Surfer & wave speed (measurement and technical considerations)

Last time I tried to post Larry’s speed document it got all jumbled. This time I am just posting links to Part 1 and Part 2.

http://www.rodndtube.com/surf/info/SurferSpd/SurferSpeed_vs_WaveSpd_Pt1.pdf

 

http://www.rodndtube.com/surf/info/SurferSpd/SurferSpeed_vs_WaveSpd_Pt2.pdf

 

For those like me whose brain hurts when I see formula, you’ll need to take precuations.

 

Bob

 

 

 

so if Vmax for a 25 foot wave, crest to trough, is 35 mph, this rather puts the kibosh on a certain ranter who, with his GPS-enabled buddies, were posting egregious speeds here a year or so ago.  The ranter was banned... the GPS posters maybe oughta been to.  If you believe LG

Here is Larry’s conclusions.

 

"Let’s try some examples:

For a 2-meter (true height) breaking wave, how fast could a surfer go on a fast section?
Vsurfer,km/hr = 20.40513366* SQUAREROOT (2) = 28.85721676 Km/hr

For a 3-meter wave, Vsurfer, km/hr = 20.40513366*SQRT(3) = 35.34272823

 For a 4-meter wave, Vsurfer, km/hr = 20.40513366*SQRT(4) = 40.81026732"

 

The links in the first post (above) contains the calculations used to generate these examples…

 

Bob

wheres roy?

 

Here are some numbers for the superbank here in australia.

 

Distance from the take off at Snapper rocks to the end of the rocky groyne at Kirra is 1.2 kilometres, in 2004 surfers were getting rides this long.

Average of the peel angle about 40 degrees.

The time it takes for a line of swell  to get from Snapper to the Kirra groyne is 2 minutes.

Speed is distance times time which works out that the wave peeled at a speed of 36 kilometres per hour or 10 metres per second.

Doing a bit of geometry you can work out that the swell was travelling at 24 kilometres per hour, 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.  

 

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?

 

The waves were 5 feet the day I timed the swells.

Regarding the peel angle.

If a line was drawn between the take off point at Snapper and the groyne at Kirra, the angle between it and a swell line would be 40 degrees.

But the waves do not peel exactly along this imaginary line, the first part of the wave peels faster at say an angle of 35 degrees then as the wave comes into Rainbow Bay the peel angle changes to about 45 degrees, then the peel angle decreases and the wave gets faster as the swell travels into Greenmount and across to Kirra groyne.

I remember seeing a photo of one of the days when the wave was peeling all the way, in the photo I counted 17 right handers, this photo was taken from the hill at Kirra.

17 right handers means 16 gaps between the lines of swell.

I checked my surf diary and on the day the photo was taken the Tweed wave rider bouy was recording an average wave period of 7 1/2 seconds.

7 1/2 seconds times 16 gaps equals 120 seconds equals 2 minutes, which is the time it takes a swell line to travel from Snapper to Kirra. 

 

 

Question 1:   How do you define the swell line? Is it a line
paralleling the alignment of the crest of the wave (which seems to be
the case)? Or is it a line perpendicular to to the alignment of the
crest of the wave (i.e. the direction of movement of the wave fronts toward shore)?

 

Question 2:  Your comment about a
decreasing peel angle being associated with a increasing speed, and
vice-versa, is contrary to Larry’s discussion that says an increasing peel
angle results in increasing speed, and vice-versa. Which claim is
correct?  Or (as I suspect) is your ‘peel angle’ the complement of
Larry’s ‘peel angle’ and we have a conflict of definitions of ‘peel
angle’?

 

Question 3:  You wrote:

I agree that the wave ‘peeled’ at a speed of 10 m/sec, assuming that by ‘peeled’ as used here means the motion of the point at which the curl breaks moves at a speed of 10 m/sec, or 36 km/hr (in a fixed coordinate system). But I don’t understand how you obtained the values of 12 and 24 km/hr for the wave and curl speeds. Also I can’t figure out how you can satisfy the Pythagorean Theorem with these numbers:

    (12)^2 + (24)^2  = 144 + 576 = 720

    sq_root (720) = 26.8 kilometers/hour, not 36 km/hr

Could you please provide a little more detail on how you obtained the values you presented?

 

FWIW, the speeds that I obtained are:

Speed of the surfer and of the motion of the break point of the curl (since they move together) = 10.0 m/sec

Speed of the advance of the wave crests towards shore: 7.66 m/sec

Speed of the motion of the breaking point along the wave crest: 6.43 m/sec

 

Note 1. These values could change a bit pending on the correct definition of ‘peel angle’.

Note 2. The Pythagorean Theorem is satisfied for the speeds I obtained.

Thanks.

mtb

 

In the early to mid 1960's, Bob Shepard was riding a pintail balsa gun, on the North Shore, that he had made.    The board was fitted with a boat speedometer that used a pitot tube pickup, at the base leading edge of the fin.   Many of us thought we were going close to 45 mph, on 15 foot Sunset, and 20 foot Waimea. But Bob was reporting 24/25 mph on the drop at Sunset, and 27/28 mph on the drop at Waimea.   We sure thought we were going faster.   Just shows that you can't always trust your senses.  The boards in use at that time were typically 10 to 11 feet long, and in my case 10' 5'' x 38 1/2 pounds.   Those big paddle in guns would really move, when driven by a significant wave.  Point is, we are going slower than we think we are.

 

Thanks for the info on the speeds Bob Shepard measured when riding large waves. In the early 70’s, the Omni board (Surfer Mag,V11,#6) used the same method (or an equivalent approach) to measure speeds obtained on smaller waves. I don’t remember the details of the speeds obtained, but my recollection is that they were in the teens to low 20’s.

These methods measure the speed of the board “through the water”. Some other methods (e.g. gps) measure the peak speed “over the bottom”. The speeds obtained with the two approaches are not always the same (for example when dropping into a wave, the speeds measured with the pitot set-up include the vertical component of speed, while the GPS measurements do not).

Over time, I’ve made over 70 measurements of peak speeds attained during a ride at several locations (primarily Swamis) with a GPS and riding waves ranging from a few feet to double overhead+. The median speed of the set of results was a speed of 19.7 mph and a mean speed of 20.4 mph. The standard deviation was 3.3 mph. The four fastest speeds recorded were 27.6, 27.9, 28.0, and 31.0 mph (by a kneeboarder, bodyboarder, surfer on a short gun, and a bodyboarder, respectively). Two of these speeds were recorded at the Wedge (bodyboarder, 15+ ft peak, 1.5X overhead), the other two at Swamis. The three slowest speeds were by a bodyboarder, a board surfer, and a kneeboarder at 14.5, 15.2, and 16.1 mph, respectively.

A quick question - are we talking speed over water? or Geographic?

I ask because a sailboat registering 4kts boat speed with its onboard knotmeter can simultaneously read 8kts on its GPS due to current, swell, etc. Since the surf zone is a far more dynamic environment as far as current, swell and craft direction; the variance could be much greater.

 

 

My referance is to speed over the water ONLY.  

 

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

 

Bob

 

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
 
 

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

If you want to see a REALLY fast wave, take a look at Maalaea's "Freightrains" on a good day. Maui, "The Valley Isle" funnels the tradewinds down the middle of the island, between two big mountains (Haleakala, on one side, is 2 miles high),  so don't expect very many glassy days there. But, those amplified tradewinds blow directly into the face of the waves...making for some pretty thrilling rides!

Take a look at these two videos:

www.youtube.com/watch?v=eMwxJO_to-Y&feature=related

And, while you're there, you might also want to watch "Maalaea 05 By Buzzy Kerbox".

 

This is truly "Pedal-to-the-Metal" surfing! Got a fast board?                    Larry Goddard