.56M

4.78M

6.00M

R.D.R = C.D.R + 0.30 MTS

C.D.R = R.D.R – 0.30 MTS

R.D.R = 36.80 MTS

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

ONE STRAIGHT LENGTH = 168.78/ 2 =

84.39 MTS

TOTAL LENGTH OF THE STRAIGHTS

= 400 – 231.22 = 168.78 MTS

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

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

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

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

CIRCUMFERENCE OF CURVES

= 231.22 MTS

STRAIGHT LENGTH = 84.39 +

84.39 = 168.78 MTS

RADIUS OF CURVES = 36.80 MTS

LANES MARKING

1.21-1.23mts

1.21-1.23mts

1.21-1.23mts

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

PYTHEGORUS THEOREM FORMULA

PA

O

3 MTS

4 M

TS

4 M

TS

3 MTS

BI- SECTOR METHOD

84.39MTS4

2.2

6M

TS9.7

6M

TS

36

.50

MT

S

400 MTS STANDARD TRACK

Method of pegging the nails

for accurate measurement

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

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

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

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

= 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 ^

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

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 ^

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

400 M STANDARD TRACK

0.0mts

7.04mts

14.70mts

22.37mts

30.03mts

37.70.mts

45.36mts

53.03mts

TO BE CONTINUE......