In the operators manual it states that the climbing angle is the angle of back t

Question

In the operators manual it states that the climbing angle is the angle of back to front tilt that the vehicle experiences; the sidehill, angle is the angle of right left tilt. For purposes of maintaining equilibrium, we mus insist that you do not exceed maximum angle of 36 degrees climbing or descending nor 30 degrees sidehill, or the vehcle will tip backwards, forwards, or sideways respectively. also, the power output is limited to 1000 newton-meters/sec. and the rover weighs 500 newtons on the moon. before taking the rover out on the lunar surface, you want to do some laboratory tests. your laboratory is equipped with a test hill with height function: H(x,y)=1/1+x2 +2y2 , where x,y, and H are all in units of say , hundreds of meters. the boundary of your test arena coincides with the H=1/17 contour.

Find the direction that the rover would have to head to go straight uphill, together with the slope in that direction at the points (1,0),(0,1) and (sqrt2/2,sqrt2/2). Find the sidehill slope that the rover would experience while climbing stright uphill at each of these points. Convert these slopes into climbing and sidehill angles at these points.

Explanation / Answer

The information about weight and power output appears to be unnecessary for answering the questions. For H it looks like you mean H(x,y)=1/(1+x^2 +2y^2) I don't know any way of sketching contours here, but: If H=1/n then 1 + x^2 + 2y^2 = n x^2 + 2y^2 = n - 1 This is an ellipse centred on the origin and passing through (± sqrt(n - 1), 0) and (0, ± sqrt [(n - 1)/2]). Thus when H = 1/17, it is an ellipse passing through (± 4, 0) and (0, ± 2 sqrt 2). When H = 1/10, it is an ellipse passing through (± 3, 0) and (0, ± 3/sqrt 2). When H = 1/5, it is an ellipse passing through (± 2, 0) and (0, ± sqrt 2). When H = 1/2, it is an ellipse passing through (± 1, 0) and (0, ± 1/sqrt 2). The direction of maximum slope at (x, y) is given by (?H/?x , ?H/?y) evaluated at (x, y). ?H/?x = - 2x/(1 + x^2 + 2y^2)^2 ?H/?y = - 4y/(1 + x^2 + 2y^2)^2 The maximum slope at (x, y) is given by sqrt[(?H/?x)^2 + (?H/?y)^2] evaluated at (x, y). Evaluating these is just algebra which I'll leave for you to do. The slope is the tangent of the angle to the horizontal. Or the angle is the arctan of the slope. The sidehill angle in the direction of maximum slope is zero. (This seems intuitively obvious to me.)

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