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Sine bar with gage pin to measure part features

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  • Sine bar with gage pin to measure part features

    I just finished a homework assignment on the use of the sine bar, and I found one concept rather confusing. The following picture shows a part on a sine bar with a gage pin, and it says that 1/2 the diameter of the pin must be subtracted from the length of the part, which is actually 6.000" with a 0.200" pin:



    I think I understand the concept, but the drawing is confusing because it appears that the actual length of the part has changed to 5.900". I think the proper way to understand this is that the height of the center of the gage pin is 1.7148"-0.1000" or 1.6148", and the part is moved that same distance from the center along its angle. Thus, the height of the angled flat of the part from the lower left corner would be 6*sin(34) + 1*cos(34), or 4.1842". The gage pin pushes the block up by 0.2*sin(34) or 0.1118". The location of the lower left corner of the piece will be at the center of the gage pin + 0.1*sin(34) - 0.1*cos(34), or 1.6148 + 0.0559 - 0.0829 or 1.5878, so the height of the surface from the surface plate should be 5.7720"-0.1118" = 5.6602, which is the answer.

    This seems extremely complicated, and perhaps the way the lesson explained how to get the answer is a little less so. Here is the lesson:
    http://enginuitysystems.com/files/CA...Bar_800225.pdf

    I could not find anything other than the ToolingU reference describing this process. I'm not really comfortable with using the surface plate and height gages and gage blocks anyway, and this sine bar makes sense for angles but not so much for this procedure. Does anyone here actually do this? Or is there an easier way?
    http://pauleschoen.com/pix/PM08_P76_P54.png

    Paul: www.peschoen.com
    P S Technology, Inc. www.pstech-inc.com
    and Muttley www.muttleydog.com

  • #2
    From only what is on the pix, I don't see any thing the 1.000000000 dimension even has any bearing.
    Just what is suppose to be measuring for ???? There is a LOT of information missing here.
    ...lew...

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    • #3
      What are you trying to measure?
      Is this 'Common Core' math?

      Comment


      • #4
        The drawing shown above is utter nonsense. Later drawings (in the coursework linked to) clearly and correctly show the pin is removed before placing the work on the sine bar. The 1" dimension is also incorrectly positioned, and using the same line thicknesses for part and dimension lines is confusing to the reader, and apparently also to the author.

        The calculations are being done using the 'virtual' pin centre as a reference - this is where the 5.9" comes from - it is the distance from the edge of the part to the centre of where the pin was. Similarly the physical 1" dimension is reduced to 0.9" in the calculation. It's a useful shortcut for a simple part as shown.

        Personally, for a more complex part, I would calculate the height of the sine bar inside corner and refer everything to that.

        Cheers

        .

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        • #5
          When using a sine bar like that which has two pins of the exact same size (all do from what I've seen), you DO NOT have to use the diameter of the pin in any way for calculating the height of the gauge blocks to get the "side opposite" dimension. That is because both pins are sitting on the same surface at 0؛. The side opposite dimension is the height required to get the angle correct, and you can put the part anywhere along the surface of the hypotenuse and it's the same angle.

          A gauge pin put below it has NO effect on the angle, it only affects the distance/location of the surface you are to machine relative to the sine bar. In the illustration, the part is to have a 1.0000 dimension form the base of the part to the point where the 34؛ angle starts. With that tolerance, it's impossible to measure without a high quality optical comparator or (better), a coordinate measuring machine. So without those tools a reference point is needed and that's where the gauge pin comes in. Assuming the 90؛ corner of the part sitting against the pin is actually a true 90؛. it will hit the tangent point of the pin, which is equidistant from the center of the pin regardless of the angle it contacts. This is why they are calculating with a correction for the diameter of the pin, to get the triangles needed to complete the trigonometry.

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          • #6
            It's a std 6", after the addition of the pin the 6" becomes 5.900", the drawing is a bit odd in that respect, but it illustrates the principle, a 0.200" pin is common but I preferred to use a bigger pin as it's easier to clock off, I found the indicator difficult to zero myself, I think that it's a standard set up exercise, later there's the compound sine table, that gave me a headache!
            Cosine error!
            Mark

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            • #7
              It's just sad that any value of the excersize is lost in pathetic presentation.

              If it's complicated it's of no use.

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              • #8
                A simple explanation that the exercise was to locate the internal corner of the sine bar support would have made this whole thread un-necessary . :-)
                ...lew...

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                • #9
                  If they are just setting the part to the correct angle then why do they need the pin and those dimensions other then the 34 degree one. You set the sine bar to the correct angle with the gauge blocks and the part is then at the correct angle. If I'm wrong then would someone "esplain" please.
                  The shortest distance between two points is a circle of infinite diameter.

                  Bluewater Model Engineering Society at https://sites.google.com/site/bluewatermes/

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                  • #10
                    This looks like a common scholarly but non-practical method of demonstrating several procedures in one lesson, none of which are well described. What is missing and perhaps is included in the lecture (is there a lecture??) is the why of it all. I agree with the OP that the lesson is unclear though I did follow it, finally.

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                    • #11
                      Normally with an inclined plane and a linear dimension a tooling ball is handy too
                      Mark

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                      • #12
                        Here is the measurement that they are TRYING to explain.



                        They are trying to measure and verify the 1.0000" distance that is marked B - C. As you know, that is difficult and if the four place dimension is to be believed, measuring it to tenths is a real challenge. So, they are going to use the sine bar to measure the distance I have shown as "d" which is a line from corner A perpendicular to the angled line C - K. From that dimension and some trig you can calculate the one inch dimension. Yes, that really is what their convoluted procedure is designed to do. Their explanation sucks.

                        Someone asked for a better way. It is not possible to put a caliper or micrometer on point C so a direct measurement is not practical (note below). You could flip the part around so that line C - K rests on the surface plate and try to take an height measurement to corner A, but that is also problematic, especially if you want tenths accuracy. Any burrs would introduce large and unknown errors. Even if you had ground all burrs off and had a completely sharp corner, just the act of lowering the height gauge on it would certainly change it by a few tenths, again wrecking the measurement.

                        I would suggest making a pin like this, with the two flats meeting at the center of the pin:



                        With the part resting on line C - K and the gauge pin on corner A, you can easily measure the distance to the top of the gauge pin. Then subtract the radius of that gauge pin and you have distance d. If the gauge pin is accurately made, you can use this method with an accuracy of tenths. The math involves only one quick subtraction so it is far less prone to error. If you have a number of parts to check, it will go a lot faster.

                        Note: OK, I know some of you are going to say you CAN mike the distance to corner C. Well, with some skill and a bit of technique you certainly can. I have done this for parts with loose tolerances. +/-0.005" is certainly possible. 0.001", well perhaps with both skill, good technique, and luck. But tenths as the stated dimension seems to imply? Not really. So please don't burn me on this point. PLEASE.
                        Last edited by Paul Alciatore; 11-19-2014, 06:45 PM.
                        Paul A.

                        Make it fit.
                        You can't win and there is a penalty for trying!

                        Comment


                        • #13
                          Well, they are using the pin to establish the height above the surface plate of a reference point that is 0.1000" from each of the corner surfaces of the sine bar. They then REMOVE that pin and measure from the surface plate to the angled surface. The center height of the pin is subtracted from that measurement.

                          The complicated trig is to determine how that measured/subtracted dimension relates to the 1.0000" distance B - C that they are actually trying to verify.


                          Originally posted by Lew Hartswick View Post
                          A simple explanation that the exercise was to locate the internal corner of the sine bar support would have made this whole thread un-necessary . :-)
                          ...lew...
                          Paul A.

                          Make it fit.
                          You can't win and there is a penalty for trying!

                          Comment


                          • #14
                            Sounds like the old saying: Those that can do, those that can't teach. I have found that people who really know what they are doing can explain things in a simple manner. When they don't really know then the explanation gets really complicated. At my last job there was another engineer who explained the mass balance in a biological treatment system and it was so clear that I was sure he knew exactly what he was talking about. Ironically the boss fought with him all the time, even as the guy kept him from making huge expensive mistakes. We are both gone now as you can only take so much of that.

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                            • #15
                              How about first verifying the angle using the sine bar in the usual way, and if that is within tolerance (or can be made to be so), then rest the part with the angled feature flat on the surface plate. Then all you need to do is measure the height of the corner A, or dimension d, which should be 1.0000*cos(34) + 6.0000*sin(34) = 4.1842.

                              My mind is a bit clearer now and that seems like a good solution that is much quicker and easier. Otherwise, there should probably be a way to figure the position (height) of the sine bar corner for any angle, and perhaps it should have a small relief at the intersection of the two surfaces so that a square shape will fit completely onto the surfaces.

                              This is one example of some of the unnecessary and confusing information in the homework assignments. These "classes" are provided by www.toolingu.com and consist of 10-20 "lessons" with accompanying voice reading of the text, and occasional "interactive exercises" with sample problems. After the lessons have been completed there is a final exam of 10-20 questions. In this case there were 11 questions and I missed one about the gage pin and another because of a stupid sin/cos error. So I got only 81.8 on this exam. I could take it over since 1 day has passed and I'd ace it, but only 70% is required to "pass", so I could have missed 4 and still be OK. There are typically enough "stupid simple" questions to guarantee a passing grade unless one is clueless.
                              http://pauleschoen.com/pix/PM08_P76_P54.png

                              Paul: www.peschoen.com
                              P S Technology, Inc. www.pstech-inc.com
                              and Muttley www.muttleydog.com

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