# Surface finish requirement for ball links.

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#### Owen_N

##### Well-Known Member
This is a subtopic of Wobble Plate engines.

convention: 10^3 = 10 to the power of 3.
- scientific notation- this deals with decimal places.
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I have included a detail drawing of the ball link.
it has a 10mm ball one end and a 12mm ball the other end, and a 6mm shaft.
all ends and sockets will be split, and retained in sliding housings.

My initial proposal is soften drill, braze, reharden, but these parts will reciprocate at 6000 rpm, and may need a precisely ground bearing surface
Sockets can be leaded bronze, which seems to be good for 150 KPa, which is adequate.

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Are there other options which will work better as bearings?
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I need to make something special, an there is a 45 degree main swing angle.
this the need for 12mm and 6mm as my main dimensions.

The ones with the hole through the middle only go to 17 degrees in total. (heim or rose types)

The pullout load is only centrifugal?? - that is pretty small for a wobble-type layout.
peak load may be similar to 50mm rotation at the peak piston tangential speed.

I wrote that down elsewhere. I think peak acceleration load, x 100g was 1.5kN, or 1.5 x 10^6 m/s(sq) or 153,000 G
rms velocity would be 66mm x 600 = 39.6 m/s, so peak v would be x 1.4 = 55.44 m/s
radial acceleration? A = v(sq)/r 61,472 m/s(sq)
so 10g would have a pull of 10 x 10^-3 x 61.4 x 10^3 = 614 N , or about 61.4 kgf.

If the bearing area = (25-9)x pi x 10^-6 = 50 x 10^-6 sq mt,
p = f/a = 614/5x 10^-5, or 128 x 10^5 or 12.8 MPa, which is reasonable.

A typical bearing load is more like 30 MPa.
Peak tangential velocity at the 12mm ball?
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Sweep speed is 45 degrees in half a cycle normal sweep would be 360 degrees, in 1/600s
therefore actual rpm = 6000/4 = 1500 rpm.
r = 6mm,
rev/s 150; or 1500 rpm.
v = Wr = 942 x 6 x 10^-3 = 5.65 m/s
peak = x 1.4 = 8 m/s.
This seems fairly quick for a plain bearing.
Will this cope with mist/splash lubrication?

What do you think?
Compare this with a con-rod-crank engine wrist pin.
At R = 3x r, this is about 18 degrees each side, and we are 22.5.

Not a huge difference, so a 12mm needle roller would handle it.
360/36 = 10, for a full cycle we swing 72 degrees at 6000 rpm this is
360/72 x 6000/ (360/72) = 1200rpm.

I will look up bronze bearing sliding speeds.
V max = 3.8 m/s
P max = 28 MPa
PV max = 2.63
we are: V = 8 m/s, maximum;

area = around pi x 6(sq) x 10^-6 = 1.13 x 10^-4 sq mt (113 x 10^-6)

P = F/A = 8 x 10^3/ 1.13 x 10^-4
= 70.8 MPa - this is high.

PV = 70.8 x 8 = 566.4

P seems a bit high?

if combustion pressure = 70 bar, and area = 7.21 x 10^-4 then F = PA = 70 x 10^5 x 7.21 x 10^-4
= 505 x 10 = 5.05 kN, mechanical peak = about 1.5 kN
Compare this with the load on a 10x15 needle roller.

From the SKF catalog, this is around the peak static load,
but the average pressure is much lower.

I would expect this size plain bearing is OK in a small 4-stroke engine.
the active area is around 15x10, or 150 x 10^-6
and P = say 6 x 10^3 / 1.5 x 10^-4 = 40 MPa.

This is fine, as the high loading is over a very short period and the average is quote low if spread over 2 complete rotations.
This gives a "dynamic" lubrication effect, where oil is pulled in and squeezed out, then recovers over quite a long period.

It is quite a lot more than the "official" figures.
The actual bronze is not a lot different, being possibly phosphor bronze or sintered bronze.

This could be why Duke were talking about being to use more rpms with their all-needle-roller scheme,

Spherical joints do not seem to be as good as cylindrical beatings, for similar sizes.
(for velocity x pressure)

I can combine a better overall bearing surfaces if I split twist, side rocking, and rotation.
Twist can be dealt with by the bottom 10mm ball, but surface pressure can be vey high there.
This is a full 45 degree swing as well.- 22.5 degrees each way.
<Edit>
If I use a more realistic force of 6.5kN,
and a diameter of 20mm,
and a diameter factor of 0.83 for curvature,

then pressure becomes 30 MPa,
and peak velocity becomes 9.42 m/s.
PV = 282

compare the alternate cylindrical bearing of 15x10, P = 6500/150 x 10^-6 = 43.3 MPa, and V is :
effectve rpm = 1200
v= 754x5x 10^-3 = 3.8 m/s
PV = 165
282/165 = 1.7
We are getting into the same ballpark. PV is half the previous value.

The 10x15 bearing would be good for a 45mm bore, or 1590 x 10^-6, P = 7.0MPa, F= 11.13 kN
This increases load by a huge factor.

piston 30mm = 707 x 10^-6, at 7.0 mpa, = 4.95 kN, x 2.25
so P (bearing) can be 2.25x 43.3 or 97, and PV becomes 370
I will draw up a 20mm ball option.
however, a 20mm ball is quite heave- at 8000 den, 4/3 pi r cu - same as the I value calc??
vol = 3.14 x 10^-6 cu mt, mass = 25.12 x 10^-3 kg, or 25 g. This doubles the outward pull from my last example.
(due to centrifugal effects)

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Last edited:
Continuation:
This question is now no longer required, as the next revision is back to cylindrical pins and rods.

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