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  1. #49
    Join Date
    Dec 2009
    Location
    Baltimore
    Posts
    1,205

    Re: The Arc on Joe's ball.



    Didn't the Ravens hire a numbers guru. Maybe he could chime in.




  2. #50

    Re: The Arc on Joe's ball.

    Quote Originally Posted by zckattk View Post
    This is the greatest and funniest thread I have ever seen. Love it.
    I think some of us are bored,something to do

    Quote Originally Posted by dandrews View Post
    Ran a couple of calculations. Using 3.2 seconds as the hang time (which is what I calculated with a stop watch, but someone else just verified) and the 59 yards traveled estimate, I calculated an initial velocity of the throw at 23.02 meters/second (which is roughly 52 mph). Using that, I got a 42.92 degree launch angle. Also, the peak height came out to be 12.5 meters (13.7 yards).

    If anyone has any theories as to how to use these numbers to get a distance of a curve, feel free. I have no idea.
    I just use something like Sketchup,it does all the figuring.
    but I used the 54 yds distance.
    Comes up with an arc of 202' 3 3/8".
    (a little over 67 1/3 yds.)
    I dunno,I could have messed it up.









  3. #51
    Join Date
    Jul 2011
    Location
    Right Where I need to be
    Posts
    1,340

    Re: The Arc on Joe's ball.

    OMG Thread of the WEEK. LOL




  4. #52

    Re: The Arc on Joe's ball.

    Quote Originally Posted by Baltimoreboy View Post
    I think some of us are bored,something to do



    I just use something like Sketchup,it does all the figuring.
    but I used the 54 yds distance.
    Comes up with an arc of 202' 3 3/8".
    (a little over 67 1/3 yds.)
    I dunno,I could have messed it up.





    Hellz Yeah! Thank you!




  5. #53

    Re: The Arc on Joe's ball.

    So assuming the vertical displacement is negligible, we have a distance traveled of 49 meters in a time of 3.2 seconds.

    We can calculate vertical velocity by using

    0 = (initial velocity)(3.2) + 1/2 ( -9.8) (3.2^2)

    This comes out to an initial vertical velocity of 15.68 m/s

    We can then approximate max height using

    0 = 15.68^2 - 2 (-9.8)(max height)

    This gives us a max height of ~ 12.544 meters or about 36 feet

    Initial horizontal velocity can be calculated using

    49 = x(3.2) + 1/2* -.1 * (3.2^2)

    This gives us an initial horizontal velocity of approximately 15.475 m/s

    Using the calculated stuff here, we can use the following:

    Integral from 0 to 3.2 of
    sqrt((15.47 - .1t)^2 + (15.68 - 9.8t)^2)

    Which gives us an arc length of 74.551 meters, or approximately 81 yards.

    Pretty awesome. Note these are rough figures, since we can't really be completely precise with the information we have
    Last edited by CptJesus; 01-15-2013 at 05:43 PM. Reason: Messed up integral




  6. #54
    Join Date
    Jan 2009
    Location
    Detroit Michigan
    Posts
    1,906
    BaltimoreBoy - Do yourself a favor and email that to ESPN/sports science with emphasis on whether Moore could have made that play. That would be a sweet segment and I can't recall an episode where they focused on ball distance / height.
    “Great minds discuss ideas. Average minds discuss events. Small minds discuss people.”

    –Eleanor Roosevelt




  7. #55

    Re: The Arc on Joe's ball.

    Quote Originally Posted by Sirdowski View Post
    BaltimoreBoy - Do yourself a favor and email that to ESPN/sports science. That would be a sweet segment and I can't recall an episode where they focused on ball distance / height.
    I'm not sure if my calculations are off or theirs are, but we're getting 2 different values.

    Also:




  8. #56
    Join Date
    Jan 2009
    Location
    Detroit Michigan
    Posts
    1,906
    Quote Originally Posted by CptJesus View Post
    I'm not sure if my calculations are off or theirs are, but we're getting 2 different values.

    Also:
    HAHAHA. funny stuff.
    Last edited by Sirdowski; 01-14-2013 at 05:09 PM.
    “Great minds discuss ideas. Average minds discuss events. Small minds discuss people.”

    –Eleanor Roosevelt




  9. #57

    Re: The Arc on Joe's ball.

    Quote Originally Posted by CptJesus View Post
    I'm not sure if my calculations are off or theirs are, but we're getting 2 different values.

    Yea,
    82' high and an arc length of 380 yds sounds a little off




  10. #58

    Re: The Arc on Joe's ball.

    This thread will end poverty and end all wars.




  11. #59

    Re: The Arc on Joe's ball.

    Quote Originally Posted by Baltimoreboy View Post
    Yea,
    82' high and an arc length of 380 yds sounds a little off
    I messed up my integral, forgot the square root. The arc lenght is 31.63 meters, or approximately 34 yards.




  12. #60

    Re: The Arc on Joe's ball.

    Quote Originally Posted by CptJesus View Post
    So assuming the vertical displacement is negligible, we have a distance traveled of 17.98 meters in a time of 3.2 seconds.

    The ball traveled about 54 yds
    (49.3776) meters.

    Don't understand the rest




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