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- https://phys.libretexts.org/Courses/Joliet_Junior_College/JJC_-_PHYS_110/02%3A_Book-_Conceptual_Physics_(Crowell)/2.03%3A_Conservation_of_Energy/2.3.02%3A_Numerical_TechniquesOn the other hand, it's possible to go too far with this idea: if we drop straight down for the whole vertical distance, and then do a right-angle turn to cover the horizontal distance, the resulting ...On the other hand, it's possible to go too far with this idea: if we drop straight down for the whole vertical distance, and then do a right-angle turn to cover the horizontal distance, the resulting time of 0.68 s is quite a bit longer than the optimal result, the reason being that the path is unnecessarily long.
- https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Classical_Mechanics_(Tatum)/19%3A_The_Cycloid/19.07%3A_The_Brachystochrone_Property_of_the_CycloidBut, if O and P are connected with a cycloidal wire, the time taken to go from O to P does not increase with distance. (See the isochronous property of the cycloid discussed in Section 19.5.) Thus, as...But, if O and P are connected with a cycloidal wire, the time taken to go from O to P does not increase with distance. (See the isochronous property of the cycloid discussed in Section 19.5.) Thus, as you increase the distance between O and P, the time taken to travel by any route other than the cycloidal one must take longer than the cycloidal route.
- https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Graduate_Classical_Mechanics_(Fowler)/02%3A_The_Calculus_of_Variations/2.12%3A_The_BrachistochroneThe problem is to curve the wire from A down to B in such a way that the bead makes the trip as quickly as possible. d x=-a \tan \dfrac{\theta}{2} d z=2 a \tan \dfrac{\theta}{2} \sin \dfrac{\theta}{2}...The problem is to curve the wire from A down to B in such a way that the bead makes the trip as quickly as possible. d x=-a \tan \dfrac{\theta}{2} d z=2 a \tan \dfrac{\theta}{2} \sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} d \theta=2 a \sin ^{2} \dfrac{\theta}{2} d \theta=a(1-\cos \theta) d \theta Now adding the \(\theta\) back in, this circular motion move steadily to the right, in such a way that the initial direction of the path is vertically down. \(\begin{equation}
- https://phys.libretexts.org/Bookshelves/Conceptual_Physics/Conceptual_Physics_(Crowell)/03%3A_Conservation_of_Energy/3.02%3A_Numerical_TechniquesOn the other hand, it's possible to go too far with this idea: if we drop straight down for the whole vertical distance, and then do a right-angle turn to cover the horizontal distance, the resulting ...On the other hand, it's possible to go too far with this idea: if we drop straight down for the whole vertical distance, and then do a right-angle turn to cover the horizontal distance, the resulting time of 0.68 s is quite a bit longer than the optimal result, the reason being that the path is unnecessarily long.
