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    (Original post by PerigeeApogee)
    Hello,

    So I'm getting frustrated here.}\text{ I've got a problem where the answer is an angle.} \text{ The final answer is given by the exam solutions as a degree.

    However, when I perform the working with all variables in degress, I get the wrong answer, and only get the right answer when I perform the working in radians and then convert to degrees at the end. Clearly, there's something amiss here.

    It's probably a school boy error. But here it is:

    \tan(\alpha_p) = \frac{W + Q(l-R_d \sin(\psi))}{U - Qh+R_d \Omega \cos(\psi)}

     \text{Solve for } \alpha_p \text{ given that:}

     V_f = 120 \, m/s \text{, where } V_f = \sqrt{U^2+W^2}

     U \text{ and } W \text{ are related by } tan(\alpha) = \frac{W}{U}

    \alpha = 10^o \text{,}
    Q = -5^o/s \text{,}
    l = 8m\text{,}
    h = 5m\text{,}
    R_d = 4m\text{,}
    \Omega = 30\text{rev/min} = 180^o/s \text{,}
    \psi = 0^o

    \text{Now, working purely in degrees:}

    V_f^2 = U^2 + U^2\tan^2(\alpha) \to U = \frac{V_f}{\sqrt{1+\tan^2(\alpha  )}} = \frac{120}{\sqrt{1+\tan^2(10^o)}  } = 118.18 m/s

    W = U\tan(\alpha) = 118.18\tan(10^o) = 20.84m/s


    \tan(\alpha_p) = \frac{20.84 + (-5)(8-4\sin(0^o))}{118.18 - (-5)(5)+4(180)\cos(0)} = \frac{-19.16}{813.18} \to \alpha_p =-1.35

    \text{But the answer is } \alpha_p =8.725^o

     \text{However, working in radians then converting to degrees: }

    \alpha = \frac{\pi}{18}rad \text{,}
    Q = -\frac{\pi}{36}rad/s \text{,}
    l = 8m\text{,}
    h = 5m\text{,}
    R_d = 4m\text{,}
    \Omega = 30\text{rev/min} = \pi rad/s \text{,}
    \psi = 0rad

    V_f^2 = U^2 + U^2\tan^2(\alpha) \to U = \frac{V_f}{\sqrt{1+\tan^2(\alpha  )}} = \frac{120}{\sqrt{1+\tan^2(\frac{  \pi}{18})}} = 118.18 m/s

    W = U\tan(\alpha) = 118.18\tan(\frac{\pi}{18}) = 20.84m/s


    \tan(\alpha_p) = \frac{20.84 + (-\frac{\pi}{36})(8-4\sin(0))}{118.18 - (-\frac{\pi}{36})(5)+4(\pi)\cos(0)  } = \frac{20.14}{131.18} \to \alpha_p =0.152 rad = 8.728^o

    So, like I said, working in radians then converting works. Working in degrees throughout doesn't.

    What's the problem here?
    Well I haven't checked but it looks like you have rounded and then used your rounded figure in future calculations.
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    One problem I can see is that (l-R_d\sin(\psi)) is a length. You then multiply it by an angular speed Q to get a speed (or angular speed, I don't know). But the important thing is that Q(l-R_d\sin(\psi)) has to have units m/s because you then add it to W which has units m/s. This can only make sense if Q is in radians.
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    (Original post by Mr M)
    Well I haven't checked but it looks like you have rounded and then used your rounded figure in future calculations.
    Chaos theory?
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    I'm, guessing here, but in your formula, Q is supposed to be in radians/s.
    If you convert it to degrees/s, you are making it 57.3 times bigger. But nothing else in your formula is factored upwards. So it won't give a valid result.
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    (Original post by PerigeeApogee)
    How?

    Radians do not have dimensions, neither do degrees. They're just ratios of lengths.

    Q is an angular rate, so its units are s^{-1} regardless of whether you measure it in radians or degrees per second. So it should work either way, no?
    Radians per second and degrees per second are different units.

    Radians have the nice property that if you have an angle in radians and you multiply it by a length then you get the length of an arc. If you try this sort of calculation with degrees then you don't get a length at the end, you just get something else. That's why when doing this sort of calculation, you need to use radians.
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    (Original post by PerigeeApogee)
    But presumably whatever you do on the RHS will ultimately be fixed when you perform the inverse tangent later on.

    i.e., you might factor it upwards by 57 when you use degrees, but since you perform the inverse tangent with your calculator in degrees mode, this is equivalent to keeping Q in radians/s and performing inverse tangent with your calculator in radians mode. No?
    I wasn't meaning use of degrees or rads in your calculator. You would just use a different mode there.

    Suppose you have the expression:\tan\alpha=\frac{W+Q}{U-Q}
    Now let W=U=100 and Q = 1 radian, then

    \tan\alpha=\frac{100+1}{100-1}=\frac{101}{99}\approx 1

    Now use degrees,

    \tan\alpha=\frac{100+57.3}{100-57.3}\approx\frac{160}{40}=4

    Vastly diffent results depending on whether the angle on the numerator and denominator is in degrees or rads.

    I think that's the kind of thing that you have done in your own calculation.
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    (Original post by PerigeeApogee)
    Where did I round?
    All the two decimal places stuff (unless you couldn't be bothered to type the whole calculation into this thread).
 
 
 
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