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  1. #1
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    Fit tolerance calculation

    I'm designing two cylindrical parts that fit together tightly with hand force. My understanding is this would be a Locational Clearance fit. The nominal diameter is 94mm, and I am finding that a tolerance of ±0.35mm is what should be expected for this diameter. If the smaller cylinder has an OD tolerance of 94mm +0.0/-0.35, then my calculation tells me the ID of the larger cylinder needs to be 94mm +0.35/-0.0, resulting in a fit range of +0.70 to 0.0 so that the lower end of the range isn't a negative value which would enter the light press fit category. But if the end result is near the upper limit of 0.7mm, wouldn't that be a loose fit?

    I'm hoping somebody can give me some clarity on this... thanks..

  2. #2
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    Re: Fit tolerance calculation

    hy begin with you id diameter, because id's are more demanding then od's ( chip evacuation and cilindricity is harder to achieve when cuttin id )

    then start crafting the od, don't remove it from spindle, and use the id to check; if it does not fit, then cut a bit more the od

    kindly
    Ladyhawke - My Delirium, https://www.youtube.com/watch?v=X_bFO1SNRZg

  3. #3
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    Re: Fit tolerance calculation

    If both parts are specced as having an extreme tolerance that is the same,they wont fit together-94.00mm will not go into 94.00mm as I'm sure you know and 0.7mm isn't a particularly tight fit.Its your design and its your choice of tolerance you are also in possession of all the other relevant information such as whether both parts are made of the same material,if not is differential thermal expansion likely to lead to problems?Will you be doing the machining yourself or asking a machine shop to make it for you?

  4. #4
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    Re: Fit tolerance calculation

    I am having a shop make the parts for me. The inner part is machined and the outer part is spun. I already know the inner part has an OD of 94+0.0/-0.35 so I am trying to figure the tolerances to indicated for the spinning mold that will determine the ID of the outer part. You had mentioned 94mm will not go into 94mm. Do you mean at the lower end of the fit tolerance range where the clearance/interference is 0.00? I was under the impression that 0.00 would be a locational clearance fit, meaning tight but can be assembled by hand.

    I simply might have to accept the 0.00 to 0.70 range and hope the fit generally ends up on the tighter end of the spectrum. But when parts are commercially produced, the companies obviously don't take this approach. My big question is how can a specific fit such as locational clearance be assured when the clearance/interference tolerance range spans multiple fit ranges?

  5. #5
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    Re: Fit tolerance calculation

    If the outer part is spun you have very little chance of holding a high precision tolerance without a final honing operation.If I really need to point out that 94.00 mm will not fit into 94.00mm then you really ought to spend a bit of time working with precise assemblies.Even with a perfect mirror finish it won't be happening.If you specify a 0.01mm gap and can accurately measure the components to ensure that such a gap exists,you have a chance.Unless a cold draught blows through the workshop door and shrinks the outer component a bit.Its entirely your decision what sizes and tolerances to apply to your designs and the greater the desired level of accuracy-the higher the price tends to be.

  6. #6
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    Re: Fit tolerance calculation

    hy birdtrick let's assure a play of :
    ... 0.7±0.02 : dimensions for the parts may be id±0.01, and od±0.01 ( id & od : 93.65 & 94.35, or 93 & 93.7, or 94.2 & 94.9; they are all around 94 )
    ... 0.02+0.04 : dimensions for the parts may be 94-0.005, and 94+0.015+0.035
    ... 0.35±0.36 : dimensions for the parts may be 94-0.01 and 94+0.7

    I simply might have to accept the 0.00 to 0.70 range and hope the fit generally ends up on the tighter end of the spectrum.
    there is no ' hope ' when it comes to tolerances & plays, it's all about the technical drawing

    the basic brick is the technical drawing; once mastered, it does not matter what language you speak

    is not only about shape, dimensions and conditions : all those speak togheter, saying how the part has to be crafted / kindly

    ps : ' hope ' appears beyond microns; before them, is ' accurcy '; precision grinders are on a thin red line inprocess grinding gaugess may shift that line 1 more digit, then is blurry
    Ladyhawke - My Delirium, https://www.youtube.com/watch?v=X_bFO1SNRZg

  7. #7
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    Re: Fit tolerance calculation

    I started trying to figure this out by looking at the ANSI fit classes to determine which I was looking for. It seemed that Locational Clearance (somewhere LC1-LC3) is what I would want since it's described as being the last fit that is still possible to assemble and disassemble by hand before the next class, being Locational Transition. In ISO terms the LC fit I want is H7/h6 for both hole and shaft basis. My next step was to look at the numbers given for fit range in this class, and the bottom end was always 0.0. The numbers first move into the negative range in the Locational Transition class. So my thinking was to indicate a range that would calculate to a 0.0 fit at the bottom end. But am I correct in understanding from what you say that in the real world a 0.0 fit cannot be assembled by hand?

    What I have realized through the process however, is that these tolerance ranges are for the final parts produced in quantity. For the spinning mold I don't believe it would be unrealistic to call out a very fine tolerance of say 94mm +0.1/-0.0 or even tighter than that perhaps.

    How far off track am I on all of this? I am very appreciative of the input and am getting a much better understanding from it.

  8. #8
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    Re: Fit tolerance calculation

    H7/h7 is pretty common ... spare parts are crafted towards the limits, so to ensure maximum play, in order to avoid the case when they won't fit ( when dimensions are toward 0 )

    But am I correct in understanding from what you say that in the real world a 0.0 fit cannot be assembled by hand?
    i believe you need to 'feel' that, in order to 'understand' : try to get your hands on a shaft o20±0.01, and 5 sleeves : o20-0.02-0.01, o20-0.01, o20+0.01, o20+0.01+0.02 and o20+0.02+0.03, then try to assamble them

    What I have realized through the process however, is that these tolerance ranges are for the final parts produced in quantity.


    tolerances are applied to all types of jobs :
    ... uniques & prototypes
    ... 0 zeries
    ... small - high series, and mass production

    is not needed to choose a tolerance class from the ansi/iso table; those are for your reference; don't choose, but decide
    Ladyhawke - My Delirium, https://www.youtube.com/watch?v=X_bFO1SNRZg

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