Surge force

In my dirty and unpaid job of making a design for a future mainlineable and usefull and environmementally sound tourist locomotive I have come in doubt.
Cox state that a 9f gives 6.77 tons hammerblow at 5 rps.
It is 40% balanced.
Surging is thus 10.2 tons fore and aft.
At 8.4 rps or 90 mph it is then more than 30 tons plus minus.
Was couplings and crew able to stand that?

 

A good question. To give a proper answer to this I would need to sit in a darkened room for some time. However I think the answer would involve the following.....

Both crew and coupling don't sense the force - they sense the displacement.
Displacement is proportional to the acceleration and the duration of the acceleration.
Acceleration is based on F=ma. Whilst F is a large number, m (mass of loco) is a very large number - so a is small.
Duration is also small - if at 5rps, it means 0.1 seconds later the force is acting in the opposite direction.
Therefore displacement is small.

For the coupling - in simple terms it is a spring - so Force at coupling is proportional to displacement - if displacement is small then force at coupling is small.

 

The austerity locomotives had no surge balance so if some of the regular crew can tell us at what speed they become intolerable for normal humans (non heroes) we can judge where limit is as mass of pistons must be more or less like 9Fs ,Brittanias etc.
By the way has the measurements of speed ,power etc for the Tornado 100 mph been published?

 

Not published, but looks like they have the green light for 90mph, Hopefully there will be some information forthcoming at the convention, but it will be notebooks at the ready as this info isn't likely to be 'Published' as such

 

Also not likely to be of much interest I suspect. My understanding was that the exercise wasn't really about power output but about riding, stability, loads on the track etc. rather than the actual power output of the locomotive. If they do have a speed/horsepower curve I suppose it would be of some interest in demonstrating what sort of timings could be achieved, but I thought the exercise was mostly about
"will it stay on the track"
"will the track survive"
"how well do our computer models match what happens in real life"
All absolutely vital information, but of limited interest to people not running trains.

 

Indeed...

 

I have found some material from well-paid doctors in nice surroundings giving advice to how much vibration the lower classes is supposed to endure for how long.
1.5 m/sec/sec is a no no for sitting transport workers .For an 8 hour working day safe level is less than half.
Austerity locomotives are unbearable, running light at 25 mph,according to Sagging Dragon.
If You have a very rigid coupling to a solid train of 1000 tons it help a lot but needs a lot of power.
120 mph for a free running two cylinder locomotive must have been an expirience for life.
Mass of pistons and locomotive plus tenders are comparable and unplesantness goes up with second power of rpm
Unpleasantness of A Saint at 120 is thus 10 times worse than Austerity at 25 and on Union complaining level,if Saint was unbalanced
If 50 % balanced only five times.
For an unbalanced austerity at 25 versus a balanced 9f at 90 it is about 6 times .
9f six times more unpleasant than austerity at 25.

 

9f and austerity are from Riddles time and I asume that piston masses are equal .400 kg each side
An austerity is 78 tons mass and tender 55.
Two 400 kg lumps stroking 711 mm ,90 degree out of phase correspond to a single mass of 800kg traveling 500mm.
25 mph is 11 meter per second and roughly 2.5 rps or omega equal 15 radian per second.(Very roughly).
If tender is loose coupled to loco, 800 kg moving 500 mm total will make rest of locomotive including cab move 5mm total 2.5 times per second.
When wife died I throwed all my high school textbooks out and i cannot remember how we calculate acceleration if we have 5mm aplitude at 2.5 Hz.
Please help

 

s = ut + 1/2at^2

5mm movement back and forth 2.5 times per second, i.e. over the course of 0.4s. To a rough approximation, assume it accelerates at an even rate for the first 2.5mm; then equivalent deceleration for 2.5mm; then the reverse. It achieves that in 0.4seconds, so the 2.5mm acceleration takes 0.1s.

0.0025 = 0 + 1/2 * a * (0.1)^2

a = 0.025/(0.5 *0.1*0.1)

a = 0.5m/s^2

1G = 10m/s^2 more or less.

So to a first approximation, I make the acceleration 0.05G.

(If you modelled it on the exact forces, which vary cyclically over one revolution, the peak would be slightly higher. But not I think an order of magnitude different).

Tom

 

Thank You for seing if workers in Denmark are softer than Enthusiasts in England
The absolute all body limit is 1.5m /sec/sec that we will get at 43 mph for free locomotive.
This is a prison for boss limit.

Errata

1 hour limit 1.4m/sec/sec and 8 hour limit 0.5 m/sec/sec

 

So that knocks the two cylinder compound on the Head ?

 

Not at all but I must know what I am doing.
The low pressure piston transmits more or less same work as high pressure and at much lower temperature,ergo same mass.
I have ten outside crankpins that can do something like the funny beams on the Nord 2-8-2 tank locomotive

http://thierry.stora.free.fr/pics/141tc1_1.jpg

http://gallica.bnf.fr/ark:/12148/bpt6k6482666t.r=?rk=21459;2

It ran nearly 120kmh uphill .
Wheels were 5 feet 2 i think.

 

With those limits, one wonders how Tom Kristensen makes a living in Denmark!

Tom

 

He is tolerated as long as he do his ugly stuff in foreign countries.
Our feminist society sees motor racing just as sinfull as coal burning.
Shaking