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Getting Closer - Young46 "Wind Aim" Tutorial

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Tue, Nov 29 2022 11:25 AM (6 replies)
  • RickinWaSt
    1,909 Posts
    Wed, Jun 17 2020 10:39 AM

    Young46 recently (May 22, 2020) put up a tutorial "Wind Aim". He provides a calculation {More about this below.} and 3 examples. He uses the final number to move the grid on the green. Every shot he takes, within 2 yards of the pin! Yea, he's an above average skilled player. However, it is the calculation getting him close.


    I’ve gone back numerous times to the video. I still come away unclear about a small portion of his calculation. Maybe someone can help me out?  


    Young46 presents this formula…


    100 ÷ Distance = A (100 ÷ 166 = 0.602)


    A x 25 = B (0.602 x 25 = 15.06)


    This is where my understanding drops off at 1:15 seconds. Young46 converts the 15.06 to 15 yards. He points out that 15 yards is also the number of yards in ½ grid. 


    He converts the 15Y to 55%, then 60%, then 8.4 yards, ending up with 0.56%. He multiples this number by 4.5 and arrives at 2.52. This refers to the number of grids away from the pin. Boom, he hits a shoot within 2 yards of the pin. 

    Thanks in advance for any clarification!


  • siva82430
    1 Posts
    Mon, Jul 5 2021 8:24 AM



    How are you .  can you send me the video link . did you get the formula how it works now :)

  • Gepetto1958
    3,966 Posts
    Mon, Jul 5 2021 11:01 AM

    too much for me ;-)

  • MarchieB
    1,475 Posts
    Tue, Jul 6 2021 7:37 AM


    GRIP IT N' RIP IT !!!!!

  • AlejBaran
    948 Posts
    Fri, Jul 16 2021 10:50 AM


  • AlejBaran
    948 Posts
    Sat, Jul 17 2021 11:44 PM

    I will try to clarify the doubt.

    In the example mentioned; The shot data is: 166 yds +1 foot and a headwind of 14 m/h at 45°approx (blowing around 7:30 if we see it on the sphere of a clock).

    THE FIRST CONFUSION MAYBE GENERATED when he calls an entire grid at 4.5 cells and assumes, which seems correct, that it measures 25 yards per side when "viewed" from a distance of 100 yds, others will call this 1/2 grid (indeed, he said this).

    In his first calculation at 0:38, he performs the calculation to find out what the apparent size of the grid is from this distance; the equation is:

    X = (166yds/100yds)*25yds---> X=15yds and this is the "apparent size of the complete grid" (4.5 cells) seen from 166 yds.

    Then comes the part where he calculates the drift due to the wind. AND IN 1:12 IS WHERE THE SECOND CONFUSION MYBE IS GENERATED; He says "... now obviously this is about a half wind I would call this about 55% so we know that's about 15 yards ..." He says 15 yds instead of 14 yds (or MPH) which is the speed of the wind (it's just a little mistake on his side); if you look at his calculator he performs the operation 14*0.6=8.4; "... this is about 60% we'll say 55, 60 so we know it's 8.4 yards" which is the lateral drift due to the wind.

    Finally at 1:28 he performs the following calculation: if 1 grid is equal to 15 yds, then 8.4 yds will be x's parts of 1 grid which (8.4*1)/15=0.56 grid; then if 1 grid is 4.5 cells, 0.56 grid will be x's cells ---------> (0.56*4.5)/1= 2.42 cells.

    All of this can be summed up as:

    1- (100/Dist)*25 = A

    2-Drift/A = B (Drift, it is the lateral deflection, in yards, that you must calculate, due the wind)

    3- B*4.5 = # (number of cells that must be moved)

    Video link:

  • birdiesfly1
    14 Posts
    Tue, Nov 29 2022 11:25 AM

    What are the calculations for the "drift"?  That is the piece/explanation that is missing here.