This Reply #2 is for "pinnypinny", who posted Comment #4 on Sep 21, 2010,
and for "mtb" who posted Comment #8 on Sep 23, 2010 - 7:12 PM.
I discussed my preferred method of measuring "Peel Angle", and gave my
reasons for doing so, in "Part 1" of my essay on Surfer Speed vs.Wave Speed and
Peel Angle", which was posted on Swaylocks by Bob Green, and also in Rod
Rodgers' "Waveriding Essentials" portion of the "WaveRidersPage". See:
www.rodndtube.com/surf/info/WaveRidersPage.shtml
In my "Part 1" essay, you might want to review lines 152-179. Also, in
"Part 2", take another look at my derivation of the "Maximum Makeable Peel
Angle" for waves of any size. That is summarized in lines 136-168.
On Sep 21, "pinnypinny" stated that the Tweed Waverider buoy on the day in
question was showing an "Average Period" of 7 1/2 seconds. That just means that
about 8 waves an hour were passing the buoy. If that is the Mean of all the
waves present, that probably wasn't the Swell Period. What did the buoy give for
the Swell Period? If you were surfing really nice waves, it probably was a
ground swell that you saw stacked up as nice lines coming in from way out far
off shore. Wind waves don't look like that.
What we need to know is: how long does it take ONE wave to peel off all the
way down the coast from Snapper Rocks to the Kirra Groin? Using the time it
takes for 16 wavelengths to come in to Snappers doesn't tell me that. If it
actually DID take 120 seconds on that particular swell direction and swell
height, then, of course it, and any surfer on it, was averaging 1200 meters in
120 seconds = 10 m/s.
If my "Maximum Makeable Wave Peel Angle" calculation is valid, then the
Wave Propagation speed is found by the formula: Vwave = Vsurfer x Cos B, where
the Cosine of Angle B = SQRT(1.28/3.28), or Cos B = 0.6246950476, and the Max
Makeable Peel Angle is 51.34019175 degrees, measured away from the
'straight-off', or Wave-Propagation Direction.
Note: If you prefer to use the peel angle as measured away from the swell
or crest of the wave, then the Wave Speed = Vsurfer x Sin A, where Angle A is
the complementary angle of B, that is, 90-B degrees, which is 38.65980825
degrees. The result is exactly the same: Sin A = 0.6246950476.
So, the Wave Propagation Speed in the breaker zone = Vsurfer x Cos B, or
Vsurfer x Sin A. Then, we get Vwave = 10m/s x 0.6246950476 = 6.246950476
m/s.
That Wave Speed is equivalent to 20.49524434 fps, or 13.97403023 MPH, or
22.48902171 Km/hr.
Remember, "pinnypinny" said the wave speed was "24 km/hr. That would be
48.2 degrees for Angle B.
Now, we can determine the water depth, d, where the wave breaks:
Since wave speed, in ft/sec, or Vwave = SQRT(gd). Solving for d, we get
(Vwave)^2 = gd, so...
d = (Vwave)^2 / g.
If I use my generic value of g = 32.13550135, then d,ft = (20.49524434
fps)^2 / 32.13550135
and get d = 13.07137039 feet, the breaker depth.
Now, we can find the Breaker Height, Hb: Hb = d,ft / 1.28, or, Hb =
10.21200811 ft (True Height).
Hmmm...The surf height was given as "5 ft". Could it be that he was using
"Local Scale", where "5" means the waves 'looked like' about 8 ft (or 3 ft
overhead)? If you include the trough, those '8 ft' waves are actually more like
10 ft, or TWICE Local Scale. If so, then his 10-meter-per-second rides were
about '5.1' Local Scale, or actually about 10.2 ft true total height.
Also, in "pinnypinny"'s summary, he wrote that "the swell was travelling
at 24 km/hr, which means that a surfer trimming ahead of the curl is travelling
at a speed of 12 kilometres per hour on a swell travelling at 24 kilometres per
hour".
Huh?
If the surfer is only going HALF as fast as the wave, the wave has passed
him up! But, if he is 'trimming ahead of the curl', he obviously HAS to be going
faster than the wave speed. Indeed, it was already determined that the surfer on
those waves was going 36 km/hr. So, his statement that the surfer was only going
12 km/hr was must've been an unintentional error. We forgive you; it IS kinda
mind-numbing after a while!
The essays that I wrote and Bob Green posted on Swaylocks, and which also
got added to Rod Rodger's website in the WaveRiders Page, explained the
trigonometry of wave and surfer motions. The cosine of the Peel Angle is the
ratio of Wave Speed / Surfer Speed. If you can measure or calculate any 2 of
those 3 elements, you can determine the unknown 3rd element. Simple in
concept...difficult in the execution.
Stay tuned...more replies to your 'Comments' are in the works.
Larry Goddard