connecting rods weighing question

[Kevin Johnson]

any thoughts

https://theultimatetooling.com/en/produ ... -balancer/

I wrote a brief email to Rossi & Kramer asking them if they wanted to provide more information on their well-crafted machine. It is not specifically mentioned in their products on their webpage.

http://www.pegrk.com

Buongiorno,

Stiamo predisponendo il video che potrete vedere su YouTube; appena pronto Vi informeremo.

Cordiali saluti.

Antonio Daminato

"We are preparing the video that you can see on YouTube; as soon as [it is] ready we will inform you."

[Kevin Johnson]

any thoughts

https://theultimatetooling.com/en/produ ... -balancer/

I wrote a brief email to Rossi & Kramer asking them if they wanted to provide more information on their well-crafted machine. It is not specifically mentioned in their products on their webpage.

http://www.pegrk.com

Buongiorno,

Stiamo predisponendo il video che potrete vedere su YouTube; appena pronto Vi informeremo.

Cordiali saluti.

Antonio Daminato

"We are preparing the video that you can see on YouTube; as soon as [it is] ready we will inform you."

[Dave Koehler]

It would appear that it can only compare total rod weight which makes it a coffee table pretty.

[Kevin Johnson]

It would appear that it can only compare total rod weight which makes it a coffee table pretty.

No, it compares the center of mass, which is different. Two rods can have different total rod weights and the same relative center of mass.


A simple digital scale will compare the total rod weight.

[Dave Koehler]

If it can't do big end and small end then it is useless for shop use. Yes, I understand the center of mass idea. It's just not feasible for day to day use.

[Kevin Johnson]

If it can't do big end and small end then it is useless for shop use. Yes, I understand the center of mass idea. It's just not feasible for day to day use.

I think that you are correct that it is not for everyday production shop use as it assumes previous steps.

Modern sinter-forged rods are much more uniform in morphology and relative positioning of machining.

The original Porsche 928 sinter-forged rods were very accurate/reproducible in overall surface contour (morphology). The bore to bore distance from the big end to the small end was also very accurate. What was not accurate was the relationship between the two -- the pair of borings was randomly translated in the plane. As the amount of material machined away in each boring was very nearly the same, the weight or force due to the gravitational acceleration of that mass would also be nearly the same. The center of mass definitely would not be.

Each rod could be scanned and an analysis performed with very specific machining in 3-space specified with simultaneous attention paid to stress patterns. The cost of the time involved would far exceed the cost of a new bespoke rod created with finer manufacturing control.

I looked up the Shadograph patents previously (they are about seven or eight decades past) but pulled the research from the thread because it was distracting from considering the specialized use of particular manual tools.

[David Redszus]

It would appear that it can only compare total rod weight which makes it a coffee table pretty.

No, it compares the center of mass, which is different. Two rods can have different total rod weights and the same relative center of mass.


A simple digital scale will compare the total rod weight.

A simple knife edge balance will allow the determination of center of gravity. Combined with a scaled total rod weight, the exact weight of both big end and small ends of the rod can be determined.

[Kevin Johnson]

...

A simple knife edge balance will allow the determination of center of gravity. Combined with a scaled total rod weight, the exact weight of both big end and small ends of the rod can be determined.

I think with the device that the rod can be flipped and indicate center of mass (gravity) that is translated in the plane normal to the bores and coincident with the longitudinal central axis of the rod. The difficulty is that the operator would need to carefully hold a mental model. The mental model is analogous to the model obtained by scanning. People think about things differently. Orientation on the Z-axis would still be missing, though. The values would be projections on that coincident plane.

[Nut124]

A simple knife edge balance will allow the determination of center of gravity. Combined with a scaled total rod weight, the exact weight of both big end and small ends of the rod can be determined.

To me this does not sound right w/o making some far reaching assumptions about the distribution of mass in the rod. I think I can see a scenario where two rods have the same total wt and center of gravity yet have drastically different end weights.

[modok]

I don't think that is possible


two possible problems with the fixture would be....
-it assumes the rods are all the same length.
-locates off the side of the rod, not the bore.

[digger]

A simple knife edge balance will allow the determination of center of gravity. Combined with a scaled total rod weight, the exact weight of both big end and small ends of the rod can be determined.

To me this does not sound right w/o making some far reaching assumptions about the distribution of mass in the rod. I think I can see a scenario where two rods have the same total wt and center of gravity yet have drastically different end weights.

If the total mass is the same and the end weights are the same the c.of.g is the same. It’s a statically determinate problem

[Dave Koehler]

A simple knife edge balance will allow the determination of center of gravity. Combined with a scaled total rod weight, the exact weight of both big end and small ends of the rod can be determined.

To me this does not sound right w/o making some far reaching assumptions about the distribution of mass in the rod. I think I can see a scenario where two rods have the same total wt and center of gravity yet have drastically different end weights.

If the total mass is the same and the end weights are the same the c.of.g is the same. It’s a statically determinate problem

Isn't that assuming they are reasonably close to begin with? There might be a better chance of that today but oem rods of the past could be wildly different. The small end could be so heavy in a set that it affects the big end once corrected. In those cases where the small end is ridiculously overweight I rough them end within 1.5-2. grams. Then I do the big end. Come back to the small end for a final trim.

IF I dive in and correct one end or the other to final and move to the other end I usually get bit. The other end can become too heavy. Kind of chasing your tail kind of thing. On those old rods I find it easiest to first do a quick check on all of them, look for surprises and form a plan of attack.
Then there is the really old antiques, tractors and diesels. The same issue is magnified significantly.

[Nut124]

If the total mass is the same and the end weights are the same the c.of.g is the same. It’s a statically determinate problem

I did some math and you are correct. It is a static problem. It however does not take into account how the mass is distributed in the rod as long as the CG and total mass are the same. One could have a rod where the ends are heavier and the beam is lighter and it would measure the same as a rod with a heavy beam and lighter ends.

Is there a way to measure for this mass distribution? Or does it matter? Inertia test?

[Kevin Johnson]

A simple knife edge balance will allow the determination of center of gravity. Combined with a scaled total rod weight, the exact weight of both big end and small ends of the rod can be determined.

To me this does not sound right w/o making some far reaching assumptions about the distribution of mass in the rod. I think I can see a scenario where two rods have the same total wt and center of gravity yet have drastically different end weights.

If the total mass is the same and the end weights are the same the c.of.g is the same. It’s a statically determinate problem

The GKN rods (first large scale production engine use of sintered rods) could have two rods with coincident small end borings relative to rod morphology but with one boring pair rotated around the centre of the small end. This would shift the positioning of the big end bore through an arc. This would yield two rods with the same big and small end weights but varying centre of gravity (due to the centre of mass following the same radian sweep but along a shorter arc.

Aside:
The general problem led to the need for matching serial numbers on the cap and beam. However, the 1000 possible numbers (000 through 999) were still not sufficient to avoid mismatched caps and beams -- likely due to dumping parts in a common bin. I say this because a customer took Hi-Res pictures of a well worn original rotating assembly with matching serial numbers visible but mismatched cap and beam. The positioning of the rod was by happenstance on the 2/6 rod journal. The slight misalignment was not caught during the hand assembly in the Porsche factory nor when the un-resized rod was reassembled with new bearings during the rebuild. The misaligned edge of the bearing shell sheared the oil film instead of allowing the hydrodynamic wedge to propagate. If the engine had been gently driven in and the bearing edge worn (instead of being raced hard almost immediately) -- as presumably it was when the car was new -- this would not have been noticed.

Anyways, small things can make a difference.

[David Redszus]

If the total mass is the same and the end weights are the same the c.of.g is the same. It’s a statically determinate problem

I did some math and you are correct. It is a static problem. It however does not take into account how the mass is distributed in the rod as long as the CG and total mass are the same. One could have a rod where the ends are heavier and the beam is lighter and it would measure the same as a rod with a heavy beam and lighter ends.

Is there a way to measure for this mass distribution? Or does it matter? Inertia test?

The distribution of mass along the rod center line can be determined by use of CAD drawings.
But, what would that tell you and how would you make use of that info?

If two rods weigh the same but have different centers of gravity, they are not the same.
Similarly, two rod with identical centers but different weights, are not the same.

The issue is to find a simple quick way to match rod total weights with identical gravity centers.

Small end mass x distance = large end mass x distance

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