Jean in Newcastle Posted October 9, 2014 Share Posted October 9, 2014 He's just started Derek Owens' Physics. He's a 2e aspie and this is a huge factor here so recommendations for him to just figure this out etc. are not going to help me. (You can think it though!) I need to be able to translate what is being asked of him into Ds' Aspie speak. So even though it has been 35 years since I've done any physics, I need to understand this. So please be gentle and type slowly! (And yes, I know we can e-mail the instructor but I want to be able to understand the problem first before we do that.) The lab is the first one on Measurement and Graphing. Ds measured several cylindrical objects with a string and a ruler. He used inches. Or more technically, he measured to the nearest quarter inch. He measured the diameter and the circumference. Question 1: His diameters do not totally match the circumference. For example he measured 13 3/4 in. circumference and 4 1/4 diameter. Is it ok that he uses fractions or should he write 13.75 in. and 4.25 in.? Question 2: If you use the formula for finding a diameter from a circumference, it doesn't come out to 4.25 but 4.37. is that ok since his ruler only went to the nearest 1/4 inch? This is just for starters. Further questions depend on your answers to the above two questions! Quote Link to comment Share on other sites More sharing options...
Twigs Posted October 9, 2014 Share Posted October 9, 2014 I can't help you with how close your answers need to be. My 1st suggestion would be to get a ruler that can measure 1/16 of an inch. Or to use metric with mm. Is a specific type of ruler required (to the nearest 1/4 inch)? Only using a ruler to the nearest 1/4 inch introduces a lot of error. Best wishes. Quote Link to comment Share on other sites More sharing options...
Jean in Newcastle Posted October 9, 2014 Author Share Posted October 9, 2014 No, 1/4 isn't required. I think the 1/4 thing was imprecision on his part and I'm going to correct it. Quote Link to comment Share on other sites More sharing options...
Jean in Newcastle Posted October 9, 2014 Author Share Posted October 9, 2014 OK - We've figured out diameter and circumference. We've plotted them accurately. We've done a slope calculation. Now we're supposed to do a percent error calculation. The instructions say that there is a "correct answer for the slope, an actual value that you would get if your measurements were perfectly accurate." Is this where getting smaller and smaller measurement units comes in? The instructions say "if you know what the value would be, continue to step 8 and calculate your percent .error." If you don't know, you're supposed to ask the instructor. I just spent over a half hour with dh (from 11:30 to midnight!) and we can't figure out what the actual value would be esp. since there were four things plotted and it seems like there should be 4 actual values? Any insight? I'm going to bed but I'll check here in the morning. Quote Link to comment Share on other sites More sharing options...
Arcadia Posted October 9, 2014 Share Posted October 9, 2014 Circumference = (pi)D Gradient = pi If you measure in mm, the best fit line might give a gradient close to pi. If you have IKEA nearby, grab some of their paper rulers to use. Assuming your D is accurate, then calculate circumference for every D Then find the difference between practical circumference vs theoretical circumference. After that divide by theoretical circumference to get percentage error. Example of calculating percent error http://astro.physics.uiowa.edu/ITU/glossary/percent-error-formula/ ETA: To be honest, I did more percent error calculations for statistics (integrated math) than for the sciences in school. Quote Link to comment Share on other sites More sharing options...
Jean in Newcastle Posted October 9, 2014 Author Share Posted October 9, 2014 thank you. Quote Link to comment Share on other sites More sharing options...
Posted October 9, 2014 Share Posted October 9, 2014 No, 1/4 isn't required. I think the 1/4 thing was imprecision on his part and I'm going to correct it. I haven't see this text, so I'm making a bunch of guesses here. Assuming that the lesson here is about measurement, error and significant digits, it might be more useful to have him start with the 1/4inch ruler. Have him measure the cylinder with it, and he should notice that whatever he is measuring isn't exact to the quarter inch. So, he'll have to make an estimate, and know that this measurement isn't 100% accurate. But that's ok. No measurement in science ever is. He might say, "we should switch to the millimeter ruler", and yes, that might be more precise, but still not 100% accurate -- there's some amount of error in every measurement. What is important is that he have a feel for how precise the measurement might be. You could ask him "You made an estimate for this measurement, but is there a lower and upper bound, that you are very confident it is between?" So, if he measured a diameter at 5 1/4 inches, he might say, "Yes, I'm sure it is between 5 and 5 1/2 inches". Then you could do the same thing with the circumference. You should then point out that the calculated value for the circumference (pi * best estimate of diameter) probably doesn't exactly match the best estimate of the measurement of the circumference. But that's ok, because you knew the estimates weren't perfect. The measured value for circumference, however, should be between the calculated value of circumference for the lower bound of diameter and the upper bound. My guess is that the point of this lesson isn't to empirically derive PI, but to teach about measurement and error. Quote Link to comment Share on other sites More sharing options...
Jean in Newcastle Posted October 9, 2014 Author Share Posted October 9, 2014 I agree about the point of the lesson. But there is a formula for a percent error calculation and I'm trying to figure out what numbers to plug into it! Quote Link to comment Share on other sites More sharing options...
Mukmuk Posted October 9, 2014 Share Posted October 9, 2014 What Arcadia wrote is accurate. Gradient is pi. % error calculation: 1. Take two points from the graph you've drawn. Calculate slope. (Circumference 2 - Circumference 1) / (Diameter 2 - Diameter 1). This should be in the blanks just below the graph. 2. % error = ( | slope - pi | ) / pi. X 100% I'm copying this from ds' worksheet. Hope it works. Eta: ds goofed on this one in that he reversed the axis (vision issues). His answer is actually inverted. The formula above should work though. Quote Link to comment Share on other sites More sharing options...
Jean in Newcastle Posted October 9, 2014 Author Share Posted October 9, 2014 Thank you. I don't know why it took 2 hours to wrestle him into doing what I had already done, (though I made him do his own) but it did. His final argument was "But why do I have to do it your way?" (His way was super complicated and ignored all formulas and what Derek Owens told him to do.). My answer was a shouted "Because Arcadia and MukMuk told you to!" Dh was almost rolling on the floor laughing. :glare: :lol: The joys of doing physics with a very bright but very stubborn Aspie. . . . Again, thank you all very much. Quote Link to comment Share on other sites More sharing options...
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