Back to basics - Elmer's #25 Wobbler

[ReFlad]

Arnold,
I always enjoy your builds.  Add another "Marvelous" to the completed column!

Ronald

[chucketn]

Arnold, can you please elaborate on "I couldn't find the template I made way back, and with the luxury of the DRO on the mill, I just calculated the port hole locations (just some basic trigonometry) and set about drilling, reaming and milling. "

I'm also building this little engine, and short of building the drill jig, can't work out how to locate the port holes in the upright.
I do have a dro on the mill.

Chuck

[propforward]

Attention - the drawing I originally posted here was incorrect. Revised one will be posted shortly.

[propforward]

I screwed up with my first layout of the holes when I drew it in autocad. So much for the lazy approach.

Here is a revised sketch of the hole layout, for plotting on a mill.

Apologies for the mess up. 

The yellow dashed outline represents the drill jig, just to show how it all comes about. The circle in the center of the bearing hole represents the pin part of the jig.

[arnoldb]

Ronald, thank you; my apologies for the late reply 

Chuck, thanks for asking - I forgot to include that.

Calculating things with CAD is a quick way to do it - like Prop did.

I'm a bit old-fashioned in some instances, so I used good old maths to do it.  It's good mental exercise as well.

OK, what follows may be a bit lengthy, but it allows one to calculate the port positions for just about any center-pivoting wobbler engine - or even help design one.  If the basic principles of operation of this type of wobbler is understood, it's fairly straight forward.

First a Crap-o-Cad picture.  This resembles the important bits of any center-pivoting wobbler engine's layout. - Fig 2 is an enlargement of the blocked section of Fig 1, and is the port layout relative to the pivot pin  It might look intimidating at first sight, but it's actually quite simple and is a large part of the important dimensions on which a wobbler engine works:


The definitions of the letters I added to the image:
t:  Length of the crank throw from the center of the crankshaft to the center of the crank pin.
D: The length from the center of the crank shaft to the center of the cylinder pivot pin.
C: The center line of the piston rod where the cylinder and piston rod is at maximum angle during operation.
theta: The actual angle when the cylinder is at it's maximum pivot
P: Center of the pivot point of the cylinder
L: The distance the center of the port hole in the cylinder is away from the center of the pivot point.
A: is the arc the center of the cylinder port hole follows along the port face
Pc: the point at which the cylinder port hole is central - this corresponds to top and bottom dead center for the piston.
Pb: is the center of one of the port holes in the block.  There's a caveat (in my opinion) to this location that I'll get around to later on.

In Fig. 2, x and y are the values we're interested in for locating the port holes in the block.
x is the distance from the center of the pivot point to the center of the port hole in the block.
y is the offset from the center line to the center of the port hole in the pivot block.

OK, now to get down to things.

In Fig. 1: To locate the coordinates of the port center holes in the block, the maximum angle that the crank would pivot the piston rod is needed.  This is also the maximum angle at which the cylinder will pivot (angle theta).  This is along line C, and maximum deflection happens where line C intersects with the circle the center of the crank pin makes while the crank rotates. At this point, a rectangular triangle is formed with a 90^o angle between lines "C" and "t" to the center of the crank shaft. Calculating angle theta is simple - it's the sin^-1  (arcsine) of the ratio of the throw length "t" to the distance between the cylinder pivot center and the crankshaft center "D". - (Fig 1).

On to Fig. 2, and the caveat I mentioned.  This layout is pretty much optimal for a wobbler engine running on compressed air with symmetrical ports and identical port sizes on the block and cylinder - like Elmer used for all of his wobbler engines.  Personally, I have an untested hunch that things may be different for optimal use of live steam in wobblers, but I have to play around with that a bit first...
To calculate the holes for the port holes in the cylinder block, it's once more on to some simple trigonometry.  The "x" and "Y" values for the port center hole in the block is easy to calculate using the formulas next to the Fig 2 block.  I only showed one coordinate in the image for one port, but the other one is dead easy - just add the same amount of "y" above the center line at the same "x" setting for the top port.  For a double-acting wobbler, use the same amount of x to the right of the pivot point "P", with equal offsets for Y above and below the center line.

To use the above, all that's needed to calculate things is the values of D, t and L

Trigonometry does not care about units of measure, so the above can be used to calculate the values needed in both metric and imperial.  Just stick to one system; they can't be mixed up during calculations.  If you do it in imperial, choose to work in either inches or thou - don't mix them. 
You can also choose to work in either degrees or radians for the angles - just stick to one of these and don't mix them.  Most people will work in degrees - I'm sometimes inclined to work in radians; that stems mostly from my programming background.

So to put all the above into values for Elmer's #25.  I'll do it in imperial, and as the port offsets are relatively small, and we need a bit of accuracy, I'll do it in thou, with angles in degrees.
The drill jig provides most of the dimensions - except for the crank throw which is easily obtained from the crank web layout.  From the plans:
D = 1 ¹/₄ " = 1.250" = 1250 thou
t   = ¹/₄"    =  250 thou
L  = 3/8"    = 375 thou

Angle theta is = arcsin (t/D)
                         = arcsin (250/1250)
                         =  11.537^o

to find x:     x = L cos theta
                       = 375 . cos 11.537
                       = 367.42 thou

Same for y: y = L sin theta
                        = 375 . sin 11.537
                        = 75 thou

Depending on how close you want to work, round the values for x and y.  I tend to try and work accurate to 0.01mm - or roughly 0.5 thou, so x~= 367.5 = 0.3675"  and y~=75 thou = 0.075"

The same, but in metric (mm)
D = 31.75mm
t   = 6.35mm
L  = 9.53mm

Angle theta is = arcsin (t/D)
                         = arcsin (6.35 / 31.75)
                         =  11.537^o

to find x:     x = L cos theta
                       = 9.53 . cos 11.537
                       = 9.34mm

Same for y: y = L sin theta
                        = 9.53 . sin 11.537
                        = 1.91mm

Some comments about using the above in the design of a wobbler:

Most wobblers of the center-pivot type follow a design where the ratio of the crank throw to the distance between the crank center and the pivot center is at a ratio of between 1:3 to 1:5.  The #25 is on the 1:5 ratio - I think Elmer used the longer ratio to enable him to make a longer piston - thus doing away with a cross-head.

A longer ratio makes for a smaller angle of change on the cylinder, and hence the port holes in the block move closer together, thus restricting the port hole sizes.  It does, however add some much-needed space between the cylinder and the crank to add a packing nut and cylinder head if one is making a double-acting wobbler.

A shorter ratio on the other hand allows for bigger port holes - and this is useful for some better performance (air/steam can enter and exhaust more easily), but leaves less room between the crank and cylinder, thus making it more difficult to make a double-acting wobbler, or in the case of a single acting one, preventing the use of a longer cylinder like Elmer used for the #25 and thus necessitating a cross-head of some form.

For an engine with equal-sized ports both on the block and in the cylinder, the maximum port size can easily be worked out from the value gotten for "y" in the calculations above.  it must be _just_ smaller than "y" to allow a bit of dead space when the piston is at top or bottom dead center - otherwise blow-by can happen where the port hole actually connects the inlet port to the exhaust port.  Depending on how accurate one can work, the port sizes for the #25 can easily be opened up to 1.8mm instead of the 1.6mm (1/16") Elmer specified.

OK, I must be boring everyone to death by now, and MEM can't do with dead members, so I'll get off the soap box.

Chuck, I hope your question is answered - if not, please give a shout  - I got a bit carried away...

Kind regards, Arnold

PS: Marv, if you find some incorrect terms/descriptions, please mention those; when I learned this part of maths, it was all done in Afrikaans, and I haven't had much practice translating my maths terms to English.

[mklotz]

Wow.  I'm pleased.  Someone who eschews CAD and still does math.  Good onya.  As you say, use it or lose it.  The only way to develop math skills is to do problems - even if those problems have already been solved or can be solved by some mechanical means, e.g., CAD or dedicated program.  I wish I saw more of our members doing what you've done here.

Don't worry about the Afrikaans.  Math is perhaps the most universal language, moreso even than music.  I see nothing amiss.

Notation-wise it's probably better to use arcsin, arccos, etc. rather than the sin to the -1 notation.  It came out OK in your text but, on my screen, this showed up:

Angle theta is = sin -1 (t/D)

which is confusing.  The ^-1 notation can also confuse mathematical novices who think you're referring to 1/sin, 1/cos, etc.

[chucketn]

Thanks, Arnold, for the explanation. I will have to study it for a bit to understand it. Math was never my strong point.

Chuck

[Captain Jerry]

Math is fine but you still have to get the values right.  I think that 3/8" = 375 thou and if I am right,  Y = 75 thou or .0750"

If the dimension is really critical, more than one approach is a good cross check


Jerry

[arnoldb]

Marv & Jerry, thanks for the proofreading  - I fixed up things according to Marv's suggestion, and fixed the 3/8" conversion 

Chuck, don't worry - it's less complicated than it looks. 

Like I mentioned, I do like doing maths, and for things like this it's by far the quickest way for me to get the answers I need.  It'll land me in trouble with environmentalists one day, as I seem to regularly chuck a square on a hippopotamus as well.
Prop showed the CAD method. 
You can also just use plain marking-out with layout dye, a compass and a rule - and draw the lines and circles on the workpiece to find the correct intersections.  Punch and drill.  It's less accurate than all the calculating boo-hah and positioning with handwheel readings or a DRO system, but it works.
Or use Elmer's drill jig the way he intended it to be used.

So that's four different ways to get the same result, and there are more ways.  Choosing one depends on what equipment you have available, and how you like going about things.

There are no "right" or "wrong" ways to do things in model engineering (well, as long as it's safe)-  use whatever means suits you, and your method of working. 
We are all as different as the engines we build, and even though I've only been at this hobby for about 4 years, I've come to realize and respect that each person doing it have their own reasons for doing and expectations from the hobby. 
By sharing the different ways of doing things on the forum, we all get to learn from each other.  But it's up to each individual to choose whatever methods they like to use.  It's supposed to be personal fun - not punishment.

Kind regards, Arnold

[DaveH]

Arnold, jou Afrikaans wiskunde is baie goed 
 
DaveH

[chucketn]

Thanks for your coments, again, Arnold. I haven't used much of the math I learned in High School over 45 years ago. Just lucky, I guess, or not. If my Dad was still alive, he'd have given me a NCIS head slap for having to ask, and then sat down and explained it to me.

Chuck

[arnoldb]

Thanks Dave  - Jou Afrikaans is nie sleg vir 'n soutie nie 

Chuck, it's a pleasure 

Kind regards, Arnold