1. Mathematical Vector Addition

4. Mathematical Addition of Vectors The process of adding vectors can be accurately done using basic trigonometry. If you follow each step carefully, you will break down each vector into it's x and y componets and determine the magnitude and direction of the resultant vector. We have provided you with a "Vector Worksheet" to help you organize your work.

5. Let's say that we are adding three vectors A, B, and C. STEP #1 - Deconstruct each vector into it's X & Y components. A) To do this, you must first find the Theta angle to the x axis for each vector.0 90 0 180 0 270 0 30 Km/hr @ 45 0 60 Km/hr @ 135 0 60 Km/hr @ 315 0

6. 0 90 0 180 0 270 0 60 Km/hr 30 Km/hr -x +y 30 Km/hr @ 45 0 60 Km/hr @ 135 0 60 Km/hr @ 315 0 = 45 o = 45 0 B. Now calculate the X & Y components of each vector treating the magnitude of each vector as the hypotenuse of a right triangle X component = Magnitude x cos y component = Magnitude x sin -x = (60)(cos45) = -42.4 y = (60)(sin45) = +42.4

7. 0 90 0 180 0 270 0 60 Km/hr 30 Km/hr -x +y -42.4 +42.4

8. 0 90 0 180 0 270 0 60 Km/hr 30 Km/hr -x +y +x +x = (30)(cos45) = +21.2 30 Km/hr @ 45 0 60 Km/hr @ 135 0 60 Km/hr @ 315 0 = 45 o = 45 0 B. Now calculate the X & Y components of each vector treating the magnitude of each vector as the hypotenuse of a right triangle X component = Magnitude x cos y component = Magnitude x sin -x = (60)(cos45) = -42.4 y = (60)(sin45) = +42.4 +y = (30)(sin45) = +21.2

9. -42.4 +42.40 90 0 180 0 270 0 60 Km/hr 30 Km/hr -x +y +x +x = (30)(cos45) = +21.2 + 21.2

10. 0 90 0 180 0 270 0 60 Km/hr 30 Km/hr -x +y +x +x = (30)(cos45) = +21.2 +x -Y 30 Km/hr @ 45 0 60 Km/hr @ 135 0 60 Km/hr @ 315 0 = 45 o = 45 0 B. Now calculate the X & Y components of each vector treating the magnitude of each vector as the hypotenuse of a right triangle X component = Magnitude x cos y component = Magnitude x sin -x = (60)(cos45) = -42.4 y = (60)(sin45) = +42.4 +y = (30)(sin45) = +21.2 +x = (60)(cos45) = +42.4 -Y = (60)(sin45) = -42.4

11. +21.2 +42.4 -42.4 +63.6 -42.4 +21.2 +42.4 -42.4 + 21.2 +63.6-42.4 +21.2 0 90 0 180 0 270 0 60 Km/hr 30 Km/hr -x +y +x +x = (30)(cos45) = +21.2 +x -Y STEP #2 - List and add all x components and y components. Including all signs. These sums are the components of the resultant vector.

12. +21.2 +42.4 -42.4 +63.6 -42.4 +21.2 +42.4 -42.4 + 21.2 +63.6-42.4 +21.2 0 90 0 180 0 270 0 60 Km/hr 30 Km/hr -x +y +x +x = (30)(cos45) = +21.2 +x -Y STEP #3 - Convert the resultant components into navigational vector notation. To do this, first use the Pythagorean Theorem to determine the hypotenuse. +21.2 Resultant = √21.2 2 +21.2 2

13. +21.2 +42.4 -42.4 +63.6 -42.4 +21.2 +42.4 -42.4 + 21.2 +63.6-42.4 +21.2 0 90 0 180 0 270 0 60 Km/hr 30 Km/hr -x +y +x +x = (30)(cos45) = +21.2 +x -Y 29.9 The magnitude of the resultant is 29.9 Km/hr. Now, find the Theta θ of the resultant using the inverse tangent formula. tan -1 = l y/x l = θ R

14. +21.2 +42.4 -42.4 +63.6 -42.4 +21.2 +42.4 -42.4 + 21.2 +63.6-42.4 +21.2 0 90 0 180 0 270 0 60 Km/hr 30 Km/hr -x +y +x +x = (30)(cos45) = +21.2 +x -Y θ R =45 o 29.9 NAV R = 45 o 29.9 Km/hr45 o Since the θ R = 45 o, the navigational direction of the resultant vector is NAV = 90 - θ R = 45 o.

15. The resultant vector = 29.9 Km/hr @ 45 o Congratulations! You have successfully calculated a vector addition! Use this guide to get the navigational angle in other quadrants. Quadrant #1: NAV = 90 - θ R Quadrant #2: NAV = 270 + θ R Quadrant #3: NAV = 270 - θ R Quadrant #4: NAV = 90 + θ R