Application of Trigonometry in Biomechanics

1. Application of Trigonometry in Biomechanics Find a distance or displacement given a set of coordinates Separate muscle force into a component causing movement and a component affecting joint stability Analyze projectile motion (baseball, discus, javelin, basketball, etc.) Find the net force acting on an object or body segment Analyze the effect of the weight of a body segment or outside force on movement

2. Basic Distance Formula • Objects in biomechanical analysis are often located as coordinates on a graph. • The displacement of these objects from one position to another can be calculated using trigonometric functions

3. Basic Distance Formula • GIVEN: The X and Y coordinates, in meters, of an object as it travels from one position to another are (2, 4) and (5, 8) • FIND: The distance the object travels from the first to the second position.

4. Y m (X2, Y2) (X1, Y1) X m DIAGRAMS: Coordinates (m) d

5. Y m (X2, Y2) (X1, Y1) X m d = √(X2 – X1)2 + (Y2 – Y1)2 DIAGRAMS: Coordinates (m) d FORMULA:

6. The solution is based on the Pythagorean Theorem. • For any right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the remaining two sides, or, C2 = A2 + B2 where C is the hypotenuse and A and B are the other two sides. In this case our distance, d, is the hypotenuse of the triangle (side C), side A is equal to X2 - X1 , and side B is equal to Y2 - Y1 . Therefore, C = A2 + B2 or (X2 , Y2 ) C B A (X1 , Y1 ) d = (X2 – X1)2 + (Y2 – Y1)2

7. Y m (X2, Y2) (X1, Y1) X m d = √(5 m – 2 m)2 + (8 m – 4 m)2 = 5 m DIAGRAMS: Coordinates (m) d SOLUTION:

8. A quarterback is standing on his own 20 yard line and is 25 yards from the right sideline. He throws the ball to a receiver that is on the opponent’s 35 yard line and is 2 yards from the right sideline when he catches the ball. A) What was the horizontal distance covered by the ball before it was caught?B) If the ball was in the air for 3.5 seconds, what was the average forward or horizontal velocity of the ball during flight? Note that in this case the football field itself is used as a coordinate system.

9. 20 yds 35 yds 25 yds 2 yds 100 yds 50 yd line quarterback receiver C B A

10. Given: X1 = 20 yds (This is the quarterback’s position relative to the goal line.) X2 = 100 yds – 35 yds = 65 yds (This sets all of the measurements relative to the same reference (or goal) line.) (This is the receiver’s position relative to the goal line.) Dt = 3.5 sec Quarterback is 25 yards from sideline (Y1 = 25 yds ) Receiver is 2 yards from sideline (Y2 = 2 yds) • Find: • The horizontal distance traveled by the football (df) • The average horizontal velocity of the football (vf)

11. You could also represent the coordinates on a standard X-Y graph. Distance from the goal line would be x-coordinates and distance from the sideline would be y-coordinates. Y yds (20, 25) 25 C B (65, 2) A 2 0, 0 X yds 65 20

12. C B 25 yds A 2 yds Diagram and Derived Information p1 p2 • A = 65 yds – 20 yds = 45 yds • B = 25 yds – 2 yds = 23 yds • C = horizontal distance traveled by the football = df

13. A2 + B2 Formulas A) df = C = B) vf = df /Dt Solutions A) df = df = 50.54 yds = (50.54 yds)(3 ft/yd) = 151.62 ft B) vf = 151.62 ft / 3.5 sec vf = 43.32 ft / sec (45 yds)2 + (23 yds)2