[newbie math question] Speed of the terminator?

Is there an easy way to figure out the speed of the terminator at a given
latitude? I can do the basic math easily enough, but once one brings axial
tilt and whatnot into the picture my brain explodes.

(Basically, I need to figure out where someone would need to be to be able
to outrun the dawn in a typical passenger car.)

- David Prokopetz.

On Thu, 1 Jul 2004 14:31:31 -0600, "David Prokopetz"
wrote:

Is there an easy way to figure out the speed of the terminator at a given
latitude? I can do the basic math easily enough, but once one brings axial
tilt and whatnot into the picture my brain explodes.

(Basically, I need to figure out where someone would need to be to be able
to outrun the dawn in a typical passenger car.)

Start with the simple case - at the equinox - and figure that the
Earth rotates through it's circumference once in 24 hours, The speed
of the terminator stays pretty close to the speed of the Earth's
rotation, so at the equator, it's just the circumference of the earth
divided by 24. At other latitudes, it's very close to that speed times
the cosine of the latitude. At dates away from the equinox, there
will be progressively larger corrections to be made as you get further
from the equinox and to higher latitudes. Even so, unless you are in
a region where the sun doesn't rise or doesn't set, these corrections
will probably be small compared to the speeds you already computed.

Al Moore

On Thu, 1 Jul 2004 14:31:31 -0600, "David Prokopetz"
wrote:

Is there an easy way to figure out the speed of the terminator at a given
latitude? I can do the basic math easily enough, but once one brings axial
tilt and whatnot into the picture my brain explodes.

(Basically, I need to figure out where someone would need to be to be able
to outrun the dawn in a typical passenger car.)

Start with the simple case - at the equinox - and figure that the
Earth rotates through it's circumference once in 24 hours, The speed
of the terminator stays pretty close to the speed of the Earth's
rotation, so at the equator, it's just the circumference of the earth
divided by 24. At other latitudes, it's very close to that speed times
the cosine of the latitude. At dates away from the equinox, there
will be progressively larger corrections to be made as you get further
from the equinox and to higher latitudes. Even so, unless you are in
a region where the sun doesn't rise or doesn't set, these corrections
will probably be small compared to the speeds you already computed.

Al Moore

"David Prokopetz" wrote in message ...
Is there an easy way to figure out the speed of the terminator at a given
latitude? I can do the basic math easily enough, but once one brings axial
tilt and whatnot into the picture my brain explodes.

(Basically, I need to figure out where someone would need to be to be able
to outrun the dawn in a typical passenger car.)

- David Prokopetz.
In northern hemishpere summer time anywhere north of the artic circle

Geezerguy

"David Prokopetz" wrote in message ...
Is there an easy way to figure out the speed of the terminator at a given
latitude? I can do the basic math easily enough, but once one brings axial
tilt and whatnot into the picture my brain explodes.

(Basically, I need to figure out where someone would need to be to be able
to outrun the dawn in a typical passenger car.)

- David Prokopetz.
In northern hemishpere summer time anywhere north of the artic circle

Geezerguy

Your going to need a road near the pole, subsonice jet down to about 50
degrees N/S, or supersonic closer to equator.
25,000 miles at equator, 24 hours, just over 1,000 mph
Need cosine of latitude under .6 to get in down to subsonice, around
about Dudson bay, Engalnd, Norway, Russia.

David Prokopetz wrote:

Is there an easy way to figure out the speed of the terminator at a given
latitude? I can do the basic math easily enough, but once one brings axial
tilt and whatnot into the picture my brain explodes.

(Basically, I need to figure out where someone would need to be to be able
to outrun the dawn in a typical passenger car.)

- David Prokopetz.

Your going to need a road near the pole, subsonice jet down to about 50
degrees N/S, or supersonic closer to equator.
25,000 miles at equator, 24 hours, just over 1,000 mph
Need cosine of latitude under .6 to get in down to subsonice, around
about Dudson bay, Engalnd, Norway, Russia.

David Prokopetz wrote:

Is there an easy way to figure out the speed of the terminator at a given
latitude? I can do the basic math easily enough, but once one brings axial
tilt and whatnot into the picture my brain explodes.

(Basically, I need to figure out where someone would need to be to be able
to outrun the dawn in a typical passenger car.)

- David Prokopetz.