How do you calculate the length of this piece?

  • #1
http://chestofbooks.com/home-improv...r-Boys/images/Making-the-Model-Engine-292.png
I want to know how you would determine how long the rocker arm in this picture is.

http://chestofbooks.com/home-improv...r-Boys/images/Making-the-Model-Engine-297.png
Here is all the parts so you know what I mean.
It obviously moves the slide valve, but mathematically speaking, how do you know the length of it and where the fulcrum would be to move it within 3 1/2 ''.
 

Answers and Replies

  • #2
SteamKing
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It looks like the overall length of the rocker arm is 7 inches, from the label on the diagram. This is not the center-center distance of the pins though. I don't think there is sufficient information to determine completely all the dimensions required to describe the function of this mechanism, however.
 
  • #3
Baluncore
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The slide valve is operated from an eccentric on the flywheel / crankshaft.
The offset of that eccentric sets the movement of the eccentric rod at twice the offset.
The movement required of the slide valve is known from it's design.
The three pins on the rocker arm need to be spaced in proportion to those movements.

The “Stephenson valve gear” on a steam engine is an adjustable version of your rocker arm that gives control of valve opening, phase and so reversal of the engine rotation.
http://en.wikipedia.org/wiki/Stephenson_valve_gear
 
  • #4
The slide valve is operated from an eccentric on the flywheel / crankshaft.
The offset of that eccentric sets the movement of the eccentric rod at twice the offset.
The movement required of the slide valve is known from it's design.
The three pins on the rocker arm need to be spaced in proportion to those movements.

The “Stephenson valve gear” on a steam engine is an adjustable version of your rocker arm that gives control of valve opening, phase and so reversal of the engine rotation.
http://en.wikipedia.org/wiki/Stephenson_valve_gear

I see, but is there a way to calculate how long it would need to be and where the pin would need to be for the slide valve to move a certain distance? Sorry if it sounds confusing.
 
  • #5
Baluncore
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There are too many possibilities to give you a single equation.
The diagrams are insufficiently detailed to read or know the sizes.
You need to know values for …
The actual movement required by the slide valve.
The centre offset of the eccentric specified, e.

Going by the dimensions given in your diagram. The flywheel axis is 2.25”? above the base.
The rocker arm is close to vertical, with a fulcrum about 0.5” above the base.
Rocker arm length is 7” between centres?, so slide valve axis is 7.5” above the base.
Steam box is 3.5”? long externally, so probably 3” internally.
The slide movement is therefore probably about 1.5”.

Where y is the length up the rocker arm from the fulcrum to the eccentric rod pin, and e is the offset of the eccentric from the flywheel axis. The rod moves twice the eccentric offset, e.
1.5” / 7” = (2 * e) / y
1.5 / 14 = e / y, therefore; e = y * 1.5 / 14, or; y = 14 * e / 1.5

From the diagram, y will be about 2.25” – 0.5” = 1.75”
Therefore the eccentricity must be about e = 1.5 * 1.75 / 14 = 0.1875”
 
Last edited:
  • #6
There are too many possibilities to give you a single equation.
The diagrams are insufficiently detailed to read or know the sizes.
You need to know values for …
The actual movement required by the slide valve.
The centre offset of the eccentric specified, e.

Going by the dimensions given in your diagram. The flywheel axis is 2.25”? above the base.
The rocker arm is close to vertical, with a fulcrum about 0.5” above the base.
Rocker arm length is 7” between centres?, so slide valve axis is 7.5” above the base.
Steam box is 3.5”? long externally, so probably 3” internally.
The slide movement is therefore probably about 1.5”.

Where y is the length up the rocker arm from the fulcrum to the eccentric rod pin, and e is the offset of the eccentric from the flywheel axis. The rod moves twice the eccentric offset, e.
1.5” / 7” = (2 * e) / y
1.5 / 14 = e / y, therefore; e = 1.5 / ( 14 * y), or; y = 14 * e / 1.5

From the diagram, y will be about 2.25” – 0.5” = 1.75”
Therefore the eccentricity must be about 1.5 / ( 14 * 1.75) = 0.061”

Perfect answer, thank you.
 
  • #7
Baluncore
Science Advisor
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I edited the final arithmetic...
 
  • #8
I edited the final arithmetic...

What is the equation from? I'm no mechanical engineer, just trying to understand this machine.
1.5” / 7” = (2 * e) / y
1.5 / 14 = e / y, therefore; e = 1.5 / ( 14 * y), or; y = 14 * e / 1.5

That specifically.
 
Last edited:
  • #9
Baluncore
Science Advisor
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The rocker arm is a lever. For small movement about the fulcrum, the movement will be proportional to the distance from the fulcrum.

The eccentric rod moves (2 * 0.1875”) = 0.375” and is attached at 1.75” from the fulcrum.
The slide valve moves 1.5” and is attached 7 inches from the fulcrum.
Those two ratios will be the same;
0.375” / 1.75” = 0.2142857
1.5” / 7” = 0.2142857

Where the eccentricity is e, and the eccentric rod is attached at y,
the slide valve stroke length is s, and the rocker arm length between centres is r.
Then we can equate the proportions or ratios by writing (2*e) / y = s / r

Cross multiply and we get 2 * e * r = s * y
Knowing three of the four variables lets us solve for the unknown.
s = 2 * e * r / y
y = 2 * e * r / s
r = s * y / ( 2 * e )
e = s * y / ( 2 * r )

Note that the arc followed by the rocker arm pins will be slightly higher at slide valve mid-stroke than at the ends of the stroke. Since the movement is small this should not be a problem, but you should build it so the variation is symmetrical above and below the slide valve rod guide. The eccentric rod does not have the same problem because it is effectively pinned at both ends.

The slight bending of the slide valve shaft could be resolved by using a slot at either end of the rocker arm where a pin could move axially in the arm. The alternative would be a short link at the top of the rocker arm to the slide valve rod.
 

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