Linear stack with inclined surfaces and profile tolerances

[CTengIS]

Hi,
I'm trying to brush up on stack-up calculations.
I prefer to make up my own exercises that are closer to real cases than textbook examples. So I came up with this v-block and shaft idea (attached pdf) with the "v" surfaces having a profile tolerance specified.

To make it interesting the profile value is different on each side of the "v".
I represented the values as "n" and "m".

The stack-up is for dimension "S" so it's only for the vertical direction. It's a 70° v-block so each profile zone is at 35° to the stack direction.

I'm having a hard time figuring out how exactly those "n" and "m" values are resolved in the vertical direction. It's easy enough to obtain numbers when experimenting with the CAD model, which I did, but I'm trying to figure out the math.

Since I experimented with the model, I can tell you that the resulting vertical variation from these profiles is NOT 0.5m/sin35°+0.5n/sin35° and neither 0.5m•sin35+0.5n•sin35.

Interestingly, it does closely approximate (0.5m/sin35°+0.5n/sin35°)/2, but I'm not sure why.
The values I tested are 0.1 for n and 0.2 for m and I got a variation of +/-0.131 as an impact of the 2 profiles on the height of the shaft.

Can you provide insight on this?
Is there some known method to take care of such tolerances as part of a linear (1-D) stack-up analysis for manual calculation?

 

[3DDave]

Why do you think that 0.5*m/sin(35 degrees) is incorrect? Sine = Opposite/Hypotenuse and the Opposite side is 0.5*m (or 0.5*n, for the other side.)

Rearrange and, knowing the half-angle is 35 degrees, Hypotenuse = Opposite/sin(35) with the Hypotenuse being the vertical translation of the surface to correspond to the offset in the normal direction.

That said, they aren't cumulative. For example if one side is negative exactly the same as the other is positive, the cylinder moves to the side. If the change is not equal then some vertical and some horizontal displacement will take place.

I'd have to spend some time with a pencil and paper to calculate the parametric equations to located the vertical position of the cylinder.


Reformated for readability; bold to indicate a requested goal

I'm trying to brush up on stack-up calculations.

I prefer to make up my own exercises that are closer to real cases than textbook examples. So I came up with this v-block and shaft idea (attached pdf) with the "v" surfaces having a profile tolerance specified.

To make it interesting the profile value is different on each side of the "v".

I represented the values as "n" and "m".

The stack-up is for dimension "S" so it's only for the vertical direction. It's a 70° v-block so each profile zone is at 35° to the stack direction.

I'm having a hard time figuring out how exactly those "n" and "m" values are resolved in the vertical direction. It's easy enough to obtain numbers when experimenting with the CAD model, which I did, but I'm trying to figure out the math.

Since I experimented with the model, I can tell you that the resulting vertical variation from these profiles is NOT 0.5m/sin35°+0.5n/sin35° and neither 0.5m•sin35+0.5n•sin35.

Interestingly, it does closely approximate (0.5m/sin35°+0.5n/sin35°)/2, but I'm not sure why.

The values I tested are 0.1 for n and 0.2 for m and I got a variation of +/-0.131 as an impact of the 2 profiles on the height of the shaft.

Can you provide insight on this?
Is there some known method to take care of such tolerances as part of a linear (1-D) stack-up analysis for manual calculation?

 

[3DDave]

Word description:

The nominal V has a point where both sides share an intercept. Take this as (0,0)

Adding the profile allowance shifts the Y-intercept of each side up or down by an amount as calculated above.

The circle lies on the intersection of two lines with offsets from the shifted profile lines equal to the radius of the circle as calculated above.

See

https://en.wikipedia.org/wiki/Line–line_intersection#Given_two_line_equations

 

[CTengIS]

Thank you 3DDave.
Sorry for making a long-winded post, I broke it down to paragraphs similar to your suggestion.

It's not that I think that 0.5*m/sin(35 degrees) is incorrect, I meant that this expression (and 0.5*n/sin35° for the other side) are not the direct answers to the contribution of each profile tolerance to the stack in the vertical direction. As you noted they don't add up to the resultant displacement.

Thank you also for providing the link and the insight about line intersections.

Adding the profile allowance shifts the Y-intercept of each side up or down by an amount as calculated above.

The circle lies on the intersection of two lines with offsets from the shifted profile lines equal to the radius of the circle as calculated above.

Could you please clarify where is "above"? Do you mean in the wikipedia article? I'm not sure which calculation to look at.

 

[3DDave]

Rearrange and, knowing the half-angle is 35 degrees, Hypotenuse = Opposite/sin(35) with the Hypotenuse being the vertical translation of the surface to correspond to the offset in the normal direction.

 

[pmarc]

CTengIS,

I would suggest starting with a simpler case where both profile tolerance values on the v are equal and equal to the larger of the original two (say 0.2), and assume there is no other variation in the stack-up (i.e., the other contributors have zero tolerance).

If you do that, you should notice that the extreme vertical movement of the ball from nominal is 0.5×0.2/sin35 and not two times that.

The result you got through testing actually supports the observation. The (0.5m/sin35°+0.5n/sin35°)/2 after assuming m and n are equal and equal to 0.2, will become: (1×0.2/sin35°)/2 = 0.5×0.2/sin35°.

Long story short, I think in the stack where the two profile values are different (and 0.2 is the greater of the two), the final effect of the two profile contributors should not be greater than 0.5×0.2/sin35°.

 

[CTengIS]

Thank you guys.
I haven't yet wrapped my head around the math that leads to the results of how much the cylindrical shaft translates vertically under the influence of the profile tolerances of the inclined supports, but I will get it eventually.

pmarc, you mentioned a "ball" in your last post, which makes me think - what if the problem changed to include a ball instead of the shaft? the ball could be fully located by 3 planar surfaces (instead of 2) on the block, which would contact the ball at one point by each surface, 3 points total. The 3 surfaces could be designed at any angle to each other and to datums A, B, and C. The whole set of 3 supports could be at some crazy orientation so that neither of the surfaces nor any line elements on them are parallel or perpendicular to the vertical direction of stack up, the right-left direction, or the normal to the view plane direction. Could it then still be solved by similar math/trigonometry as now, and would analyzing the tolerance stack up in the vertical direction still be considered a 1-D analysis, or would it become a 3-D stack problem because of the consideration of how variation in 3 axes impacts the translation at the vertical direction?