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Mathematical Vector Addition
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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.
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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
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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
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0 90 0 180 0 270 0 60 Km/hr 30 Km/hr -x +y -42.4 +42.4
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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
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-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
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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
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+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.
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+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
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+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
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+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.
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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
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