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Need help for toboggan speed

  1. Aug 8, 2008 #1
    i need help, i need to calculate the slope of a hill to reach 40mph on a toboggan? how do i do this with a fluctuating toboggan weight...meaning what is the slope i require for different weights? also how would i adjust for dips in the hill?

    can anyone help? this is not might area and would appreciate something simple that i could use.

  2. jcsd
  3. Aug 9, 2008 #2


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    Welcome to PF!

    Hi zoltar ! Welcome to PF! :smile:

    Acceleration is what matters, and as Galileo proved (on a slope, by the way), all weights have the same acceleration under gravity.

    So the weight doesn't matter. :smile:
    Not following you … what sort of dips? :confused:
  4. Aug 9, 2008 #3
    It's not so simple.

    It depends on the friction of the toboggan on the snow - which depends on the design of the toboggan, and the quality and temperature of the snow.

    And then it also depends on the air resistance of the toboggan and its rider(s).

    So there isn't a simple formula that will give the right answer. You're probably better conducting an empirical experiment (trying it out for real, and seeing what results you get).
  5. Aug 10, 2008 #4
    It is relatively simple if you assume no air resistance/friction etc. But it will depend on how much space you have and how quickly you want to reach your target speed.

    [tex] \frac{1}{2}m (\Delta v)^{2} = mg \Delta h[/tex] (kinetic energy gained = gravitational potential energy lost)

    [tex] \Delta h[/tex] [tex]= \frac{v^{2}}{2g}[/tex]

    [tex] = \frac{(40*1.6)^{2}}{2*9.8} = 209m[/tex]

    So you will need a height change of 209m to achieve 40mph. Now you can calculate, using right-angled-trigonometry and knowledge of the space available, the angles you'd need to achieve this. Note this is just an approximation of course.
  6. Aug 10, 2008 #5


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    One should be careful with units here. The numerator would have units if (km/hr)2 and the denominator units of m/s2, so one must apply conversion factors 3600 s/hr and 1000 m/km. In doing so, the above expression becomes,

    [tex] = \frac{(40*1.6)^{2}}{2*9.8*3.6^2} = 209 / (3.6)^2 m = 16.1 m[/tex]

    Also, the change in kinetic energy is proportional to [tex]\Delta(v^2)[/tex], not [tex](\Delta{v})^2[/tex].
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