Dtyjlui

5 The Outside Online article How to Fuel for a Solo, Unassisted Antarctic Crossing starts with "Colin O’Brady thinks it’s possible—but just barely—to haul enough calories to traverse the continent. Here’s how," It mostly discusses "human fueling" (calories) but later says: That means starting out with a sled weighing 375 pounds (Worsley’s, in contrast, weighed 330 pounds). Unlike Worsley, he’s not bringing cigars or a bottle of Royal Brackla Scotch whiskey to toast his progress. In fact, he admitted, “I’m not bringing a second pair of underwear.” That ruthlessly pragmatic approach contrasts with the swashbuckling tradition of polar exploration pioneered by the British—but this particular challenge, O’Brady believes, simply doesn’t allow any margin for whimsy. The starting load of food was about 2...

0 if $U_{j}^{n},V_{j}^{n}$ are approximations for $u_{j}^{n} = u(x_j,t_n)$ and $v_{j}^{n} = v(x_j,t_n)$ How does applying the fourier transform defined by: $U_{j}^{n} = frac{1}{2pi}int_{-frac{pi}{Delta x}}^{frac{pi}{Delta x}} hat{U}^{n}(k)e^{ikjDelta x}dk$ to $frac{U_{j}^{n+1} - U_{j}^{n}}{Delta t} = frac{1}{Delta x^2}(U_{j+1}^{n} - 2U_{j}^{n} + U_{j-1}^{n})$ give $hat{U}^{n+1}(k) - hat{U}^{n}(k) = frac{Delta t}{Delta x^2}(-4sin^2(frac{kDelta X}{2}))hat{U}^n(k)$ I know that applying the fourier transform we get $frac{1}{2pi} int_{-frac{pi}{Delta x}}^{frac{pi}{Delta x}} frac{hat{U}^{n+1}(k) - hat{U}^{n}(k)}{Delta t} e^{ikj Delta x} dk =frac{1}{2pi} int_{-frac{pi}{Delta x}}^{frac{pi}{Delta x}} hat{U}^n(k)e^{ikj Delta x} frac{e^{ikDelta x} -2 + e^{-ikDelta x}}{Delta x^2} dk $ but im stuck after here ...