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Wind strengths and distance chart...

Thu, Sep 19 2024 5:40 PM (128 replies)
  • bneyla1
    143 Posts
    Wed, Apr 24 2013 5:44 PM

    I divide yardage by 10, and multiply by half the wind.

     

    so 170 yard shot with 10mph wind is 1.7 X 5 = 8.5. I use a calculator...

    Also I usually add a little more for wind in my face and take off a little for wind at my back. I also hit most of my shots with full backspin

  • bubbadork
    984 Posts
    Thu, Apr 25 2013 4:22 AM

    170 divided by 10 isn't 1.7....

  • courteneyfish
    15,796 Posts
    Thu, Apr 25 2013 4:48 AM

    I guess. 

  • ct690911
    7,205 Posts
    Thu, Apr 25 2013 6:57 AM

    bneyla1:

    I divide yardage by 10, and multiply by half the wind.

     

    so 170 yard shot with 10mph wind is 1.7 X 5 = 8.5. I use a calculator...

    Also I usually add a little more for wind in my face and take off a little for wind at my back. I also hit most of my shots with full backspin

    170/10 = 1.7 ?...must be quite the calculator you use

  • ct690911
    7,205 Posts
    Thu, Apr 25 2013 6:57 AM

    bneyla1:

    I divide yardage by 10, and multiply by half the wind.

     

    so 170 yard shot with 10mph wind is 1.7 X 5 = 8.5. I use a calculator...

    Also I usually add a little more for wind in my face and take off a little for wind at my back. I also hit most of my shots with full backspin

    170/10 = 1.7 ?...must be quite the calculator you use

  • Soggyblogger
    224 Posts
    Thu, Apr 25 2013 10:06 AM

    ct690911:

    bneyla1:

    I divide yardage by 10, and multiply by half the wind.

     

    so 170 yard shot with 10mph wind is 1.7 X 5 = 8.5. I use a calculator...

    Also I usually add a little more for wind in my face and take off a little for wind at my back. I also hit most of my shots with full backspin

     

    170/10 = 1.7 ?...must be quite the calculator you use

    Zeroing in on his mistake misses the point of whether his formula has merit. Though off the top of my head, my inclination is to reject linear formulas as well as any formula that doesn't recognize different courses have different characteristics such as trees or no trees.

    I believe my eyes so when I have watched thousands of shots take off from my club and go straight for a while until they reach some elevation and then begin to significantly be moved by the wind, but once they begin to descend and then stop moving so much, I conclude trees are part of the programing. Even on treeless courses, the wind appears significantly stronger up high then down near the ground.

    Then there is the question for math and physics majors: Does a 20 mph wind affect an object flying through the air exactly twice as much as a 10mph wind? Practical experience tells me no, but I do not know the phyics for this nor the math.

  • bubbadork
    984 Posts
    Thu, Apr 25 2013 12:53 PM

    No, a 20 mph wind does not have twice the effect of a 10 mph wind (on a golf ball). In addition to the variation in drag there is a variation in lift (magnus force). For cross-winds this lift has a vector, same as the drag.

  • ct690911
    7,205 Posts
    Thu, Apr 25 2013 1:30 PM

    Soggyblogger:

    ct690911:

    bneyla1:

    I divide yardage by 10, and multiply by half the wind.

     

    so 170 yard shot with 10mph wind is 1.7 X 5 = 8.5. I use a calculator...

    Also I usually add a little more for wind in my face and take off a little for wind at my back. I also hit most of my shots with full backspin

     

    170/10 = 1.7 ?...must be quite the calculator you use

    Zeroing in on his mistake misses the point of whether his formula has merit. Though off the top of my head, my inclination is to reject linear formulas as well as any formula that doesn't recognize different courses have different characteristics such as trees or no trees.

     

    I believe my eyes so when I have watched thousands of shots take off from my club and go straight for a while until they reach some elevation and then begin to significantly be moved by the wind, but once they begin to descend and then stop moving so much, I conclude trees are part of the programing. Even on treeless courses, the wind appears significantly stronger up high then down near the ground.

    Then there is the question for math and physics majors: Does a 20 mph wind affect an object flying through the air exactly twice as much as a 10mph wind? Practical experience tells me no, but I do not know the phyics for this nor the math.

    I wasn't "zeroing in" on his mistake per se because we all make mistakes...but it was kinda funny when he introduced the statement he uses a calculator then promptly makes a simple math error in his example.

  • ckinfidel
    156 Posts
    Sat, Jul 20 2013 5:28 PM

    This is pretty cool. Needs adjustments but a good starting point.

  • RadioactiveRebel
    36 Posts
    Sun, Jul 21 2013 1:27 PM

    After playing a couple weeks, I'll add my 2 cents worth for how I play.

     

    Winds under 10mph I totally ignore. Over 10mph I generally divide by 2 and add/subtract to my distance. For crosswinds over 10 I hit a bit early or late to compensate rather than move the marker.

     

    For uphill shots I divide the amount of feet high to hole by half and add that to the yards needed. So 130yrds to hole 10 feet up I view as 135yrds and then figure wind as above. Now sometimes this falls short. Why? Because though the hole is only 10 feet higher than you, the edge of the green may be higher than that and cause you to fall short.....and a downhill slope where your ball lands may cause it to roll too far even with back spin.

     

    If I put full back spin on ball I just shave 5 yards off distance if hitting over 100 yards.

     

    Is it perfect? No. Does it work out close enough? Most of the time. Have a hole that constantly falls short? Make note of that holes special circumstances and adjust accordingly next time.

     

    Also note this is using the starter gear you get when first sign up.

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