Big brachistochrone
in [Math]
|
Prev: A Local Property of Convex Sets
Next: HISTORIC 12 POINT CHARGE AGAINST THE INCOMPETENCE OF CURRENT MATHEMATICS THEORY
From: gudi on 11 Aug 2010 15:57 On Aug 8, 7:37 am, James Waldby <n...(a)no.no> wrote: > On Sat, 07 Aug 2010 14:32:31 +0200, Thomas Nordhaus wrote: > > William Elliot schrieb: > >> On Sat, 7 Aug 2010, gudi wrote: > > >>> From an altitude H = 5000 km above equator a rollercoaster is built in > >>> the equatorial plane touching the equator. What should be its shape so > >>> that the time of descent is a minimum? (we neglect wind resistance > >>> etc., and consider only gravity). For altitudes << H we should have > >>> the brachistochrone cycloid as a special case. > > >> No need to shape it; just drop it. > > > Well, well that's just a trivial case ;-) How about a half-pipe > > brachistochrone going from point A 5000km above the ground to > > ground-zero at the antipodal point and ending at point A again going > > once around the equator? > > If earth's rotation is considered too (rather than only gravity), > would you shape a half-pipe differently for east-bound travel vs > west-bound? > > <www.paperblog.fr/1888349/espece-de-brachistochrone/> in French > seems to be considering half-pipe cases as far as I can tell from > google's translation, while problem 7.6 on page 234 of Classical > Mechanics by Rana & Joag as shown at long link below asks one to > consider the effect of rotational speed on the shape of an underground > tunnel for tautochronous motion. It appears that neither of these > references contemplate returning to the original point, or limiting > the depth of fall (and rise) to 5000km, or direction of travel. > <http://books.google.com/books?id=dptKVr-5LJAC&lpg=PA234&ots=AILtY1JJ_...> > jiw Thanks. Rana & Joag reference answers the query to the point. The differential equation describes a simple harmonic motion in 2 dimensions with a definite time period between cusps. Are the following then right? There are two "fast" min. time tracks,one below and other above the Earth. The latter is an epicycloid for motion above the earth. The orbit/track cannot be tangent to earth but normal to it at point of contact on earth's surface at zero velocity before rebound. The "apogee" velocities are 1.6 km and 0.36 km /s respectively. The quarter-time is 26.43 mins for a track initially parallel to earth and starting 5000 km above it. And 70.71 mins for a 35,785 km high geosynchronous orbit (Calculated using the pi sqrt(a/g). formula).There is no East bound/West bound orbit sense. Regards, Narasimham
From: gudi on 11 Aug 2010 16:15
On Aug 8, 7:37 am, James Waldby <n...(a)no.no> wrote: > On Sat, 07 Aug 2010 14:32:31 +0200, Thomas Nordhaus wrote: > > William Elliot schrieb: > >> On Sat, 7 Aug 2010, gudi wrote: > > >>> From an altitude H = 5000 km above equator a rollercoaster is built in > >>> the equatorial plane touching the equator. What should be its shape so > >>> that the time of descent is a minimum? (we neglect wind resistance > >>> etc., and consider only gravity). For altitudes << H we should have > >>> the brachistochrone cycloid as a special case. > > >> No need to shape it; just drop it. > > > Well, well that's just a trivial case ;-) How about a half-pipe > > brachistochrone going from point A 5000km above the ground to > > ground-zero at the antipodal point and ending at point A again going > > once around the equator? > > If earth's rotation is considered too (rather than only gravity), > would you shape a half-pipe differently for east-bound travel vs > west-bound? > > <www.paperblog.fr/1888349/espece-de-brachistochrone/> in French > seems to be considering half-pipe cases as far as I can tell from > google's translation, while problem 7.6 on page 234 of Classical > Mechanics by Rana & Joag as shown at long link below asks one to > consider the effect of rotational speed on the shape of an underground > tunnel for tautochronous motion. It appears that neither of these > references contemplate returning to the original point, or limiting > the dept of fall (and rise) to 5000km, or direction of travel. > > <http://books.google.com/books?id=dptKVr-5LJAC&lpg=PA234&ots=AILtY1JJ_...> > > -- > jiw Thanks. Rana & Joag reference answers the query to the point. The differential equation describes a simple harmonic motion in 2 dimensions with a definite time period between cusps. Are the following then right? There are two "fast" min. time tracks,one below and other above the Earth. The latter is an epicycloid for motion above the earth. The orbit/track cannot be tangent to earth but normal to it at point of contact on earth's surface at zero velocity before rebound. The quarter-time is 26.43 mins for a track initially parallel to earth and starting 5000 km above it. And 70.71 mins for a 35,785 km high geosynchronous orbit (Calculated using the pi sqrt(a/g). formula).The "apogee" velocities are 1.6 km and 0.36 km /s respectively. There is no East bound/West bound orbit sense. Regards, Narasimham |