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Identifying limit of puck height

  1. Sep 17, 2015 #1
    1. The problem statement, all variables and given/known data

    This homework isn't for a course. There is a book on hockey physics with a section discussing puck drop below intended target. When I use the book's formula modified for drag and lift, it seems to show that a 100 mph shot, taken from 30 feet from the net, can not hit the crossbar. The puck height reaches a maximum height of 46.63652463 inches with an initial puck angle of 24 degrees. Changing to 23 and 25 degrees reduce the final puck height. I guess this result is either because of the lift coefficient peaking near 25 degrees as mentioned on page 78 or because on the bottom of page 77 it is mentioned that formula 3.7 which includes a simple fit to data is only valid to within 20 percent.

    I'm wondering why the puck wont rise above 48 inches.

    I think I’ve seen high school kids, whom I doubt can shoot 100 mph, hit the crossbar from 30 feet away.

    2. Relevant equations

    Are provided in order below.

    3. The attempt at a solution


    Here is my work, in case I’ve made some mistakes.


    Formula 3.3 to determine puck drop below target without air resistance:


    gx2

    ________________


    2v2 cos2θ


    g = 9.80665 m/s2


    x = 30’ = 9.144 meters


    v = 100 mph = 44.704 m/s


    θ = 24 degrees = 0.4188790205 radians



    Formula 3.3 with drag:


    Formula 3.5 = velocity drop


    Velocity drop = -6.3 * (1/10/10/10) * 44.704 * 9.144 = -2.575272269


    Velocity at end = 44.704 + -2.575272269 = 42.12872773


    Average velocity = (44.704 + 42.12872773) / 2 = 43.41636387


    Formula 3.3 using average velocity:


    (9.80665 * (9.144 * 9.144)) / (2 * (43.41636387 * 43.41636387) * (cos(0.4188790205) * cos(0.4188790205))) = 0.2606128904 meters puck drop from target


    0.2606128904 meters = 10.16390273 inches


    Puck height would be 48 - 10 = 38 inches


    Formula 3.3 with drag and lift:


    Formula 3.7 (This is the simple fit to data):


    (4 * (1/10/10/10/10/10) * 24 degrees -7.8 * (1/10/10/10/10/10/10/10) * 24 degrees * 24 degrees = 0.00051072


    Formula 3.6 lift:


    0.00051072 from formula 3.7 * 44.704 initial velocity * 44.704 initial velocity = 1.020647166 lift


    New gravity = gravity minus (lift / mass) (from bottom paragraph of page 79):


    Puck Mass: Page 85 shows puck mass as 170 grams. However, to get the calculations in the bottom paragraph of page 79 to end up at 32 inches I used 0.138 for puck mass. Using .170097 for puck mass gave me an answer of 33.5 inches.


    9.80665 gravity - (1.020647166 lift / 0.138 mass) = 2.40465604348 revised gravity


    Formula 3.3 with drag and lift:


    (2.40465604348 gravity revised for lift * (9.144 distance in meters * 9.144 distance in meters)) / ( 2 * (43.41636387 avg velocity * 43.41636387 average velocity) * (COS(0.4188790205 puck angle in radians) * COS(0.4188790205 puck angle in radians))) = 0.06 meters puck drop


    .06 meters puck drop = 2.498475371 inches puck drop below target
     
  2. jcsd
  3. Sep 18, 2015 #2

    Simon Bridge

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    Youd have to compare the conditions irl with the model the text book uses.
     
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