Hammer Blow

Simply put, what is hammer blow (is it just the act of lots of heavy motion going round quickly?) and how is it calculated? Is there a simple formula or does it depend on the individual locomotive?

 

When i nod my head - you hit it!

 

It is the effect of any un-ballanced mass of the motion, and the force of the cylinders beating, remember, steam locos are double acting engines, so this will occur hich ever whay you are running.

 

hammer blow varies for each engine, at what speed it operates and with how much load it is bearing. it is all down to the moments induced when the unbalanced masses of the motion rotate. without wishing to sound patronising, (im doing something similar in my degree) it is definately not simple. bear in mind that it also has a lot to do with the track, its quality and the ground its laid on.

 

There is the inevitable Wiki page here:

http://en.wikipedia.org/wiki/Hammer_blow

 

Hammerblow is not the easiest of concepts to understand, going hand in hand with balancing. This latter falls into two categories: Rotational and reciprocating. We should consider rotational imbalance first as it’s the easier of the two.
Should a weight on a string be swung in a circle, it will try to fly outwards in contradiction to the force of gravity. This is due to centrifugal force. According to Newton’s First Law of Motion, a body will try to keep moving in a straight line. The weight tries to maintain its flight in the direction it is travelling at any given point in time; it tries to follow a tangent to its circular path, ninety degrees to its radial position. What causes it to follow the circle is centripetal force acting along the string, drawing the weight to the circle’s centre.
If there is an out of balance weight on a wheel, it will try to make the wheel follow an ever widening path; as the out of balance force reaches the top of the wheel rim’s path, it will try to lift it. As it comes down towards the rail, it tries to push the wheel into the rail, or the rail into the ballast. This in essence is hammerblow, and the formula is F = mw^2R, where F = the out of balance force, m is the out of balance mass, w is angular velocity in Radians per Second (the maths won’t work, but consider this as rpm), and R is the radius of the mass from the wheel centre.
It is easy to eliminate the hammerblow simply by placing another weight, in proportion to its own radius from the wheel centre (m x R must be the same for both). This will create a similar force acting in the opposite direction, and equal and opposite forces cancel each other. End of problem.
Not really, not with a steam engine, as we now must examine the reciprocating masses. These include the piston, piston rod, crosshead, and front one third of the connecting rod. If the engine has Walschaert’s valve gear, add to this the union link and lower half of the combination lever. All these are moving only in a fore and aft direction, and once moving they have momentum: as they reach the end of their stroke, they want to keep moving, despite delayed release of exhaust steam and the effect of lead steam to stop them. Ultimately, they are stopped by the crankpin preventing further movement thus: as the piston moves from front to rear, it must first stop with the wheel on back quarter, before reversing direction and travelling from back to front. The momentum causes it to continue rearwards, pushing the crankpin, wheel and axle rearwards in the axlebox, until all clearance between them is taken up. At this stage it pushes the axlebox rearwards in the hornguide, until all this clearance is used, when it pushes against the frame, causing a back and forth surge. This produces a heavy knock (which increases with mileage and general wear) which rises with speed. The process then begins in the opposite direction as the wheel turns but with the same result as the wheel and box are pulled forward as the wheel approaches its front quarter. It will become very uncomfortable for the crew, and eventually become sufficiently violent that mechanical damage will result. The WD 2-8-0s were so bad at 30 mph that the coal worked its way down the tender, on to the footplate where it worked its way forward to pile up under the firehole doors, and waited to be shovelled in!
On way to resolve the problem was to add weights to the wheel rim opposite the crankpin. So as the piston, etc. tries to move the wheel rearwards, the balance weight tries to move it forward and vice-versa. The two then balance out.
It will now be realised the we have reintroduced a rotating out of balance force: in the fore and aft direction the weight is balanced by the reciprocating masses, but since these have no vertical component, the weight will try to lift the wheel as it approaches the top quarter, and force it downwards towards bottom quarter. Pre-war, the LMS carried out an experiment with some Black Fives on oiled rails. The engines were actually moving forwards at about 10 - 18 mph, while the wheels were turning at a rate equivalent to up to 104 mph. Observers walking alongside could clearly see the driving wheels lift 2.4 inches clear of the rail once every revolution, to slam back down on to the track at the other half turn. Now that’s what I call hammerblow! Three engines were used with the reciprocating masses balanced to 66.6%, 50% and 30%. The results were: 66.6% - lift 2.4”, fore and aft oscillation normal; 50% - lift 0.4”, fore and aft oscillation moderate; 30% - lift 0”, fore and aft oscillation excessive (Stanier 4-6-0s at Work, A.J. Powell, 1983 Ian Allan, ISBN 0 7110 1342 X).
There is therefore a conflict between reciprocating imbalance leading to mechanical damage, and rotational imbalance leading to hammerblow and track damage. It was always difficult to achieve an acceptable compromise with a two-cylinder engine as the two sides were 90 degrees out of phase; neither helped balance the other. Multi-cylindered engines fared better; a three cylinder engine always had one cylinder at least partly balanced by the other two, while a four cylinder type (at least, one with its cranks set at 180 degrees) always had two pistons moving in opposite directions, giving equal and opposite forces and balancing each other. This was most effective where all cylinders drove on to the same axle, and C.J. Bowen Cooke so designed the Claughtons for this very reason, and Sir Nigel Gresley did likewise with his three-cylinder types. But all cylinders driving one axle imposes severe stresses on the crank axle, which most designers chose to avoid. This meant that the reciprocating masses acted on their respective wheels, since there was only the coupling rod to transmit the forces from one wheel to its neighbour, but they were neutralised over the engine as a whole.
It is interesting to consider the BR Standards; all bar one were two cylinders only. This reduced costs, weight and maintenance, but was one reason why the Britannias were not always popular with crews, who considered them rough riding. Those two big cylinders drove very heavy reciprocating parts, such that passengers several coaches back complained of excessive surging. The men compared them with pre-war, multi-cylinder types: the A3s, Royal Scots and Castles which were generally preferred.

 

Thanks for the replies, it makes more sense now. I was aware that the motion would have some effect, but I assumed (whoops!) that the pistons, piston rod and crosshead wouldn't have any effect. I was considering a simplified investigation into it as part of my Physics A-Level, but it sounds a bit complicated for a 2 week investigation.

 

Best person to answer this is Mr Riley the amount of paperwork and engineering calculations to get 76079 over Barmouth bridge has to been seen to be believed. Hammer blow believe me he knows inside out.

Ian over to you!!

 

Presumably the formula was derived from F=ma and a=Rw^2? So wouldn't this give the rotational force of the wheel, as opposed to the actual force acting vertically, either down into the rail or lifting the loco off the track?

Thanks

 

Yes, it is the force applied at all angles of rotation, but part of it is balanced in the fore and aft directions by the opposing reciprocating masses (that's the point of the balance weights!), but there are no opposing forces in the vertical plane, hence the hammerblow.

So this rotational force is constant at all angles, but minus the reciprocating forces in the horizontal plane, gradually increasing as the angle changes until it reaches maximum in the vertical plane, then falls away again towards horizontal.

 

Oh, sorry, yeah, that makes sense ](*,)

 

From http://www.railway-technical.com/st-glos.shtml#H
definition of Hammer Blow

The force exerted by the thrust of the connecting rod on the crank and transmitted to the rail with each revolution of the driving wheel. Rotating masses must be balanced but since this is only the wheels and the connecting rods, this is reasonably easily done by balance weights.


Reciprocating masses such as pistons, piston rods, etc are much more difficult to balance. They are balanced at the wheel centres and on the crank axle itself. In fact the design of the crank axle may be inherently self-balancing to some extent. It is not desirable to balance 100% of reciprocating mass because this would result in the load on wheels dynamically changing during rotation (and this is exactly what hammer blow is). It was common practice to balance 60% of reciprocating mass but this was found to cause hammer blow, so was reduced to 30% on two cylinder engines. The balance is spread unequally over all coupled wheels. Four-cylinder engines, because of their cycle, are self-balancing, so balance weights are not used. Once again, adding weights to them actually causes hammer blow. Engines must not be operated under power without connecting rods as the unbalanced forces can actually destroy the track. See Wheel Balancing.

 

???

There were a couple of points in this explanation which were slightly iffy, but condensing a complex subject into just a few lines does cause some problems. I am though trying to figure out just how engines can be operated under power without connecting rods!

It was not unknown to remove a single rod from a crippled loco to allow it to work forward - very slowly - to clear the track, sometimes with the train still attached, but all of them???

 

I attended a lecture by S C Townroe (spelling?) once. He showed a photo of a Drummond Black Motor 0-6-0 running as a 2-2-2 on a branch line train on a busy summer Saturday when nothing else was available. She'd had a spot of bother and that was the only way to get her to run.

 

well perhaps you should argue that with the people that wrote it. i said it was from www.railway-technical.com i have not altered it.

 

Yes, I had realised that, hence the wording of my reply didn't mention you as originator, just "this explanation..." No offence to you was intended, and I hope none taken.

As to " ...attended a lecture by S C Townroe (spelling?) once. He showed a photo of a Drummond Black Motor 0-6-0 running as a 2-2-2 on a branch line train on a busy summer Saturday when nothing else was available. She'd had a spot of bother and that was the only way to get her to run." Yes, but these are coupling rods which transmit torque from the driving to coupled wheels, but the connecting rods transmit the piston's force to the cranks, so the engine won't go without them! The coupling rods, by the way, have an entirely circular motion so are easily balanced; the front third of the connecting rods, by contrast, are counted as reciprocating and included with the pistons, piston rods, etc as reciprocating masses. This was one of the issues I have with the original explanation.

As for running a loco without coupling rods, this was looked at by Jim Markland in Foxline's 'More Recollections of a Bolton Engineman' when a 4F was delivered to Horwich works in a similar condition. The driver was amazed at how free running it was, and "the driver's one regret was that he was not able to try her with three or four bogies behind." No mention of excessive hammerblow, though!

 

Unconverted Bulleids dont do hammer blow, so whats the problem????

 

If that is the case explain how the balance is applied? Did Mr. Bullied discover a new form of steel with zero mass for the various components both rotating and reciprocating.

Minimised perhaps, eliminated doubtful - despite what is published in various erudite journals.

 

The reciprocating masses of three-cylindered engines are to some extent self-balancing, in that their relative movements cancel each other out and so produce close to Nil resultant force. This has to be tempered by any difference in weights of those masses - particularly those of the inner cylinder, which are often shorter and lighter - and the cylinders' angles of thrust being similar. But with care, it should be feasable to drastically reduce, if not actually eliminate, weights within the wheels as counter-balance.

There still remains the issue of rotational imbalance: crankpins, coupling rods, etc. These still need to be balanced within the wheel. The rotating masses acting on each wheel are usually calculated and a corresponding balancing weight added within that wheel.

I cannot remember where I read this - it was many years ago - but Mr Bulleid tried a different way. With the cranks at 120 degrees, there was some rotational crossbalancing from one side to the other, and he made use of this. He calculated the rotational imbalance across the whole axle, then split the balance weight equally between the two sides. Obviously the weights would no longer be oposite the crank pins.

The idea was that this reduced the actual weights need to to achieve balance, but I would have thought that a 'couple' would have been produced.

I've occassionally wondered at the logic of this - I don't doubt it worked, by the way! - so would welcome further comments from people who know these engines better than I.