Function 83 Posted November 21, 2013 (edited) Hello everyone I have a question about something I really can't stand in physics: approximates (the rules which say that if you have a sum or a difference, the result may not be more accurate than the less accurate term; if you have a product or a quotient (ratio?), the number of significant figures may not be higher than the number of significant figures of the factor with the less significant figures) (I can't stand these approximates for their inaccuracy ) To exercise this, I've used a self-made example: A bullet is being shot at a height of [math]276.457 m[/math], under an angle [math]\theta[/math] of [math]0.524 rad[/math] and with an initial velocity of [math]87.78\frac{m}{s}[/math]. How much has the bullet dropped after it has flown [math]74.5m[/math] in a one-dimensional way? Using 'the formula' of the trajectory of a projectile: [math]y=y_0+x\cdot \tan{\theta}-\frac{gx^2}{2v^2\cos^2{\theta}}[/math] [math]y = 276.457m + 74.5m\cdot 0.578 - \frac{9,81\frac{m}{s^2}\cdot 74.5\cdot 74.5}{2\cdot 87.78\cdot 87.78 \cdot 0.866\cdot 0.866}[/math] Lowest accuracy is 3 significant figures in numerator and 1 significant figure in denominator [math]=276.457 + 43.1 - \frac{5.44\cdot 10^4}{1\cdot 10^4}[/math] Lowest accuracy is 1 significant figure [math]= 276.457 + 43.1 - 5[/math] Lowest accuracy is one unit [math]= 3\cdot 10^2 m[/math] Is this correct? Thanks. -Function Edited November 21, 2013 by Function 0 Share this post Link to post Share on other sites

swansont 6980 Posted November 22, 2013 [math]y = 276.457m + 74.5m\cdot 0.578 - \frac{9,81\frac{m}{s^2}\cdot 74.5\cdot 74.5}{2\cdot 87.78\cdot 87.78 \cdot 0.866\cdot 0.866}[/math] Lowest accuracy is 3 significant figures in numerator and 1 significant figure in denominator No, there is no term in the denominator with only 1 significant figure. 2 is an exact number, not a measured variable. It essentially has an infinite number of significant digits — the number is exactly 2. IOW, it's 2.00000000000… 0 Share this post Link to post Share on other sites

Sensei 893 Posted November 22, 2013 (edited) I have a question about something I really can't stand in physics: approximates Experimental data will be even worser than calculated. You have no idea (and can't have) what is f.e. wind speed and its direction (it's variable, and not even constant while single bullet flight).. The longer distance (so time of flight is higher) the more influence of wind speed. Especially at the end of flight, when bullet has been already decelerated. Experiments are repeated dozen/hundred/thousand times, and at least Median of results is used (so the highest and the lowest results are ignored). Edited November 22, 2013 by Sensei 0 Share this post Link to post Share on other sites

gabrelov 3 Posted November 22, 2013 ^^^^ Yes that is the best way. When doing even basic science experiment you have to do more trials and record it. Then you get the mean. You have to consider other variables also specially if you don't have all the necessary equipments and location. Computation is whar we call theory since it may not exist in real life but proven in mathematical terms. 0 Share this post Link to post Share on other sites

John Cuthber 3637 Posted November 22, 2013 The calculation of errors adn uncertainty is a lot more complicated than just counting significant figures. http://www.bipm.org/utils/common/documents/jcgm/JCGM_100_2008_E.pdf 0 Share this post Link to post Share on other sites

Function 83 Posted November 22, 2013 The calculation of errors adn uncertainty is a lot more complicated than just counting significant figures. http://www.bipm.org/utils/common/documents/jcgm/JCGM_100_2008_E.pdf Well, yes, I know... But this topic was mainly about significant figures... About uncertainty; I found out that the 'uncertain' result of a measurement is equal to the answer with no uncertainties plus or minus the most extreme uncertainties: e.g. the measurement of the surface of an A4-sheet of paper without using significant figures: I measure a width of [math]21\pm 0.1 cm[/math] and a length of [math]29.6\pm 0.1cm[/math] The surface is (if I'm correct) i.c.: [math](621.6\pm 5.07)cm^2\left(=21\cdot 29.6\pm (2.1+2.96+0.01)\right)[/math] 0 Share this post Link to post Share on other sites

John Cuthber 3637 Posted November 22, 2013 Usually, but... Imagine I'm measuring the length of something then calculating something that depends on the tenth power of that length (it's an odd idea but I'm just illustrating the point) I measure it as 10cm +/- 1 cm That's two sig fig. Now I calculate the answer as 10 ^10 cm^10 (whatever that might mean) So I know the real length was between 9 and 11 cm So the real answer is between 9^ 10 i.e. 0.349 X 10^10 and 11^ 10 i.e. 2.59 X10 ^10 But those are not even right to one sig fig So you can't rely on the output of a calculation having the same number of sf as the inputs. It's usually easier for complicated calculations to work with the relative error, rather than the number of sig fig. So I measured the length to 1 part in 10. That's a 10% error. The clever bit is that you can combine the relative errors together to get a reasonable estimate of the error on the outcome. So, to take the case of the A4 paper. 21 +/- 0.1 one part in 210 or about 0.5% error and 296 +/- 0.1 is about 0.33 % error So the error margin on the area is about 0.8% So the area is 621.6 +/1 0.8% (or about 5 cm^{2)} Imagine a square bit of paper instead. I measure the side as being 1 +/- 0.01 cm (so that's right to 1%) The side is between 0.99 and 1.01, so the area is somewhere between the square of those numbers 0.9801 and 1.0201 And a cube with that length of side would be between 0.97... and 1.03... So, I measured the side as being within 1% and the area is within 2% and the volume within 3%. For small errors you can say that the error in the calculated value is the error in the measured value multiplied by the power to which it's raised. (people who like calculus can prove this) This gives you a pretty good way of estimating the error on a calculated value without having to do much calculation. Even better- it lets you calculate how poorly you can measure each of the input values in order to get a required accuracy at the output. 1 Share this post Link to post Share on other sites

Function 83 Posted November 22, 2013 (edited) Thanks! One last thing: are trigonometric numbers (sin, cos, tan) (as in the example) considered mathematical numbers, like the 2 in message #2? Or do they have an influence on the sig digs? I'd say they have an influence, for they are based on a physical quantity (angle)? And how about the discriminant of a quadratic equation? Also here, I'd say it has an influence? Edited November 22, 2013 by Function 0 Share this post Link to post Share on other sites

timo 554 Posted November 22, 2013 That depends on how many significant digits you take into account when actually calculating something. If you calculate on a computer you can probably assume something like 15 digits. In practice, that is pretty indistinguishable from arbitrary precision (your meter scale would have to have an accuracy in the order of one proton diameter for this to matter). If you approximate pi=3 then of course you have less relevant digits. 0 Share this post Link to post Share on other sites

Function 83 Posted November 22, 2013 (edited) Well, it's very simple: I want to know if trigonometric numbers & discriminants influence the signifucant figures in exercises in which the physical quantities are given (exactly). Here's a document in which I made 2 exercises myself. In that document, I assumed that discriminant & trigonometric numbers do influence the number of significant figures. Does someone feel like looking into it? (Don't mind the Dutch title. Note: "Stel" = "let ... be ..." ) Thanks! The document: http://www.mijnbestand.nl/Bestand-7R3UUY7Q8T3P.docx Edited November 22, 2013 by Function 0 Share this post Link to post Share on other sites

John Cuthber 3637 Posted November 23, 2013 It's complicated. The trig functions have errors that are related to their first derivatives. How good is your understanding of calculus? 0 Share this post Link to post Share on other sites

Function 83 Posted November 23, 2013 Depends.. Just say what you have to say 0 Share this post Link to post Share on other sites

John Cuthber 3637 Posted November 23, 2013 These people say it better than I would http://www.lhup.edu/~dsimanek/scenario/errorman/calculus.htm http://en.wikibooks.org/wiki/General_Engineering_Introduction/Error_Analysis/Calculus_of_Error 0 Share this post Link to post Share on other sites

Function 83 Posted November 23, 2013 That's a lot of text to read What would you do with the tan and cos and the discriminant? 0 Share this post Link to post Share on other sites

John Cuthber 3637 Posted November 23, 2013 In those cases I'd try changing the values a bit and see what happens. It will depend on the value, as well as the function. The simple approach would be that, near maxima and minima of functions, small changes in the input don't affect the output. However if the graph of the function is steep at some value then the effect of a small change from that value will be large. 0 Share this post Link to post Share on other sites

Function 83 Posted November 23, 2013 (edited) In those cases I'd try changing the values a bit and see what happens. It will depend on the value, as well as the function. The simple approach would be that, near maxima and minima of functions, small changes in the input don't affect the output. However if the graph of the function is steep at some value then the effect of a small change from that value will be large. Suppose the latter is true; what do you do? Something that has the lowest result or the highest? Or the closest to the experimental result? Imagine you're making a physics test. Would you take the trig. numbers & discriminant into account (let them influence the sig figs)? Edited November 23, 2013 by Function 0 Share this post Link to post Share on other sites

John Cuthber 3637 Posted November 24, 2013 What you try to do is a better experiment. Redesign the experiment so that you are not using a steep part of the graph (easier said than done but it is, for example, the reason why the litre is defined as the volume of water at the temperature of maximum density. Because it's a local maximum (near 4C) small changes won't matter. As for what I'd do in a test: I'd follow the instructions I had got in the class or the textbook. They test shouldn't include this sort of thing if you were not told about it in the books. 0 Share this post Link to post Share on other sites