Why round numbers in physics if you want to be precise?

My friend at school asked today why do we round the decimal places if we want to be precise in physics. I think he's talking about significant figures. Because the rule is that we use the least amount of significant figure for the answer but if we want to be precise why don't we just leave the decimal as it is why round it. I didn't get that good of an answer and I didn't quite understand some of the answers that people put down so I'm asking this again because I don't know how to extend


the deadline of the question.





Thank you very muchWhy round numbers in physics if you want to be precise?
it depends up on the no.of significant figures the values you use to solve an equation has.........for eg:- when you divide 4.5 by 7 you get an answer equal to 0.64285714............but you round up the answer to 0.64 because the numerical 4.5 has only 2 significant figures and hence the answer must also have 2 significant figures......





Further the rounded numbers are very very small and negligible and hence would not make an answer dis satisfactory.....





Wasiq KhurshidWhy round numbers in physics if you want to be precise?
accuracy is important. however if accuracy is to be meaningful you can only be as accurate as your least accurate result.


for example if you are using a value of 2.3, and in an equation you get a result of 9.87653 then the other decimal places mean very little, as you are only using 2 figures in an essential part of the equation.
We want to be as precise as possible, but exactness is very rare in physics. Our ability to measure things is limited in precision. All measurements, including most of the universal constants, include error bars around the known value.





When you are told to round a number because of significant figures, it is because the digits are not significant. They are garbage. Random numbers. Meaningless.





A simple example: F = ma, or a = F/m. Suppose you are told the mass of an object is 7 kg. Do we really know that it is 7.000000000 kg, down to the mass of an individual atom? No, we don't. If I say ';7 kg'; with only two significant figures, I mean ';approximately 7 kg, but it might be anywhere from 6.5 to just under 7.5';.





Suppose I said I was exerting a force of 10 N on this object. Again, do I mean 10.000000000 N? No, I do not. My ability to measure or calculate this force is limited. Maybe it's from a magnet whose strength I only know to 1% precision. Maybe it's a mass I only know to 5%. So ';10 N'; might mean ';9.7-10.3 N';.





OK, so now I calculate a = 10/7 = 1.42857143 m/sec^2. But it's really 10 +- 0.3 divided by 7 +- 0.5. So a could be as low as 9.7/7.5 = 1.2933333 or as high as 10.3/6.5 = 1.58461538.





Since it could be anywhere from 1.29 to 1.58, a value of 1.42857143 doesn't really make any sense as a ';precise'; answer. I don't know that the value really has an 8 in the 3rd decimal place. Heck, I don't even know that it has a 4 in the 1st decimal place. All of those digits beyond the 4 are just noise. They could be anything at all and still be consistent with my input values.
Let's say you want to do an experiment to determine the magnitude of the acceleration of gravity at the Earth's surface. You know it's essentially constant (and you're neglecting air resistance), so you are going to use the constant acceleration equation of motion, which says





d(t) = d0 + v0 * t + 1/2 * g * t^2





where d(t) is the height at a given time t, d0 is the initial height, v0 is the initial vertical velocity, and g is the acceleration of gravity. Let's solve that for g.





g = 2*[d(t) - d0 - v0 * t] / t^2





In order to get your measurement, you're going to get a bowling ball, put it on the roof of a building, and have your buddy push it off. Meanwhile, you'll be on the ground with a stopwatch.





Let's say you've got a meter stick to measure the height of the roof (the initial height of the ball, equal to d0), and it's marked in millimeters. You come up with a measurement of d0 = 10.231 m above the ground. And of course, when the ball hits the ground, its height is exactly 0 m (infinite precision).





You don't have to measure the initial velocity, because your buddy isn't giving it any. We can assume it's exactly v0 = 0 m/s (infinite precision).





Let's plug those into our equation.





g = 2*[0 - 10.231 m - 0 m/s * t] / t^2


g = 2*-10.231 m / t^2


g = -20.462 m / t^2





Now, the only other thing you have to measure to find the value of g is the time. Let's say you have a stopwatch that measures in increments of seconds. You start it when your buddy drops the ball, and stop it when the ball hits the ground. When you read it, it says the time elapsed is 1 second.





You plug that into the equation for g, and you get





g = -20.462 m / (1 s)^2


g = -20.462 m/s^2





But wait! Other people have measured the acceleration of gravity very precisely, as being -9.80665 m/s^2. What's going on here? Is something affecting your local gravitational field? Should you win a Nobel Prize for your discovery? Or maybe you've just done your experiment incorrectly and you should get an F in physics class.





Well, no, none of those things are true. You just didn't pay attention to significant figures. If you had, you would have rounded your answer to one decimal place, because of the precision of the 1 s measurement of time. Then you would have gotten g = 20 m/s^2. Actually, we should write that in scientific notation so there's no confusion about the precision. g = 2 x 10^1 m/s^2.





Now, when you compare that to the accepted value of 9.81 m/s^2, you can see that the difference in your measurement is just experimental error. And when your value is that far off, you don't gain anything by adding extra decimal places. As a matter of fact, you actually lose something, because you trick your reader into thinking that those extra decimal places are important - that your value is determined that precisely.





Reporting it as g = 2 x 10^1 m/s^2 indicates to the reader that there's a margin of error about the size of the last digit - around 卤1 x 10^1 m/s^2. If you compare that to the actual error, you'll see it's pretty close.





Let's repeat the measurement with a stopwatch that measures tenths of a second. This time, you get a time of 1.4 s. That gives you





g = -20.462 m / (1.4 s)^2


g = -10.4397959... m/s^2





This one we ought to round to one decimal place, which gives us g = -10.4 m/s^2. (Actually, to be technically correct, we ought to use significant figures at each step - (1.4 s)^2 = 1.96 s^2, which rounds to 2.0 s^2, and the division gives 10.231 m/s^2, which rounds to 10.2 m/s^2.)





So, when we say g = -10.4 m/s^2, we're indicating to the reader that the measurement is correct to the ones place.





What precision stopwatch would we need to get closer to the true value? Well, if we measured to 1/100 of a second, we'd get 9.87 m/s^2, and if we measured to 1/1000 of a second, we'd get 9.813 m/s^2. Of course, making the time more precise than that wouldn't be very useful, because then our height measurement would control the precision of the answer.





So, to summarize - you round using significant figures to indicate the precision of the value you're reporting. I hope that helps you out!