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Great Circle Question From A Reindeer.  
User currently offlineComorin From United States of America, joined May 2005, 4896 posts, RR: 16
Posted (7 years 9 months 4 days 11 hours ago) and read 1963 times:

Hello fellow weenies:

The shortest distance between two points terrestrially is the Great Circle chord. However, the above holds true only assuming that the Earth doesn't spin, there are no prevailing winds, or if the route is strictly equatorial.

In practice, and assuming no winds, is the effect due to the Earth's spin (max on Polar routes) significant on ultra-long segments at airliner speeds? I'm curious if this is taken into consideration in route planning. I did read the earlier Great Circle thread by Pilotaydin but it didn't help me.

Appreciate any insight on this.

Signed,

Santa's Reindeer  santahat 

13 replies: All unread, jump to last
 
User currently offlineZeke From Hong Kong, joined Dec 2006, 9105 posts, RR: 75
Reply 1, posted (7 years 9 months 4 days 10 hours ago) and read 1947 times:

Quoting Comorin (Thread starter):
In practice, and assuming no winds, is the effect due to the Earth's spin (max on Polar routes) significant on ultra-long segments at airliner speeds?

Short answer, no.

The aircraft is also spinning at the same rate as the ground below, just the frame of reference you are using.

If you jump up and down on the spot, you dont move off the spot because the earth rotated below you.



We are addicted to our thoughts. We cannot change anything if we cannot change our thinking – Santosh Kalwar
User currently offlineZANL188 From United States of America, joined Oct 2006, 3526 posts, RR: 0
Reply 2, posted (7 years 9 months 4 days 10 hours ago) and read 1945 times:
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Quoting Comorin (Thread starter):
Signed,

Santa's Reindeer

Yo Rudolph!

Santa doesn't "travel" on Christmas Eve. He has a line of helpers stretching from the North to the South Poles who hover. As the earth spins and brings a target, er, "chimney" into view the helper takes aim and lets fly. The helper also drops enough inflatable cookie eating Santa decoys to make the "Santa visits every home" myth believeable.

Santa saves considerable $$$ this way as he doesn't have to pay the reindeer or the helpers mileage. Sleigh maintenance expense also goes way down.



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User currently offlineComorin From United States of America, joined May 2005, 4896 posts, RR: 16
Reply 3, posted (7 years 9 months 4 days 9 hours ago) and read 1913 times:

Quoting ZANL188 (Reply 2):

 rotfl 

That's not good news for us reindeers that get paid flight time...However, as Senior Reindeer, I do get to choose my shift...  Smile

R.


User currently offlineComorin From United States of America, joined May 2005, 4896 posts, RR: 16
Reply 4, posted (7 years 9 months 4 days 8 hours ago) and read 1893 times:

Quoting Zeke (Reply 1):
f you jump up and down on the spot, you dont move off the spot because the earth rotated below you.

Appreciate the answer! - Happy Holidays, btw.

From a microscopic point of view, you actually do move off the spot, though it's a very small displacement.

I'll harness my thoughts and post again a little later on this. It's about the change in angular velocity when you jump as radius changes ( by a few feet) but your tangential velocity still stays the same - you know, the reason why it looks like rockets take off in a curved path - a Coriolis effect? If, however, you make up for this on the way down, then you would land on the same spot ( I think). If you jumped at the N or S Pole however, you would land on the same spot.

In an airplane on a North /South track, the atmosphere pushes you sideways to keep you true to ground. If you take off from the Equator, you have a built in tangential velocity of about 1100mph due West. You would have to lose this amount of energy by the time you land at the Pole (safely!). This energy is absorbed by the atmosphere on the flight.

Zeke, I think you are pointing out that these effects are so small that they are not relevant in real-life navigation - again, thanks.








,


User currently offlineVikkyvik From United States of America, joined Jul 2003, 10036 posts, RR: 26
Reply 5, posted (7 years 9 months 4 days 6 hours ago) and read 1875 times:
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Quoting Comorin (Reply 4):

I'll harness my thoughts and post again a little later on this. It's about the change in angular velocity when you jump as radius changes ( by a few feet) but your tangential velocity still stays the same - you know, the reason why it looks like rockets take off in a curved path - a Coriolis effect? If, however, you make up for this on the way down, then you would land on the same spot ( I think). If you jumped at the N or S Pole however, you would land on the same spot.

In an airplane on a North /South track, the atmosphere pushes you sideways to keep you true to ground. If you take off from the Equator, you have a built in tangential velocity of about 1100mph due West. You would have to lose this amount of energy by the time you land at the Pole (safely!). This energy is absorbed by the atmosphere on the flight.

Coriolis would have an effect on airplanes. Basically, the longer your flight path, the more effect Coriolis will have (it's also dependent on your speed).

In terms of rockets, if the rocket is blasting off vertically, then it will go straight up. For Coriolis to come into play, the object in question needs to be changing latitude (moving north or south). In other words, if you're moving directly east or west, Coriolis will have no effect on you at all. As soon as you start to move north or south, it will have an effect. It'll also have no effect at the equator, and greatest effect at the poles.

Coriolis only tends to noticably affect things that are extremely large (like hurricanes) or are travelling over long distances (like airplanes), and the greater the speed of the object, the greater the Coriolis effect.



"Two and a Half Men" was filmed in front of a live ostrich.
User currently offlineComorin From United States of America, joined May 2005, 4896 posts, RR: 16
Reply 6, posted (7 years 9 months 4 days 4 hours ago) and read 1847 times:

Vikkyvik, thanks for the usual insight!  thumbsup 

My understanding is that a rocket will only go straight up if launched at either Pole. If you launch a rocket at the equator, you have already added the tangential velocity of about 1100 mph. While you are right in saying that the rocket goes straight up, it does not go up a radial as its tangential velocity will have to change to keep that track. To an observer on earth, the rocket will appear to curve away after some time.

I think this is also why the Arianespace Spaceport is located near the Equator in Kuorou, French Guiana.

BTW, It's been a while since I've used my brain this hard, so please let me know if I'm on the right track  Smile


User currently offlineZeke From Hong Kong, joined Dec 2006, 9105 posts, RR: 75
Reply 7, posted (7 years 9 months 3 days 23 hours ago) and read 1804 times:

Quoting Comorin (Reply 4):
I'll harness my thoughts and post again a little later on this. It's about the change in angular velocity when you jump as radius changes ( by a few feet) but your tangential velocity still stays the same - you know, the reason why it looks like rockets take off in a curved path - a Coriolis effect? If, however, you make up for this on the way down, then you would land on the same spot ( I think). If you jumped at the N or S Pole however, you would land on the same spot.

Aircraft in the atmosphere are in an inertial frame of reference. That means in nil wind it take as long to fly east to west, as it does west to east. Aircraft are said to be in equilibrium in cruise, i.e. not accelerating.

If we were in a non-inertial frame of reference, all we would need to do to fly west bound is to hover, and we would automatically be doing 60 kt over the ground at the equator. This is basically what happens to a rocket, it does this by escaping the atmosphere and earths gravity, to do so it must be constantly accelerating in relation to earth.

Quoting Vikkyvik (Reply 5):
Coriolis would have an effect on airplanes. Basically, the longer your flight path, the more effect Coriolis will have (it's also dependent on your speed).

Does not effect aircraft, rockets are a different storey, they are in a non-inertial frame of reference.



We are addicted to our thoughts. We cannot change anything if we cannot change our thinking – Santosh Kalwar
User currently offlineVikkyvik From United States of America, joined Jul 2003, 10036 posts, RR: 26
Reply 8, posted (7 years 9 months 3 days 22 hours ago) and read 1783 times:
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Well I have to say, I think the following statement of mine is wrong:

Quoting Vikkyvik (Reply 5):
In other words, if you're moving directly east or west, Coriolis will have no effect on you at all.



Quoting Zeke (Reply 7):

Does not effect aircraft, rockets are a different storey, they are in a non-inertial frame of reference.

Well, in my brief Google search, I've found some differeing viewpoints on the matter (unfortunately, I don't have my "Atmospheric Environment" course book anymore). It seems that there would be some small correction for Coriolis in airplane flight, but perhaps it's negligible.

Or maybe my memory is faulty  Smile

~Vik



"Two and a Half Men" was filmed in front of a live ostrich.
User currently offlineZeke From Hong Kong, joined Dec 2006, 9105 posts, RR: 75
Reply 9, posted (7 years 9 months 3 days 21 hours ago) and read 1770 times:

Quoting Vikkyvik (Reply 8):
It seems that there would be some small correction for Coriolis in airplane flight, but perhaps it's negligible.

We dont correct for many things, e.g. gravity changes with latitude and we do not correct for it. In space, its different, the speeds are much greater.

Where accuracy was required like inside the old INS systems many corrections were required for rigidity, real wander, apparent wander, transport wander and gyroscope precession, even then the position was in error.



We are addicted to our thoughts. We cannot change anything if we cannot change our thinking – Santosh Kalwar
User currently offline113312 From United States of America, joined Apr 2005, 572 posts, RR: 1
Reply 10, posted (7 years 9 months 3 days 21 hours ago) and read 1768 times:

Remember that the atmosphere is spinning along with the surface of the earth. The issue you seem to raise is comparing shortest distance from point A to point B with the cost in fuel, time or a combination of these to cover such a span between points.

Regardless if the fuel is Jet-A or hay for the reindeer, the issue of cost is more complex than simply the ground miles covered. Generally, minimum fuel and/or time will also consider the patterns of High and Low pressure and high altitude winds to take advantage of tail wind component or minimize the effect of headwinds.

However, a total cost model should also consider additional factors in addition to minimizing the consumption of fuel. For example, you might be paying for the reindeer, crew, and engines by the hour. Flying a profile that saves 50 cents of fuel while expending 5 bucks additional hourly costs isn't cost wise. It could turn out that flying as fast as possible over the shortest (great circle) course could be the best plan for minimum cost even though it's less fuel efficient.


User currently offlineComorin From United States of America, joined May 2005, 4896 posts, RR: 16
Reply 11, posted (7 years 9 months 3 days 10 hours ago) and read 1702 times:

This is A.net at it's best - access to great answers by real professionals!  thumbsup 

Quoting Zeke (Reply 9):

Question about the 60 knots to hover at the Equator: Since the Earth is 25,000 miles in circumference, and spins once every 24 hours, the tangential velocity at the equator is 25000/24 ~ 1040 mph. Would you not need to have a similar groundspeed to hover?

Your point about the Inertial frame of reference is well taken.

Quoting 113312 (Reply 10):

You're right about the underlying reason for my question - it's about finding the minimum cost routing in an ULH scenario. So my take-away is minimum cost results from minimum flight time. Also, as Coriolis effects are negligible, then in still air the GC is the optimum route, regardless of Origin/Destination latitudes.

Again, thank you each and every poster and I will put in a good word for you with my Boss  santahat  tonight!


User currently offlineZeke From Hong Kong, joined Dec 2006, 9105 posts, RR: 75
Reply 12, posted (7 years 9 months 3 days 7 hours ago) and read 1682 times:

Quoting Comorin (Reply 11):
Question about the 60 knots to hover at the Equator: Since the Earth is 25,000 miles in circumference, and spins once every 24 hours, the tangential velocity at the equator is 25000/24 ~ 1040 mph. Would you not need to have a similar groundspeed to hover?

Yep my mistake, 15 degrees and hour, 1 degree = 60 nm approx 900 nm/hr



We are addicted to our thoughts. We cannot change anything if we cannot change our thinking – Santosh Kalwar
User currently offlineStarlionblue From Greenland, joined Feb 2004, 17044 posts, RR: 66
Reply 13, posted (7 years 9 months 3 days 2 hours ago) and read 1638 times:

Quoting Comorin (Reply 6):
My understanding is that a rocket will only go straight up if launched at either Pole. If you launch a rocket at the equator, you have already added the tangential velocity of about 1100 mph. While you are right in saying that the rocket goes straight up, it does not go up a radial as its tangential velocity will have to change to keep that track. To an observer on earth, the rocket will appear to curve away after some time.

I think this is also why the Arianespace Spaceport is located near the Equator in Kuorou, French Guiana.

Absolutely. It's much cheaper to launch from lower latitudes. The Space Coast is not in Florida for nothing. Baikonour also comes to mind. And of course Sea Launch actually moves the launch platform to the equator.

The exception is polar orbit vehicles, which are going perpendicular to the Earth's rotation, and thus do not take advantage of the Earth's rotation when launching.



"There are no stupid questions, but there are a lot of inquisitive idiots."
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