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high winds

• Ihatecheats 298 Posts

  Sun, Mar 4 2012 7:18 PM

  JimbeauC:

  newcastleb:

  how to tell how far from target 9 yds is?

  The most exact way is to switch to fore view. (lower right of screen - arrow) and switch shot type to chip. This activates the aiming grid. There is debate about how wide the grid is. I think it's 10 yards wide. 

  It kind of disappears in the rough, which is a bit of a problem. In heavy crosswinds, you'll be aiming some strange places. 

  Another problem is depending on angle of the camera, the grid can get turned into a diamond. If the hole has a rear view camera, use it. 

  Course knowledge is important. Some holes seem to have their yardage marked in error. Beth Page #1 comes to mind. With the formula, you will come up short.

  Glad I was able to help. Now go hit 'em close for tap in birds.

  JimbeauC many thanks for an enlightening post. I am a Legend and I didn't know that knowledge of the aiming grid and wedge as I always just played freehand. I shall use your useful  information tomorrow and try and get even closer to the pin.

  When people try and explain wind direction they should just say what hour it is then it would be easy to know where the wind is blowing from.

• JimbeauC 5,835 Posts

  Sun, Mar 4 2012 9:39 PM

  Ihatecheats:

  I am a Legend and I didn't know that knowledge of the aiming grid

  Mr I, (difficult to address you without the word, 'hate' although I share your sentiments.)

  Ummm,

  Eeeeeek.

  Any idea how honored I am to have helped someone of your stature. I must confess that the entire formula and method actually came from a most generous Legend at my CC. I don't know if he would want to be identified, so I shall refrain. 

  Hmm, "Knock 'em close for tap ins doesn't really fit, as I'm sure you do that already. 

  How about...

  Nothing but eagles, mayhap doubles.

• Bonez2k6 35 Posts

  Sat, Mar 17 2012 8:06 AM

  Jim could you elaborate a little more on the formula for this... I think its a+b=c (square)?.

   

  "This is trickier. In a quartering wind, the 16 MPH represents the hypotenuse of a right triangle, in this case, one with two equal sides. (Pythagorean Theorem) I never fully calculate this, but use the general idea to get in the ballpark. In this case I did, to the first decimal. 11.3 - representing, in this case, both the cumulative wind effect on distance and yaw."

  11.3 * .8 = 9.04.

   

  Bonez

   

• JimbeauC 5,835 Posts

  Sat, Mar 17 2012 10:55 AM

  Bonez,

  Not quite. Here's the formula.

  

  -----------------------------------------------------------------------

  In the original wind example, c = 16. a & b are equal.

  16 squared is 256. Because a = b, and (a squared) + (b squared = 256), both a and b will be the square root of 128 (half of 256)

  11.3 * 11.3 = 127.69 - close enough. Actually 11 is close enough for me.

  This equals the vectored wind effect. I still have to factor in the distance. Half of 160 = 80, which is the percentage effect of the wind, so I multiply 11.3 * .8 = 9 and a tad, representing the wind effect in yards. 

  --------------------------

  That's the full calculation, but just understanding the idea, I come pretty close, just knowing (in this particular example with a = b, about 2/3 is the ratio that gets me close enough. Heh, doing the above with a 60 second time limit in multi, would be daunting. I suppose someone could come up with a chart that would include the scenarios where a doesn't = b, ie wind out of the 2:00 o'clock direction.

  Hope this clarifies it.

• Bonez2k6 35 Posts

  Sat, Mar 17 2012 2:17 PM

  Thanks Jim,

   

    I'm pretty good with math but I couldn't get your solution now I understand it.

  Thanks for clarification....

   

  Bonez

• JimbeauC 5,835 Posts

  Sat, Mar 17 2012 3:18 PM

  Bonez2k6:

  Thanks for clarification

  Most welcome. Happy to help.

• Bidi 1,948 Posts

  Sat, Mar 17 2012 4:07 PM

  depending on your balls and clubs, but you can try also this way:

   

   

   

   

  5         7        10        12         15        17        20     (MPH)

   

  -----------------------------------------------------------------------------------

   

  2         3          4           5            6         7         8          (% of adjustment )

   

   

   

  150 yard w/ 17 MPH headwind ( 150 + 7% = 160.5 )

   

  150 yard w/ 17 MPH tailwind  ( 150 - 7% = 139.5 ) 

   

   

   

  22        25        27        30         

  -----------------------------------------------------------------------------------

   

  9        10        11         12         % OF adjustment ) 

   

• genorb 1,255 Posts

  Sat, Mar 17 2012 6:10 PM

  JimbeauC:

  Heh, doing the above with a 60 second time limit in multi, would be daunting. 

  Not really. In the case you described, if a=b then

  a^2 + b^2 = c^2     becomes     2 a^2 =c^2

  thus

  a = c / sqrt(2) = 0.7*c

  So you just need to multiply the given wind (16 mph in your example) by 0.7 to get the effective forward wind (which is equal to the effective lateral wind in this example because a=b). Once you know the effective forward wind, you can then use your formula to get the wind effect in yards. So that's just two multiplications.

  Regards

• gr8flbob 592 Posts

  Sat, Mar 17 2012 7:01 PM

  Ah yes, nothing quite like running some trig calculations, whilst watching the 60 sec clock (that some yahoo insisted on) run down, just after the 3rd or 4th MGD and a couple of tokes.

  Precise calculations are all well and good, but the variability that WGT tosses in makes the effort nearly not worth it. Wind < 5 or 6 mph? don't bother. 10 mph, add or subtract about 8% to club + ball rating; 20 mph, add or subtract 15% or so. Adjust accordingly for in-between.

  With wind quartering (from 1:30, 4:30, 7:30 or 10:30) reduce the adjustment to 5% for 10mph wind or 10% for 20mph. Any further to the side, I make no adjustment for distance other than adding top spin (if it's a 'head' wind) or back spin (if it's a tail wind).

  The big worry here, of course, is where to aim. More on that later ...

• JimbeauC 5,835 Posts

  Sat, Mar 17 2012 8:11 PM

  genorb:

  multiply the given wind (16 mph in your example) by 0.7

  BINGO.

  .7 is close enough to my quick calc of 2/3 or .67 to be close enough. Now we're getting the hang of doing this quickly.

  Nutshell: when calculating the vector result, you just have to know a bit about the ratios. Goodness, I could get into sines, cosines, and tangents. No need to get even near to that complexity. 

  Heh, of course we have VEM to contend with.

  Okay, I'll include this for giggles. 

  Too much calculation ??

  Actually, I intended this to land about 1 yd short. Oh well. Dead on head or tail winds behave the best.... usually.