2D and 3d motion analysis
Mazarik university
BMX trick
DROUOT Florian,
429503
BEGINING : 1 ,93 sec
HIGHIEST :
RECEPTION/ END : 2,53 sec
I pointed with paint differents generals points of the trick :
– b is the point of the begining, b(540 ;480)
– H is the highiest point of the jump, H(395;400)
– E is the last point of the jump , E(240;480)
– O is the vertical image of H on eB distance , O(395;480)
– A is the angle O^bH
About the echelle, the orange skate mesure 60cm.
D(315;400) and I(385;400)
LET'S GO !
LENGHT :
Here, the lenght of the jump is the distance between E and b, and
Eb= v((540-240)²+(480-480)²) = 300px
HIGH OF THE TRICK :
The high correponds to the distance HO, so :
HO=v((395-395)²+(400-480)²=80px
We need to know DI to convert the pixels to cm :
DI= 385-315 = 70px , so 70px = 60cm
Conclusion :
Lenght = (300*60)/70 = 257cm
High = (80*60)/70 = 69 cm
SPEED :
The time of the jump is 2,53-1,93= 0,6 sec
My speed is 2,57m in 0,6 sec. Or 2,57*1,33=3,43 m/s
ANGLE :
tan(b)= OH/Ob Ob= 540-395 = 145px
tan(b)=80/145=0.55
So b =28.8°