Hodograph: A useful geometrical tool for solving some difficult problems in dynamics

William Rowan

Hamilton

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James Clerk Maxwell, Matter and Motion (Dover, New York, 1991).

David and Judith Goodstein, Feynman’s Lost Lecture (W. W. Norton, New York, 1996).

If $sin φ=sc,$ so that the two graphs just touch each other [see Fig. 5(b)], the slope of the ground at the intersection point is equal to the inverse of the slope of $f1,$ because $[dyground/dx]0=[df2/dy]0−1=[df1/dy]0−1=[(d/dy)sc2a1+y/a]0−1=(1+y0/a)/sc.$ If we call $φ1$ and $φ2$ the angles formed by the initial and final velocities, respectively, with the horizon we have that $φ1+φ2=φc,$ and $cos φ1=1+y0/a cos φ2,$ where the last relation comes from the geometry of Fig. 4. With a little algebra we obtain $tan φ2=(1+y0/a)/sc−cot φc.$ Therefore if $φc$ is the oblique angle that satisfies $sin φc=sc,$ the slope of the projectile trajectory is smaller than the slope of the ground at the landing point, which is impossible (because this would mean that the projectile arrives at this point from inside the ground). Thus, only the corresponding obtuse-angle solution can lead the projectile to this point. The same reasoning holds for values of φ such that $sin φ$ is slightly less than $sc.$ Both intersection points then can be reached by the obtuse-angle solution, and the nearest point can be reached by the oblique-angle solution only if $dyground/dx⩽tan φ2.$

The stationary point we calculated here corresponds to a minimum, because an infinitely deep well would lead to infinite time travel.

Because all expressions are written as functions of $u,$ we try to optimize the inverse function of $u(θ);$ namely, $θ(u).$

The end points of θ are $θin=−π/2,$ and $θfin=π/2,$ because these values lead to zero velocities when substituted in $u=(1/λ)cos θ,$ the hodographic curve.