Note: This thread is 6545 days old. We appreciate that you found this thread instead of starting a new one, but if you plan to post here please make sure it's still relevant. If not, please start a new topic. Thank you!
-
Topics Being Discussed Right Now on The Sand Trap
-
- 3 replies
- 638 views
-
- 10,470 replies
- 886,181 views
-
- 11 replies
- 7,240 views
-
"5 Minutes Daily" Practice Challenge 1 2 3 4 989
By iacas, in Instruction and Playing Tips
- 5 minutes daily
- dedication
- (and 6 more)
- 17,798 replies
- 1,751,032 views
-
- 60 replies
- 22,183 views
-
Announcements
-
TST Partners
Direct: Mevo, Mevo+, and Pro Package.
Coupon Codes (save 10-20%): "IACAS" for Mevo/Stack/FitForGolf, "IACASPLUS" for Mevo+/Pro Package, and "THESANDTRAP" for ShotScope. 15% off TourStriker (no code). -
Posts
-
All great info. Thanks for the reply. -
Yea, it's more complicated than your high school projectile motion equations. I am thinking it could increase under certain conditions. A gust of wind blowing in the same direction as the spin, causing more high and low pressure on the ball in a certain way that it increases the spin? It has zero vertical velocity at its apex. So, it is all velocity caused by gravity for the vertical component. Yea, landing angle is a big thing. It is parabolic. Your apex is 90 yards in the air. A 30-yard elevated green is 1/3rd that height. At the apex, your vertical descent angle is zero, it should be horizontal. So, you are going from zero theta to let's say 45 degrees. Even if it was linear, let's say you're landing angle is close to 30 degrees. That is less than a driver and probably is significant. Yea, it depends on how you hit it. Especially for downhill shots. If you hit a flighted shot, it might react more like a normal shot because of the lower launch and lower apex relative to your position. Versus a normal shot might come in at like 70 degrees, instead of 45 degrees. -
.thumb.JPG.640dd6f31fe7f6aef91f879119bae1d9.JPG)
-
Spin will decay slightly over time, but not by a lot. The horizontal portion of the velocity will also decay due to air resistance. The vertical component will be increasing since the ball is accelerating due to gravity (albeit that the spin is creating lift, which will counteract that some). Neither of those has much of an impact of how the ball will react. The biggest difference is the vertical land angle. The angle theta prime (not sure how to show that on here) will be shallower than theta. That means the ball will stop faster at theta than at theta prime. The other thing is because there is still a horizontal component to the velocity, it will carry less far at theta prime than at theta. The effects of those two things work in opposite directions. Which one "wins" will depend on ground conditions, ball flight, spin, any necessary carry distances, etc. Fortunately the margins are fairly small so you can wing it with enough experience. The calculation of the carry distance change is what your range finder estimates when you have slope turned on. -
So, I was looking at this image and wondered what the best way is to play your approach to an elevated green versus a lowered green. Is the spin and velocity profile at θ' much different than at θ? I don't know the physics of it but to my wee brain, it would seem that at θ' the spin would be higher but velocity lower. At θ the spin would seem to be lower but velocity higher since it has more time to fall from its peak where it would be zero. Even the image below is off visually since we know the arc of the ball flight isn't consistent throughout. It's okay if you tell me I'm overthinking this. 😂
-
Recommended Posts