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London To Sheffield In 8 Minutes


madannie77

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Who needs the new high speed railway line when a trusty old "Peak" locomotive can travel at this speed.

<iframe width="420" height="315" src="http://www.youtube.com/embed/FZ_LWvQeDBE" frameborder="0" allowfullscreen></iframe>

(I could do without the music, however).

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Who needs the new high speed railway line when a trusty old "Peak" locomotive can travel at this speed. <iframe width="420" height="315" src="http://www.youtube.com/embed/FZ_LWvQeDBE" frameborder="0" allowfullscreen></iframe> (I could do without the music, however).

Great video find MA

Like the nostalgic trip down the "Dive Under" at the Sheffield

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Great video find MA

Like the nostalgic trip down the "Dive Under" at the Sheffield

Indeed: that was the part which I enjoyed the most when I first saw it :).

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Indeed: that was the part which I enjoyed the most when I first saw it :).

I was suprised how many telegraph poles were still by the line side.

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Who needs the new high speed railway line when a trusty old "Peak" locomotive can travel at this speed.

<iframe width="420" height="315" src="http://www.youtube.com/embed/FZ_LWvQeDBE" frameborder="0" allowfullscreen></iframe>

(I could do without the music, however).

Looks like "London to Sheffield in 8 minutes" is based on the well known 1950's speeded up classic called "London to Brighton in 4 minutes"

But this one is even faster

London to Sheffield is about 160 miles.

To do that in 8 minutes is 20 miles per minute

Or, in more familiar units, 1200 mph

In aircraft parlance that would be around Mach 1.85.

Nearly twice the speed of sound and nearly as fast as Concorde at full speed.

Would the train stay on the track with that sonic shock wave of compressed air that can't get out of the trains way fast enough being forced under the train between the track and the trains underside?

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Guest Martin-Bacon

Fantastic find.

Have a look at http://yfaonline.com/assetDetails.cfm?film=1007&keyword=sheffield&fromSearchValue=fromKeyword&start=1

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Loved it. I watched the video at work (in my lunch break of course) since my satellite internet at home was buffering so much I estimated the video would have run longer than the actual train journey.

Looks like the scheduled stops - Leicester, Derby and Chesterfield - were edited out.

So, Mr Maths - alias Dave - here's a question. If the scheduled stops had been 12 minutes each and had not been edited out, how much longer would the 8-minute journey have taken?

Don't worry about deceleration and acceleration, I wouldn't even know how to ask the question. :):) :)

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Loved it. I watched the video at work (in my lunch break of course) since my satellite internet at home was buffering so much I estimated the video would have run longer than the actual train journey.

Looks like the scheduled stops - Leicester, Derby and Chesterfield - were edited out.

So, Mr Maths - alias Dave - here's a question. If the scheduled stops had been 12 minutes each and had not been edited out, how much longer would the 8-minute journey have taken?

Don't worry about deceleration and acceleration, I wouldn't even know how to ask the question. :):):)

I paused it at those stations, just because I thought the right thing to do lol

It took a couple of attempts to stop over shooting the platforms

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Loved it. I watched the video at work (in my lunch break of course) since my satellite internet at home was buffering so much I estimated the video would have run longer than the actual train journey.

Looks like the scheduled stops - Leicester, Derby and Chesterfield - were edited out.

So, Mr Maths - alias Dave - here's a question. If the scheduled stops had been 12 minutes each and had not been edited out, how much longer would the 8-minute journey have taken?

Don't worry about deceleration and acceleration, I wouldn't even know how to ask the question. :):):)

OK, if you are going to simply the problem that much the answer is simple.

It would be = total travelling time + (number of stops x12)

So if there were only 3 12 minute stops ( Leicester, Derby and Chesterfield), then it would take 8 + (3x12) which is 44 minutes.

However, it is more complex and you would have to take acceleration and decelleration into account.

To accelerate instantly from 0 to almost twice the speed of sound is the equivalent of being fired out of a very high velocity gun, and to decellerate by the same amount in the same time would be like being in a head on collision with a vehicle of the same mass travelling at the same speed in the opposite direction giving a head on relative impact speed of around 2400 mph.

These sudden changes in momentum would exert such massive forces on the driver and passengers that they would be instantly killed.

If you take accelaration and decelleration into account at a more sedate and bearable rate then the total journey time with 3 stops will invariably be greater than the 44 minutes previously calculated.

In Jules Vernes original 1865 novel of "First men in the moon" his proposed method of sending them to the moon was by firing them to the moon in a large artilliary shell fired from a massive gun to give it sufficient speed to reach its target 250, 000 miles away without gravity overcoming its motion first.

Bearing in mind that once the shell leaves the end of the gun barrel it is totally ballistic, - it has no further means of self propulsion and must already have sufficient momentum to carry it the rest of the way against the retarding force of gravity it would have to accelerate from 0 to over 25,000mph within a realistic gun barrel length if it was to work. In reality, the acceleration required to do this would kill the crew before the shell had left the end of the barrel.

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I paused it at those stations, just because I thought the right thing to do lol It took a couple of attempts to stop over shooting the platforms

I've known some real trains on the Sheffield - London route, travelling at fairly low speeds below 100mph overshoot the platforms as well.

Perhaps you have a career as a train driver on this route.

It's rubbish going down to London all the time, - but it's great coming back to Sheffield again! lol

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OK, if you are going to simply the problem that much the answer is simple.

It would be = total travelling time + (number of stops x12)

So if there were only 3 12 minute stops ( Leicester, Derby and Chesterfield), then it would take 8 + (3x12) which is 44 minutes.

However, it is more complex and you would have to take acceleration and decelleration into account.

To accelerate instantly from 0 to almost twice the speed of sound is the equivalent of being fired out of a very high velocity gun, and to decellerate by the same amount in the same time would be like being in a head on collision with a vehicle of the same mass travelling at the same speed in the opposite direction giving a head on relative impact speed of around 2400 mph.

These sudden changes in momentum would exert such MASSIVE forces on the driver and passengers that they would be instantly killed.

If you take accelaration and decelleration into account at a more sedate and bearable rate then the total journey time with 3 stops will invariably be greater than the 44 minutes previously calculated.

In Jules Vernes original 1865 novel of "First men in the moon" his proposed method of sending them to the moon was by firing them to the moon in a large artilliary shell fired from a MASSIVE gun to give it sufficient speed to reach its target 250, 000 miles away without gravity overcoming its motion first.

Bearing in mind that once the shell leaves the end of the gun barrel it is totally ballistic, - it has no further means of self propulsion and must already have sufficient momentum to carry it the rest of the way against the retarding force of gravity it would have to accelerate from 0 to over 25,000mph within a realistic gun barrel length if it was to work. In reality, the acceleration required to do this would kill the crew before the shell had left the end of the barrel.

Oh no, you're not getting away with that. The 12 minutes are the actual time of the stops, as if the train were doing a normal journey. You have to work out the length of the stops at film speed, then add them on to the 8 minutes.

See if you get the same answer as me :mellow:

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Oh no, you're not getting away with that. The 12 minutes are the actual time of the stops, as if the train were doing a normal journey. You have to work out the length of the stops at film speed, then add them on to the 8 minutes.

See if you get the same answer as me :mellow:

12 minutes seems a long time for stops, especially Chesterfield

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12 minutes seems a long time for stops, especially Chesterfield

Hang on a bit, what have you got against Chesterfield?

Just because their church spire is a bit bent.

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Oh no, you're not getting away with that. The 12 minutes are the actual time of the stops, as if the train were doing a normal journey. You have to work out the length of the stops at film speed, then add them on to the 8 minutes.

See if you get the same answer as me :mellow:

I take your point, but surely a 12 minute stop is a 12 minute stop.

You are travelling on a very fast train, but unless the passengers move equally as fast (which they wouldn't, - they are mere humans) it would still take the same time, 12 minutes, to load and unload.

However if the train travels at 100mph normally and is now doing 1200 it is going 12 times faster, so at that scale factor 12 minutes would become a 1 minute stop (approximately), so 3 1 minute stops onto the 8 minute journey with instantaneous acceleration and decelleration would give a total film showing time of around 11 minutes.

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