# ENES_6 - hovercraft turns to compensate Algorithm 2 tries...

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C.J. Gorrell ENES100 1. 900(2-x)+250(-10-x)+200(9-x)=0 [900(2)+250(-10)+200(9)]=1350x .8148cm=x 900(8-y)+250(-10-y)+200(-9-y)=0 900(8)+250(-10)+200(-9)=1350y y=2.148 cm 2. I=MR^2 .5(2).18^2=.0324kgm^3 α =M/I= + .1n/.0324= + 3.086 rad/s^-2 0-initial 1-forward propeller 2- reverse propeller ϖ = ϖ o + α t= 0+ (-3.086)t ϖ 2= ϖ 1+ α 2t2= 3.086t2+ -3.086t1=0 t1=t2 θ 1= θ o+ ϖ ot+.5 α t^2=0+0+.5(-3.086)t^2 θ 2= θ 1+ ϖ 1t2+ .5( α )(t2)^2 (-1.543)t^2+(-3.086)t1t2+.5(3.086)(t2)^2= - π /4 t1=t2=.504s 3. a. Algorithm 1 looks to straddle the line. If one sensor reads black, the

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Unformatted text preview: hovercraft turns to compensate. Algorithm 2 tries to stingray across the line. It turns into the line when the inside sensor reads black and the outside sensor reads white. b. If the sample rate is not high enough that if a sensor misses reading the black line that it has no failsafe. Also momentum may carry the sensors past their ideal position changing the expected circumstances c. The lack of friction makes it harder to control, therefore momentum becomes an issue....
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ENES_6 - hovercraft turns to compensate Algorithm 2 tries...

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