beam engine

For my next project I am making a beam engine, not specified which one as yet just looking at basic theory.

Ok a first question. The engine will have a crankshaft attached to a crank disc and a crank pin. What will be the position of the crank pin centre when the piston is at either TDC and/or BDC.

From a beginners viewpoint it's easy to assume that they will be on a vertical centre line through the face of the crank disc and therefore at 90 degrees to a horizontal centre line. Is this correct?

Geoff

Hello, Geoff. I think that you will find that top and bottom dead centres (points of maximum swing of the beam) occur when a straight line is drawn through the centre points of the crank pin, the crankshaft and the con-rod / little end; i.e. when the centre line of the con-rod aligns with the crank pin centre line.

When the crank pin is 'up', it will be a little closer to the cylinder than the true vertical centre line and when 'down' will be slightly further away from the cylinder.

The angular displacement of the crankpin from the vertical cenre-line of the crankshaft will not be the same for TDC and BDC, but the difference is very small and of no account for all practical purposes.

Geoff G

Hi Geoff,

Perfect reply, thank you.

After reading your first and second paragraph I feel a little dumb because when you think about it's all fairly obvious, but not initially for me!!!!!

Thanks again. I shall press on

Geoff

Not necessarily, it depends on the exact horizontal distance between crank axis and beam pivot. If this distance equals the half beam length then Geoff is correct, but if slightly reduced so that crank is directly below the little end pin at both centres, then symmetry rules. Pedantic people like me allow for that sort of thing, very probably a waste of time!

OK, Duncan. Well done, you win with that scenario! Just goes to show what thinking a problem right through to the end can do for you.

Geoff G

Hey I'm learning all the time here

Duncan I assume when you say in your first sentence "crank axis" you mean crankshaft axis?

The example I'm looking at is as you describe, the horizontal distance is half the beam length.

Thanks for your comments

Geoff

......................I assume when you say in your first sentence "crank axis" you mean crankshaft axis?..........YES

.......... the horizontal distance is half the beam length.........In that case Geoff is right

You'll have even more fun when you come to figuring out the Watts parallel motion linkage at the other end of the beam. Enjoy!

You'll have even more fun when you come to figuring out the Watts parallel motion linkage at the other end of the beam. Enjoy!

Hi Pete, no I'm ok with that it's sorted. My main problem was establishing the optimum position for the crank pin at tdc and bdc but Geoff G put me right on that.

Atb and take care Geoff

It is taken as read that the Piston Rod is vertically below the pin at the cylinder end of the beam, at Inner and Outer dead centre, using the Watts linkage to cope with the "vibration" .(because of the arc ) at intermediate points

In the same way, that the Crankshaft axis is vertically below the pin at the other end of the beam.

The beam does not have to be equal about the pivot point. Although many were pumping engines, it was quite usual for Cornish beam engines to have asymmetric beams.

Howard..