[Steve Okula]

My club will be hosting a European Tour event in September. We have an issue with the driving range not being long enough for the Tour players to hit driver. Presently, the range fairway extends about 230 meters, but downhill (9 meter drop), and with the prevailing wind. There's a low fence at he bottom that stops most of the members' balls. Beyond the end of the driving range fairway proper there is about 40 meters of high rough grass then there is the fairway of the second hole.

There's no room to build another tee further back up the hill, but we could put one out at the far end of the range, in the rough area, so the players will be hitting in the opposite direction uphill and into the prevailing wind. The contour plan tells me that it would then be something over 270 meters long over a 12 meter rise. It's important that the balls not carry too far up the hill because other essential practice areas like putting and chipping greens might be in range.

How much does the rise add to the real length of the terrain? Is there a formula or anything to figure that out? I mean, hitting a ball 270 meters up a 12 meter slope would be the equivalent of how far over flat ground?  What is the real difference in distance between 230 down 9 meters and 270 up 12 meters?

[George Pazin]

What is this meters thing you speak of?

Elevation change should affect the longer clubs more than the shorter clubs. My guess would be 270 meters downhill would be equivalent to about 300 meters uphill, with the elevation changes you cite.

[Jason Topp]

Steve:

I will be interested in whether others have a formula.  One must exist because I know rangefinders have been built that estimate an adjusted distance that takes slope into account.

My theory is that slope should impact carry distance at a 1 for 1 ratio.  In other words - if the landing spot is 10 yards (or meters) lower than the tee, the ball will carry 10 yards farther.  My reasoning is that much of the trajectory data for drivers shows the ball landing at an angle relatively close to 45 degrees.

Of course - roll would also be affected very dramatically by the slope and further increase the difference in distance on the course.

It sounds like the tee at the bottom of the range might work but it will be close and you will want a fence to protect the other practice areas.

[JESII]

Just so you know Steve, building the tee down below in expectation of hitting uphill INTO THE WIND will only guarantee the wind that week will come from the opposite direction.

[JMEvensky]

I've always heard that Tour Pro's despise hitting to an upsloping range.It makes them think their trajectories are off.

[Richard Choi]

Ooo... math problem! In meters, no less!

Going by straight Pythagorean theorem, you get this:

c^2=a^2+b^2, which if you plug your numbers into, you get c=sqrt(12^2+270^2).

That works out to c=270.2665 meters, so it is not going to buy you much...

But you can also look at it from the other way. A really long drive is typically approaching the ground at about 30' (approach angle). If this ball hits at the top of 12m hill with a 270m drive, it is actually equivalent to a 290.785m drive on flat ground. So that is not bad.

If you modify that for a drive with an approach angle of 40', you get an equivalent drive of 284.3m.

I think you can probably assume that your 270m driving range with 12m incline will hold a drive about 280 to 290m long.

[Alex Miller]

Ooo... math problem! In meters, no less!

Going by straight Pythagorean theorem, you get this:

c^2=a^2+b^2, which if you plug your numbers into, you get c=sqrt(12^2+270^2).

That works out to c=270.2665 meters, so it is not going to buy you much...

That doesn't take into account the distance it's losing from actual hang time, that's just actual along the ground distance.

Assuming it lands at a 45 degree angle you would get around 282 meters. 270 from the actual and 12 from the 12 meters it would have travelled while falling the last 12 meters.

[Richard Choi]

Ooo... math problem! In meters, no less!

Going by straight Pythagorean theorem, you get this:

c^2=a^2+b^2, which if you plug your numbers into, you get c=sqrt(12^2+270^2).

That works out to c=270.2665 meters, so it is not going to buy you much...

That doesn't take into account the distance it's losing from actual hang time, that's just actual along the ground distance.

Assuming it lands at a 45 degree angle you would get around 282 meters. 270 from the actual and 12 from the 12 meters it would have travelled while falling the last 12 meters.

Edited! I posted new numbers above.

[Bill Brightly]

I'm sure tour players wouldn't complain about "irons only" or restricted flight balls, would they?

[Kevin_Reilly]

Pro landing angles are typically under 40 degrees according to TrackMan.  I'll let the trig guru's above update their calculations.

[Richard Choi]

Pro landing angles are typically under 40 degrees according to TrackMan.  I'll let the trig guru's above update their calculations.

Updated!

[Tim Nugent]

We did some empirical tests in Hawaii and decided that 1:1 was a good average.  Since you don't seem to have 330m, (300+30) I'd say you have a delemma on your hands.

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