Hi Bill, from what I can gather your friend’s boat speedometer would have been measuring speed through the water. Since the water in wave is also moving forward to a certain degree I think you could probably add that speed to the across the water speed. So in conclusion I think you probably were traveling as fast as you thought you were.

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

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

Yes, you did give your reasons for choosing your definition of "Peel Angle". My complaint is simply that you chose to call the angle that you defined the "Peel Angle" since the name "Peel Angle" has already been established in the technical literature. For an example, see Fig. 3 in:

To make it even more confusing, your "Peel Angle" is the complement to the "Peel Angle" as used in the literature. Using the same name for different things (esp. within the same class of things) can easily lead to confusion (and interesting discussions when two persons carry on a discussion involving peel angle when neither realizes that the other is using a different definition).

This is my reply to questions about Surfer Speed, measured THROUGH THE WATER, as
opposed to the speed over the bottom (measurable with GPS devices).

Bill Thrailkill wrote on September 23rd, 2010, 7:49 PM (Post #9) about Bob Shepard's surfboard speed experiments at Sunset Beach and Waimea Bay. Bob and his friends were surprised that they weren't going as fast as they thought they
were on 15-20 foot waves.

The speed that their surfboard is travelling 'across the water', if they are maintaining the same relative position on the wave face (just ahead of the curl), is what they want to know. That is what I called "Vcurl" in my formulas.
I have already posted a couple essays explaining the derivation of the equations I used for finding the 'Maximum Surfer Speed' and 'Curl Speed', relative to 'Wave Speed'. They are based on the Energy Budget (Potential and Kinetic)
available to a surfer on any given wave.

You guys may want to review "Part 1" and "Part 2" of the essays on the Paipo Board website that Rod Rodgers created and maintains. Take another look at:

The "Curl Speed", Vcurl = SQRT(2g x Hb), where I use the value of g = 32.13550135 ft/sec^2, and Hb=Breaking Wave Height, in feet.

Please note that Hb must be 'True TOTAL Breaking Wave Height' (counting the Trough out in front). Wave physics doesn't use "Local Scale", (i.e. half-height measure), nor does it use "Slant Height" (double true height), or any other
half-ass 'surfer bias' wave-height estimation. Gotta be the real height!

For example: If a wave is "Head High" to a surfer trimming across the wave face, and if the surfer is in a normal, relaxed stance for a wave only that high (not a low-crouching big-wave stance), then, if he is average height in stature, say
5 ft 9 inches, how high is his head above the deck of the board?

How high is YOUR head above the floor when you assume that same surfing stance? If you don't know, you should try measuring your stance with a tape measure. Let's say it's about 6-9 inches less than your fully-erect measured height.
Then, "head high" is probably about 5 ft to 5' 3" for a 5 ft 9 in. surfer.

OK...Now you're on a "Head High" wave. The lip of the wave is at least 5 feet above your feet, on the deck of the board. BUT, is your board at the BOTTOM of the wave? I don't think so...

Well, then, how far up the face of the wave ARE you? When you are trimming across the wave, cruising along where the wave is still steep enough to keep you sliding downhill, you are NOT at the bottom of the wave..right? If you were too
low, you would come gliding to a mushy stop as the board stalls. You can't even reach the bottom of the wave...the Trough..., anyway. Nor would you want to!

You are almost certainly riding the part of the wave that is Above Sea Level, even when you're low on the wave. That height, "Hasl" can be measured from the beach by the "Line-Of-Sight" method. The trough is another 15-25% lower, below
sea level. That can only be measured directly, out in the water, or estimated photographically.

OK, then...how big IS that "head-high wave" when measured from the Bottom of the wave to the Top? It's more than 5 feet. Maybe a little more than 6 ft.

I measured lots of waves at a bunch of different surf spots, the first severalyears after I came to Hawaii. I found out that what was called "double-overhead" at Ala Moana Bowls and Makaha was actually about 12-13 feet. That's about 20-30% bigger!

Now, they use "Local Scale" or Hawaiian Scale, which is actually "Half-Meters", not feet. It is about half the True Total Height, or about 3/5ths of what it 'looks like', without the Trough.

If you call 'head high' 5 ft, that's just Hasl. True height is about 20-25% higher. If you call it "3", that's Local Scale, which is 3 half-meters, or 1.5 meters, or about 5 ft Hasl.The True Total Height, then, is TWICE Local Scale, or about 6 ft. That's Hb, 'Trough to Crest' height.

If you want to know how big the waves really are, just go out and actually measure a few. It's not that hard! You might be shocked. Or, go out and measure the water depth in the lineup. The "Breaker Depth Index", or BDI, is the ratio
of water depth divided by the wave height where they break: BDI = d/Hb.

Bob Shepard measured surfboard speeds through the water at Sunset Beach (triple-overhead), and at Waimea Bay (quadruple-overhead). At Sunset, he got speeds of 24-25 MPH, and at Waimea he measured speeds of about 28 MPH. These of course are NOT GPS speeds, but actual speeds across the wave face, i.e., speed relative to the water surface, on the moving wave form. GPS measures the resultant speed of the surfboard over the bottom, where Vsurfer^2 = Vcurl^2 + Vwave^2

If you look at the Right Triangle that depicts these 3 motions above, then, if you use the "Peel Angle" A, (as measured away from the wave crest, then SIN A = Vwave /Vsurfer, and COS A = Vwave /Vcurl.

But, if you use the Break Angle B (as measured away from straight off), then COS
B = Vwave /Vsurfer.
And, Sin B = Vwave /Vcurl.

In either case, Vcurl is the speed of the curl moving across the crest of the breaking wave, at a right angle to the motion of the wave. The Resultant of the Wave Speed and the Curl Speed is the "Surfer Speed", as depicted by the
hypotenuse of the triangle. Vcurl is speed across the water, and Vsurfer is speed across the bottom. GPS measures Vsurfer, but a boat speedometer on the board measures Vcurl.

If I use the Maximum Makeable Ride Angle or Break Angle B of 51.34019175 degrees, the Tangent of B = Tan(51.34019175 degrees) = 1.25

The Tangent is the ratio of (the SIDE OPPOSITE B /the SIDE ADJACENT to B) = Vcurl /Vwave.
Tan B = Sin B / Cos B = 0.780868809 /0.624695048 = 1.25, so Vcurl = 1.25 x Vwave. You're going 25% faster 'across the water' (with the curl), than the wave itself is moving toward the beach.

The GPS speed of the surfer, Vsurfer = Vwave / COS B = Vw / 0.624695048 = 1.600781059 x Vw.
So, you can go about 60% faster than the Wave Speed, on a fast-breaking wave. If the wave breaks any faster than that, you probably won't make it very far. The section will close out on you.

If I use Bob Shepard's Sunset Beach 'over-the-water' or curl speeds of 24-25 MPH, I can calculate the true height of the breaking waves he was on. If he was going across the wave on a max'ed out ride angle, then Vsurfer = Vcurl /Sin B =
24.5 MPH / SIN(51.34019175) = 24.5/0.780868809 =31.37530876 MPH.
Then, Vwave = Vsurfer x COS B = 31.37530876 x 0.624695048 = 19.6 MPH Hb, ft = (Vsurfer, MPH /7)^2 = (31.37530876 /7)^2 = (4.482186966)^2 = 20.09 feet

We can check this:
Find the wave speed and breaking water depth, d, for a Breaking Wave Height of 20.09 ft:
If d = 1.28 x Hb = 1.28 x 20.09 ft = 25.7152 feet of water.Then, Vwave, fps = SQRT(gd) = SQRT(32.13550135 x 25.7152 ft) = 28.746666...fps
So, Vwave, MPH = (15/22) x Vwave, fps = (15/22) x 28.74666666... = 19.6 MPH. Check!
Note that Vcurl = 1.25 x Vwave = 1.25 x 19.6 MPH = 24.5 MPH

At Waimea Bay, Shepard recorded 28 MPH speeds across the water. That's Vcurl, MPH.
So, Vsurfer = 35.85749573 MPH, and Vwave = 22.4 MPH. Hb = 26.24 ft, and d,ft = 33.5872 feet.
The so-called "20-ft lineup" at Waimea is just outside of "the Boil", (where it's nearly 31 feet deep and where "Real Waimea" starts breaking). It looks like '20 ft', but is actually about 24+ ft when it breaks at the Boil. Then,
Shepard's waves would have looked like about 21 or 22 ft, without the Trough.

On Sep 24,2010, at 5:51 AM, "mtb" posted Comment #10, where he said that:Surfboard or bodyboard speeds in waves up to double-overhead at Swami's were measured at a median speed of 19.7 MPH, and a mean speed of 20.4 MPH, (GPS); and at The Wedge, in "15 ft +" surf, speeds of about 28 MPH (GPS) were observed.

Since these are GPS speeds, they are speeds over the bottom, not speeds through the water.

At Swamis, the swell direction in the winter season is not going to give you the fastest-peeling waves,
but IF the waves were really a true 10 ft from trough to crest, then they would break in about 12.8 feet of water, so the wave speed, Vwave (fps) = SQRT(gd) = SQRT(32.13550135 x 12.8) = 20.28138105 fps. Then, the wave speed, Vwave (MPH) = (15/22) x Vwave (fps) = 13.82821435 MPH.

If we use the MEDIAN surfer speed of 19.7 MPH, then COS B = Vwave / Vsurfer = 0.701939815.
Remember I'm using the Break Angle "B" as measured away from straight-off, here. The Break Angle, B then is the Angle whose Cosine = 0.701939815, so B = 45.41716546 degrees.

If you prefer to use the Peel Angle "A", (or the Greek letter "alpha", preferred by "mtb") as measured away from the CREST of the wave, then you will use the Sine function instead of the Cosine function for the relationship
between the wave speed and the surfer speed: SIN A = Vwave / Vsurfer: Thus, SIN A = 13.82821435 / 19.7 = 0.701939815, so the Peel Angle A = 44.58284354 degrees.

Note that Angle A = 90 - B = 90 - 45.41716546 degrees = 44.58284354 degrees.

A wave peeling off at 45 degrees is a pretty decent ride, but not the fastest ride possible. It leaves you a bit more cushion of available speed to use for climbing and dropping, in other words, hotdogging!

So, how fast CAN a surfer go on a true 10-ft high wave? My formula for Maximum Surfer Speed is:

Vmax,(MPH), surfer = 7 x SQRT(Hb, ft). So for 10ft, Vmax = 7 x SQRT(10) = 22.13594362 MPH.
That's for a Maximum Makeable Break Angle of 51.34019175 degrees, (away from
straight-off), or... a Minimum Makeable Peel Angle of 38.65980825 degrees (away from the wave crest).

At the Wedge, the 4 fastest surfer speeds measured in "15'+ surf" were given as 27.6, 27.9, 28.0, and 31.0 MPH. I'll use 28 MPH. If I can assume that "15 ft+" is probably around 16 feet, then...Vmax, MPH = 7 x SQRT(16 ft) = 7 x 4 = 28 MPH! That's a perfect fit!

But, what about that one 31.0 MPH reading? How big could the wave have been if it was a max'ed out ride? If Vmax,surfer(MPH) = 7 x SQRT(Hb), then Vmax /7= SQRT(Hb), so Hb = (Vmax,MPH /7)^2.

Check: If Vmax = 28 MPH, then, Hb,ft = (28MPH /7)^2 = (4)^2 = 16 ft. Check!

Then for 31 MPH: Hb,ft = (31/7)^2 = (4.428571429)^2 = 19.6122449 ft for the wave height.

That's it for now...

I'd like to say "Thanks!" to all of you guys who responded and offered valuable observations and useful criticisms. I tried to be specific and thorough in my presentation, so that other people who care about such things, but who may not
be mathematically-inclined, could follow the reasoning and process. With your contributions, I feel confident that my formulas are pretty close to being useful for determining Surfer Speed.

MAHALO!!! Larry Goddard Honolulu, Hawaii Oct 4, 2010

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

PinnyPinny, Here's what Larry had to say: "The guy didn't say how big the waves were (or looked, to him) when he was
timing the travel time of the swells from Snappers to Kirra Groin. Also, he
didn't specify how he arrived at the average peel angle for the entire ride. Was
it measured from the swell line, or was it measured (or estimated or
calculated?) as so many degrees away from straight off?
If I use his 40 degree angle, measured away from the swell lines, then I
get about 23 KM/Hr wave speed, pretty close to his estimated wave speed of
around 24 KM/HR, but then I get a breaking wave height of nearly 4 meters (about
13 ft), which would look about like 10 ft to most surfers.

Since all the calculations are linear it isn't going to make too much difference if the figures were guestimated or not. All that will happen in a linear system is that the answer will move about a bit as the guesses about underlying factors change (wave angle and so on). The speed of the surfer is not going to double or triple unless the guesses are all way off. If I have time I'll do some Monte Carlo simulation to show this.

From a practical point of view, it seems that ground swells at a frequency of 6 per minute must be further spaced than ones at 10/minute. Does this mean that they are travelling slower or faster? This has an impact because the calculation results in surfers going faster on bigger waves. In my opinion wind swells are the fastest moving swells (which have highest frequency), so one would go faster (but have shorter rides).

Like most things that are measured (and a particular bugbear of mine), averages tell little of value. Point calculations of peak (top) speed, lowest speed and speed when travelling back towards the curl all are more interesting because they inform board design (and are likely to tell us what we already know: there are trade-offs).

I'm interested in whether top speed is greater than the speed achieved during free fall in a late take-off? The latter is calculable for most wave heights, with terminal velocity playing a part because of the board being supported in air, but the former?

Hi Bill, from what I can gather your friend’s boat speedometer would have been measuring speed through the water. Since the water in wave is also moving forward to a certain degree I think you could probably add that speed to the across the water speed. So in conclusion I think you probably were traveling as fast as you thought you were.

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

Yes, you did give your reasons for choosing your definition of "Peel Angle". My complaint is simply that you chose to call the angle that you defined the "Peel Angle" since the name "Peel Angle" has already been established in the technical literature. For an example, see Fig. 3 in:

http://www.coastalmanagement.com.au/R&D/reef_surfing/

To make it even more confusing, your "Peel Angle" is the complement to the "Peel Angle" as used in the literature. Using the same name for different things (esp. within the same class of things) can easily lead to confusion (and interesting discussions when two persons carry on a discussion involving peel angle when neither realizes that the other is using a different definition).

mtb

Here is a youtube clip of the 2nd half of a superbank ride, the wave size is an example of what I call 5 feet.

The wave is peeling fast at a 35 to 40 degree peel angle and the surfer travelling at 10 metres per second.

Time of the footage 1minute and 3 seconds, which is a 600 metre ride.

http://www.youtube.com/watch?v=34ma7YQer20

Howzit pinny, What we need to see ia some video of Malaaiia (spelling) on Maui if it's a speedy wave we want to look at. Aloha,Kokua

Aloha, Kokua

Larry sent me this earlier today -

"Here is my reply #3:

This is my reply to questions about Surfer Speed, measured THROUGH THE WATER, as

opposed to the speed over the bottom (measurable with GPS devices).

Bill Thrailkill wrote on September 23rd, 2010, 7:49 PM (Post #9) about Bob Shepard's surfboard speed experiments at Sunset Beach and Waimea Bay. Bob and his friends were surprised that they weren't going as fast as they thought they

were on 15-20 foot waves.

The speed that their surfboard is travelling 'across the water', if they are maintaining the same relative position on the wave face (just ahead of the curl), is what they want to know. That is what I called "Vcurl" in my formulas.

I have already posted a couple essays explaining the derivation of the equations I used for finding the 'Maximum Surfer Speed' and 'Curl Speed', relative to 'Wave Speed'. They are based on the Energy Budget (Potential and Kinetic)

available to a surfer on any given wave.

You guys may want to review "Part 1" and "Part 2" of the essays on the Paipo Board website that Rod Rodgers created and maintains. Take another look at:

Lines 136-249 in "Part 1" in: www.rodndtube.com/surf/info/SurferSpd/SurferSpeed_vs_WaveSpd_Pt1.pdf

and Lines 108-168 in "Part 2" in: www.rodndtube.com/surf/info/SurferSpd/SurferSpeed_vs_WaveSpd_Pt2.pdf

The "Curl Speed", Vcurl = SQRT(2g x Hb), where I use the value of g = 32.13550135 ft/sec^2, and Hb=Breaking Wave Height, in feet.

Please note that Hb must be 'True TOTAL Breaking Wave Height' (counting the Trough out in front). Wave physics doesn't use "Local Scale", (i.e. half-height measure), nor does it use "Slant Height" (double true height), or any other

half-ass 'surfer bias' wave-height estimation. Gotta be the real height!

For example: If a wave is "Head High" to a surfer trimming across the wave face, and if the surfer is in a normal, relaxed stance for a wave only that high (not a low-crouching big-wave stance), then, if he is average height in stature, say

5 ft 9 inches, how high is his head above the deck of the board?

How high is YOUR head above the floor when you assume that same surfing stance? If you don't know, you should try measuring your stance with a tape measure. Let's say it's about 6-9 inches less than your fully-erect measured height.

Then, "head high" is probably about 5 ft to 5' 3" for a 5 ft 9 in. surfer.

OK...Now you're on a "Head High" wave. The lip of the wave is at least 5 feet above your feet, on the deck of the board. BUT, is your board at the BOTTOM of the wave? I don't think so...

Well, then, how far up the face of the wave ARE you? When you are trimming across the wave, cruising along where the wave is still steep enough to keep you sliding downhill, you are NOT at the bottom of the wave..right? If you were too

low, you would come gliding to a mushy stop as the board stalls. You can't even reach the bottom of the wave...the Trough..., anyway. Nor would you want to!

You are almost certainly riding the part of the wave that is Above Sea Level, even when you're low on the wave. That height, "Hasl" can be measured from the beach by the "Line-Of-Sight" method. The trough is another 15-25% lower, below

sea level. That can only be measured directly, out in the water, or estimated photographically.

OK, then...how big IS that "head-high wave" when measured from the Bottom of the wave to the Top? It's more than 5 feet. Maybe a little more than 6 ft.

I measured lots of waves at a bunch of different surf spots, the first severalyears after I came to Hawaii. I found out that what was called "double-overhead" at Ala Moana Bowls and Makaha was actually about 12-13 feet. That's about 20-30% bigger!

Now, they use "Local Scale" or Hawaiian Scale, which is actually "Half-Meters", not feet. It is about half the True Total Height, or about 3/5ths of what it 'looks like', without the Trough.

If you call 'head high' 5 ft, that's just Hasl. True height is about 20-25% higher. If you call it "3", that's Local Scale, which is 3 half-meters, or 1.5 meters, or about 5 ft Hasl.The True Total Height, then, is TWICE Local Scale, or about 6 ft. That's Hb, 'Trough to Crest' height.

If you want to know how big the waves really are, just go out and actually measure a few. It's not that hard! You might be shocked. Or, go out and measure the water depth in the lineup. The "Breaker Depth Index", or BDI, is the ratio

of water depth divided by the wave height where they break: BDI = d/Hb.

Bob Shepard measured surfboard speeds through the water at Sunset Beach (triple-overhead), and at Waimea Bay (quadruple-overhead). At Sunset, he got speeds of 24-25 MPH, and at Waimea he measured speeds of about 28 MPH. These of course are NOT GPS speeds, but actual speeds across the wave face, i.e., speed relative to the water surface, on the moving wave form. GPS measures the resultant speed of the surfboard over the bottom, where Vsurfer^2 = Vcurl^2 + Vwave^2

If you look at the Right Triangle that depicts these 3 motions above, then, if you use the "Peel Angle" A, (as measured away from the wave crest, then SIN A = Vwave /Vsurfer, and COS A = Vwave /Vcurl.

But, if you use the Break Angle B (as measured away from straight off), then COS

B = Vwave /Vsurfer.

And, Sin B = Vwave /Vcurl.

In either case, Vcurl is the speed of the curl moving across the crest of the breaking wave, at a right angle to the motion of the wave. The Resultant of the Wave Speed and the Curl Speed is the "Surfer Speed", as depicted by the

hypotenuse of the triangle. Vcurl is speed across the water, and Vsurfer is speed across the bottom. GPS measures Vsurfer, but a boat speedometer on the board measures Vcurl.

If I use the Maximum Makeable Ride Angle or Break Angle B of 51.34019175 degrees, the Tangent of B = Tan(51.34019175 degrees) = 1.25

The Tangent is the ratio of (the SIDE OPPOSITE B /the SIDE ADJACENT to B) = Vcurl /Vwave.

Tan B = Sin B / Cos B = 0.780868809 /0.624695048 = 1.25, so Vcurl = 1.25 x Vwave. You're going 25% faster 'across the water' (with the curl), than the wave itself is moving toward the beach.

The GPS speed of the surfer, Vsurfer = Vwave / COS B = Vw / 0.624695048 = 1.600781059 x Vw.

So, you can go about 60% faster than the Wave Speed, on a fast-breaking wave. If the wave breaks any faster than that, you probably won't make it very far. The section will close out on you.

If I use Bob Shepard's Sunset Beach 'over-the-water' or curl speeds of 24-25 MPH, I can calculate the true height of the breaking waves he was on. If he was going across the wave on a max'ed out ride angle, then Vsurfer = Vcurl /Sin B =

24.5 MPH / SIN(51.34019175) = 24.5/0.780868809 =31.37530876 MPH.

Then, Vwave = Vsurfer x COS B = 31.37530876 x 0.624695048 = 19.6 MPH Hb, ft = (Vsurfer, MPH /7)^2 = (31.37530876 /7)^2 = (4.482186966)^2 = 20.09 feet

We can check this:

Find the wave speed and breaking water depth, d, for a Breaking Wave Height of 20.09 ft:

If d = 1.28 x Hb = 1.28 x 20.09 ft = 25.7152 feet of water.Then, Vwave, fps = SQRT(gd) = SQRT(32.13550135 x 25.7152 ft) = 28.746666...fps

So, Vwave, MPH = (15/22) x Vwave, fps = (15/22) x 28.74666666... = 19.6 MPH. Check!

Note that Vcurl = 1.25 x Vwave = 1.25 x 19.6 MPH = 24.5 MPH

At Waimea Bay, Shepard recorded 28 MPH speeds across the water. That's Vcurl, MPH.

So, Vsurfer = 35.85749573 MPH, and Vwave = 22.4 MPH. Hb = 26.24 ft, and d,ft = 33.5872 feet.

The so-called "20-ft lineup" at Waimea is just outside of "the Boil", (where it's nearly 31 feet deep and where "Real Waimea" starts breaking). It looks like '20 ft', but is actually about 24+ ft when it breaks at the Boil. Then,

Shepard's waves would have looked like about 21 or 22 ft, without the Trough.

On Sep 24,2010, at 5:51 AM, "mtb" posted Comment #10, where he said that:Surfboard or bodyboard speeds in waves up to double-overhead at Swami's were measured at a median speed of 19.7 MPH, and a mean speed of 20.4 MPH, (GPS); and at The Wedge, in "15 ft +" surf, speeds of about 28 MPH (GPS) were observed.

Since these are GPS speeds, they are speeds over the bottom, not speeds through the water.

At Swamis, the swell direction in the winter season is not going to give you the fastest-peeling waves,

but IF the waves were really a true 10 ft from trough to crest, then they would break in about 12.8 feet of water, so the wave speed, Vwave (fps) = SQRT(gd) = SQRT(32.13550135 x 12.8) = 20.28138105 fps. Then, the wave speed, Vwave (MPH) = (15/22) x Vwave (fps) = 13.82821435 MPH.

If we use the MEDIAN surfer speed of 19.7 MPH, then COS B = Vwave / Vsurfer = 0.701939815.

Remember I'm using the Break Angle "B" as measured away from straight-off, here. The Break Angle, B then is the Angle whose Cosine = 0.701939815, so B = 45.41716546 degrees.

If you prefer to use the Peel Angle "A", (or the Greek letter "alpha", preferred by "mtb") as measured away from the CREST of the wave, then you will use the Sine function instead of the Cosine function for the relationship

between the wave speed and the surfer speed: SIN A = Vwave / Vsurfer: Thus, SIN A = 13.82821435 / 19.7 = 0.701939815, so the Peel Angle A = 44.58284354 degrees.

Note that Angle A = 90 - B = 90 - 45.41716546 degrees = 44.58284354 degrees.

A wave peeling off at 45 degrees is a pretty decent ride, but not the fastest ride possible. It leaves you a bit more cushion of available speed to use for climbing and dropping, in other words, hotdogging!

So, how fast CAN a surfer go on a true 10-ft high wave? My formula for Maximum Surfer Speed is:

Vmax,(MPH), surfer = 7 x SQRT(Hb, ft). So for 10ft, Vmax = 7 x SQRT(10) = 22.13594362 MPH.

That's for a Maximum Makeable Break Angle of 51.34019175 degrees, (away from

straight-off), or... a Minimum Makeable Peel Angle of 38.65980825 degrees (away from the wave crest).

At the Wedge, the 4 fastest surfer speeds measured in "15'+ surf" were given as 27.6, 27.9, 28.0, and 31.0 MPH. I'll use 28 MPH. If I can assume that "15 ft+" is probably around 16 feet, then...Vmax, MPH = 7 x SQRT(16 ft) = 7 x 4 = 28 MPH! That's a perfect fit!

But, what about that one 31.0 MPH reading? How big could the wave have been if it was a max'ed out ride? If Vmax,surfer(MPH) = 7 x SQRT(Hb), then Vmax /7= SQRT(Hb), so Hb = (Vmax,MPH /7)^2.

Check: If Vmax = 28 MPH, then, Hb,ft = (28MPH /7)^2 = (4)^2 = 16 ft. Check!

Then for 31 MPH: Hb,ft = (31/7)^2 = (4.428571429)^2 = 19.6122449 ft for the wave height.

That's it for now...

I'd like to say "Thanks!" to all of you guys who responded and offered valuable observations and useful criticisms. I tried to be specific and thorough in my presentation, so that other people who care about such things, but who may not

be mathematically-inclined, could follow the reasoning and process. With your contributions, I feel confident that my formulas are pretty close to being useful for determining Surfer Speed.

MAHALO!!! Larry Goddard Honolulu, Hawaii Oct 4, 2010

This post was too jumbled so I deleted it. A better formatted version is immediately below.

Bob

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

Here's what Larry had to say:

"The guy didn't say how big the waves were (or looked, to him) when he was timing the travel time of the swells from Snappers to Kirra Groin. Also, he didn't specify how he arrived at the average peel angle for the entire ride. Was it measured from the swell line, or was it measured (or estimated or calculated?) as so many degrees away from straight off? If I use his 40 degree angle, measured away from the swell lines, then I get about 23 KM/Hr wave speed, pretty close to his estimated wave speed of around 24 KM/HR, but then I get a breaking wave height of nearly 4 meters (about 13 ft), which would look about like 10 ft to most surfers.

Since all the calculations are linear it isn't going to make too much difference if the figures were guestimated or not. All that will happen in a linear system is that the answer will move about a bit as the guesses about underlying factors change (wave angle and so on). The speed of the surfer is not going to double or triple unless the guesses are all way off. If I have time I'll do some Monte Carlo simulation to show this.

From a practical point of view, it seems that ground swells at a frequency of 6 per minute must be further spaced than ones at 10/minute. Does this mean that they are travelling slower or faster? This has an impact because the calculation results in surfers going faster on bigger waves. In my opinion wind swells are the fastest moving swells (which have highest frequency), so one would go faster (but have shorter rides).

Like most things that are measured (and a particular bugbear of mine), averages tell little of value. Point calculations of peak (top) speed, lowest speed and speed when travelling back towards the curl all are more interesting because they inform board design (and are likely to tell us what we already know:

there are trade-offs).I'm interested in whether top speed is greater than the speed achieved during free fall in a late take-off? The latter is calculable for most wave heights, with terminal velocity playing a part because of the board being supported in air, but the former?

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