umer007 Posted October 5, 2005 Posted October 5, 2005 tan(θ°) = (viv) / (vih) = (1.28 ± 0.0676m/s) / (0.514 ± 0.0676m/s) = 2.55 ± 0.467 m/s θ° = 68.0 ± 3.65° This answer may not be correct to the decimal place but it shud b relatively close. Could someone plz explain to me how 3.65° becomes the uncertainty for the angle? Plz. explain very clearly and simply. Im in a hurry. THanks in advance
mezarashi Posted October 5, 2005 Posted October 5, 2005 If you look at it from a geometric point of view, You have say a right triangle, whose sides A and B are uncertain. You know that for any A or B, you can find the angle doing arctan. Worst case scenarios: You can see that if A remains constant and B gets longer, the angle will change. The same happens if A gets longer while be is constant. Using this concept, I think it is easier to imagine why there is an ambuguity in the angle of about 3.8 degrees or so.
umer007 Posted October 5, 2005 Author Posted October 5, 2005 is it possible to show how to get that 3.65 degrees mathematically? If so could you plz write it for me. Thx.
mezarashi Posted October 5, 2005 Posted October 5, 2005 As far as I know, there's no exact formula for these types of questions... I'm not sure, I think I may have seen some before, but they were not very intuitive formulas. The best way to go is to start from the fundamentals. That is, worst case scenario to worst case. This works whether you want to do multiplication, addition, tan, sin, etc. If you understood what I was talking about earlier about the geometry, you can derive this very easily as well. What is the worst case each way. For the angle to be as large as it can, you have the situation (A + B)/(C - D). For the angle to be as small as it can be, you have (A - B)/(C + D) assuming all the variables A-D are positive. You can find the margin of error then in doing tan x1 = (A+B)/(C-D) tan x2 = (A-B)/(C+D) delta x = 1/2 (x1 - x2) delta x = 1/2 ( arctan[(A+B)/(C-D)] - arctan[(A-B)/(C+D)] ) Pluggin in the variables, you will get your answer.
swansont Posted October 5, 2005 Posted October 5, 2005 As far as I know, there's no exact formula for these types of questions There's a duplicate of this thread elsewhere. You take the derivative of the formula. y=f(x) dy/dx = f'(x) dy = f'(x) dx dy is the uncertainty in y, dx is the uncertainty in x. Evaluate f'(x) at the appropriate point
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