Surfer & wave speed (measurement and technical considerations)

General Discussion
21

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

22

Good stuff!   Confirms my belief about ''true'' wave size.     Thanks.

23

Bill,

 

I’ll let Larry know about your post as he doesn’t check Sways frequently - the last thing he e-maied me was ‘If I write anything else for Swaylocks, it will be how to use my “Line-of-Sight
Method” of measuring actual wave heights…without getting wet!’. 

 

I was keener to hear his thoughts on what makes a board go fast which he has discussed with me in parts. I have spoken to a local shaper about having one of his designs replicated and Larry has been generous in answering a pile of my questions -speaking of which I’d also like to ask you about Bob Shepherd but will put this into a pm.

 

regards

 

Bob.

24

    Howzit Larry,Glad to see somebody knows how to spell the name. Wish we could find that footage of Owl from the 70's where the wave catchs up to him and just chews him up . Aloha,Kokua

25

HOWZIT, guys!

This is my 2nd attempt to post a reply to the comment by "Kokua" (posted October 5, 2010). When I finished composing my detailed reply, I clicked on "Preview comment", then ALL my text input just disappeared into cyberspace somewhere!!

This will be much shorter...

Owl Chapman got an 'epic tube ride' at Sunset Beach during the '70 -'71 Winter season. There were pictures and a story of that ride in Surfer Magazine, Vol 12, No.1 (March, 1971), on pages 36-41. Good luck trying to find a copy...they're collector's items, now...worth a hundred dollars...EACH!

I was living on the North Shore that winter, sharing a house near Pupukea with Midget Farrelly and his wife and baby  daughter. Remember Owl getting an insane tube ride at Sunset. That may be the ride that "Kokua" was referring to.

Aloha!

26

    Howzit Larry, No the video was at Malaala on Maui and I remember Sammy  Hawk  telling me they were going over there for that swell.There is a copy of the video out there in cyberland and I have seen it in the last few years but I can't remember where. Lets face it,Malaalaea has been called the fastest wave in the world and if you ahve ever seenit at it's best you would agree.Wish I could remember which movie it was in.Aloha,Kokua

27

Hi, guys!

I'll try to post this myself, rather than asking Bob Green to do it for me. He deserves a break...

The last couple days I've been evaluating the effect of using different values for the ratio of the Breaker Depth versus the Breaking Wave Height, given the same Wave Height of 25 feet.

 This ratio is known as the "Breaker Depth Index, which I'll call "BDI".

The deeper the water where the waves break, the faster the Wave Speed will be in the surf zone. Bigger waves break in deeper water, so they are moving faster.

In the open ocean, where the waves are in water that's deeper than about half of their deepwater wavelength, the Wave Propagation Speed is proportional to the Wavelength and the Period. The Wave Speed in knots, Ckts = Tseconds / 0.33.

But, in 'Transitional Depth water' that is less than about half of the open-ocean wavelength, the swell starts 'feeling the bottom', and begins slowing down. This is where the first refraction of the swell begins to take place. A 20-second swell has a deep-water wavelength over 2000 ft long, so it first starts slowing in water that's still over 1000 feet deep. At that depth, it's still moving at freeway speeds.

When the swell gets into water that's only about 1/20th of the original open-ocean wavelength, it's considered to be in 'Shallow Water'. In such shallow water, which could be about 100 feet deep for that long 20-second swell that's approaching, the wave speed is becoming mostly independent of the period, only affected now by the Depth of the water.

In 'Shallow Water', Wave Speed, Vwave = SQRT(g times d), where g=acceleration of gravity at that Latitude, and d is the water depth. But d = BDI times Hb, so, if I use feet, d,ft = BDI x Hb,ft.

I used 25 feet for the Breaking Wave Height, Hb,ft. Then, as I increased the Breaker Depth Index from unity (one, or 1) to (1 / 0.78) or BDI = 1.282051282, the Breaking Depth also increased, and therefore the Wave Speed ALSO increased.

 If the Maximum Makeable Break Angle, or Ride Angle, was calculated according to the Energy Budget formula, the Maximum Makeable Ride Angle, B (measured away from 'Straight Off' where B = 0 degrees) is: the Angle B whose Cosine = SQRT[BDI/(2+BDI)].

I discovered that as the 25-foot wave got faster in the deeper water, the Maximum Makeable Ride Angle got smaller (and the Peel Angle A got larger). Remember, A = (90-B) degrees, as it's measured away from the crest of the wave.

But, here's what's REALLY interesting: As I tried various values of BDI, the Curl Speed, i.e., "Vcurl", the speed of the surfer ACROSS THE WATER, remained the SAME! That's the speed you would measure using a boat speedometer, NOT the GPS speed, which is given in the formulas as "Vsurfer". Remember, Vsurfer is the speed relative to the BOTTOM, (or the Earth).

It seems that ONLY the Breaking Wave Height has any effect on how fast you can go!

That makes perfect sense, since it's the Height of the wave that gave you all that Energy to play with. You obtained all that Energy Budget at the Drop-In! How fast you CAN go is dependent ONLY on the Energy you have available to you!

For a 25-foot wave, I got a Maximum Vcurl speed of 27.33040833 MPH

For a 26.24' wave, I get a Maximum Vcurl speed of 28 MPH

To find Vcurl, max, in MPH, Use:

Vcurl, max (MPH) = 5.466081666 x SQUAREROOT(Hb, ft).

 

Comments, anyone?

28

 

As I understand it, your Vcurl is the speed at which the break point of the curl progresses along the length of the wave crest. If that’s the case, I don’t see how the curl speed has any limit – at least at the speeds we’re talking about in surfing. For example, a long-crested wave moving nearly perpendicular to shoaling isobaths can break virtually simultaneously along its length (I’m sure that most of us have seen long-crested curling waves break in which the curl moves along hundreds of feet down the length of the crest within a couple of seconds.)

Whether or not a surfboard can keep up is another question. You assume that the board and rider have an energy budget comprised of both kinetic energy and potential energy – and that total energy remains constant during a ride. This assumption is equivalent to assuming that there is no gain or loss of total energy during the ride, or that any loss of energy is compensated by an associated gain of energy of equal magnitude from another mechanism. It is relatively easy to demonstrate that there is a loss of energy – for example, when the board and surfer plane across a level sea surface and there is no source of energy they quickly slow to a stop (e.g. when coasting straight off out into the flats) rather than continuing at the same speed as at the end of the drop. Conversely, if the energy budget consisted solely of the potential energy available when starting to catch a wave, and there is no gain or loss of total energy during the ride, then if or when the surfer and board go up the face of the wave, they should come to a halt (relative to the wave face) as they reach the crest. Yet all of us have seen bodies and boards launched at speed into the air above the crest of the wave during a late kickout.

Attached is a graphic showing the output from a (1980’s to 1990’s) computer simulation model of surfing on a wave and how the speed achieved (steady-state conditions) is affected by how the surfer trims the board (“angle-of-attack”) and where one places the board on the face of the wave (“wave slope”). The contour lines are essentially your “Ride Angle” (smaller angle -> increasing speed). In this particular simulation, the optimum trim position (relates to AOA) is predicted to be about 11 degrees, and the optimum position on the wave is where the slope of the sea surface (measured inshore to offshore) is about 48 degrees, This resulted in a predicted speed of about 22.7 mph. Other predicted quantities (not shown) are: speed, wetted area, wetted length, and location of the surfer’s center-of-mass (AOA trim).

mtb

 

 

 

29

Hi, "mtb",

I have often wondered if a surfer, after dropping in on a wave, could continue his bottom turn all the way around 180 degrees, and then climb immediately all the back way up to the top of the wave with enough excess speed (or Momentum) to be able to launch himself up into the air ABOVE the wave, to an altitude that was even HIGHER than the top of the wave where he started.

I doubt it...

The fact they they CAN 'get big air' on waves farther down the line, where the wave is smaller, only shows that they still have enough Energy left, and excess speed on tap, to do it where the wave has lost some of its initial height.

However, if they COULD do it right after the 'drop-in-and-turn' phase of their ride, that would surprise me, especially if the wave height is undiminished after they make their their turn. Most surfers take off at the peak of the wave, where the shallow spot on the bottom precipitates the initial breaking of the wave, (i.e., in the lineup), where the top of the wave might be 20% higher than the shoulder immediately following their take-off.

The formula I originally described in the "Part 1" essay on "Surfer Speed...etc" was for GPS speed of a surfer, relative to the BOTTOM. So, "Surfer Speed" is the Resultant of two motions: Curl Speed and Wave Speed, moving at right angles to each other, where:

(Surfer Speed)^2 = (Curl Speed)^2 + (Wave Speed)^2

Note that Curl Speed is also the speed of a surfboard relative to the WATER, but only IF the board stays in the same relative position on the wave face. It is the speed you can measure with a boat speedometer on the board.

The fastest trim line on a wave is the HIGHEST line that you can maintain. Bodyboards and Paipo boards can trim WAY higher than a surfboard, and the riders that DO go up there are maximizing their speed.

The formula I derived for Curl Speed is:

Curl Speed = SQUAREROOT(2gHb),             or, Vcurl = SQRT(2g times Hb)

where, Vcurl is in ft/sec, g = 32 + (13550 / 99999) exactly, (or about  32.13550136, rounded),

and Hb = True, Total Breaking Wave Height, in feet (including the Trough).

If you know or can calculate Surfer Speed, Vcurl = Vsurfer x SIN B, (or =Vsurfer x COS A)

If you know or can calculate Wave Speed, Vcurl = Vwave x TAN B, (or =Vwave / TAN A)

The reason I posted all this stuff in the first place (on several websites) was to 'shake the bushes' and see if anybody out there was trying to measure surfboard speeds in a reliable manner. The worldwide response was not as great as I had hoped for, but what few responses I DID get were very helpful and I want to thank all you guys for your input.

By the way (I almost forgot)...

(for "mtb"): I have a question for you.

Did you use a Logarithmic Curve for the Wave Face, or a Circular function, to get the slope of the wave face? Terry Hendricks used a circular wave face shape for his calculations of board speeds. I always considered it a logarithmic curve from the trough up to at least the point where the wave goes vertical, just before breaking.

Also, on your interesting graph of "Track Angle (which I label Peel Angle "A" in my formulas) vs. Angle of Attack and wave-face Slope", there is a 'Minimum' Track Angle of somewhat >51 degrees shown at about 11 degrees AOA and 48 degrees Slope. That's very interesting to me, because MY "Maximum Makeable Ride Angle" or "Break Angle B" came out to about 51.34 degrees. That would be about 38.66 degrees for the Minimum Peel Angle.

What do you make of that? Could I be validating your graph? (or vice-versa?). Hmmmm...

Anyway, Many Thanks, again! Your input has been very helpful!

 

30

Howzit, again!

I wanted to clarify what I believed were the real-world ramifications of my "Surfer Speed, etc" formulas, which were based on a surfer's Total Energy Budget.

The formulas assumed that there was NO energy loss as the surfer continued his ride across a wave. That's patently impossible, of course. I'm looking for the Theoretical Maximum Energy here, which assumes NO LOSS.

It was also assumed that the surfer dropped in from the very TOP of the wave. That probably also is rarely the case (unless you're a paipo boarder, making a VERY late takeoff...not unusual at all!).

Then, it was assumed that the very BOTTOM of the wave is reached when making the bottom turn. Never happens! But, that was the assumption I made so I could use the ENTIRE Breaking Wave Height, Hb, in the energy calculations.

Note that I was looking for the Fastest-Possible "MAKEABLE" Curl Speed that a surfer could keep up with. That is a function of of the Wave Height and the Peel Angle. There is absolutely nothing in the formulas that put a limit on how fast the wave itself can peel across the crest of the wave. The only limit on Curl Speed is...Infinity! That's when you have a total close-out of the entire remaining portion of the breaking wave. The Peel Angle A at that moment is ZERO degrees (and the "Ride Angle", or Break Angle, B, measured away from 'straight-off, is 90 degrees.

The maximum possible energy is obtained if the surfer uses the ENTIRE wave height, and if there is NO loss of energy during the ride. In that unlikely case, what is the theoretical angle of ride CLOSEST to the wave crest that would enable the surfer to make the wave?

In the real world, the Maximum Ride Angle seems to be around 50 or 51 degrees away from straight-off (Angle B), or 39 or 40 degrees away from the crest of the wave (Minimum Peel Angle, A).

The wave can, and often will, peel off faster than the surfer can go across the wave. Ask the regulars at Maui's Maalaea Bay, or Supertubes, or Snapper Rocks...On some swell directions, the wave, or portions of it, are simply UNmakeable! You get closed out...somebody else further down the line drops in on your now-empty wave, and goes as far as HE can on it.

Some day, we will have rocket-powered boards that can go around close-out sections at will, but don't collide with another rocket-powered board. If the fuel tanks rupture, the resulting explosion will take out the entire lineup! Ha!

Let's go surfing, now...

31

Hi Larry,

You’re too fast for me!  I can’t respond as rapidly to your posts in this thread as quickly as you produce them. Hence you’re already posting a new post before I can even respond to your previous one. So I’ve decided to stop posting to this thread until I can respond to your set of posts via a single post. When I complete that task I’ll post my response here in the form of an attached MS Word *.doc file.

But first, a couple of comments relating to your most recent posting:

That’s almost correct. What is really assumed is that there is no net gain or loss of energy in the system. An obvious source of energy loss is via the various mechanisms that produce drag (e.g. skin friction, induced, form, wave train generation, etc.). But where does the input of energy into the system come from? It comes from gravity, the combined weight of the surfer and board, and the wave shape and the wave-associated motions of water.

The surfer has some degree of control over the rate of energy loss, and/or the rate of energy gain, by how he positions the board on the face of the wave, and how he trims the board for minimum drag. As long as the rate of energy loss due to drag is less than the rate of gain of energy gain, the total energy in the “energy budget” will be increasing and thus permitting increasing speeds, or allowing the board to be positioned increasingly higher on the wave, or some combination of the two. In the extreme (i.e. no energy loss to drag at all), the maximum speed that can be obtained builds indefinitely – or at least until limited by some other factor (such as the availability of wave face space).

mtb

ps. In support of this view, I call attention to the statement that you made that maximum speed is achieved by riding as high as you can on the face of a wave. That speed condition would seem to be impossible in your unchanging ‘Energy Budget’ approach (even in the absence of energy losses) since it is the potential energy in the budget that increases with increasing position of the wave face while the kinetic energy decreases accordingly. Thus one would expect the board speed to be increasing the lower one is riding across the face of the wave.

 

 

32

Hi, once again, "mtb",

Sorry about the frequency of my responses; I'm retired, and probably have more time on my hands than all you guys who still have time-consuming jobs to keep you busy.

I've been surfing since 1950, and riding paipos since 1964. My experience with the paipos has been that I can go significantly FASTER if I stay as HIGH on the wave as possible. Part of the reason, though, is that I maintain VERY close trim to minimize drag. I would guess that my angle of attack is less than 5 degrees, maybe even approaching zero degrees when at high planing speeds. My board has a planing hull, but it is also buoyant, so as to minimize wetted area and the amount of water being disturbed as I pass by.

If you go low on the wave, you will go slower. Gravity is your 'engine', but the steeper upper reaches of the wave gives you more 'drive' from that engine powering your ride. Try it yourself.

I have no trouble passing surfboards. The typical surfer on a longboard maintains excessively high angles of attack on most of his ride. All that drag keeps them from going as fast as they could. And the large amount of rocker on the board's bottom also means lots of drag.

In 1970 I told Dewey Weber and Harold "Iggy" Ige, his shaper, about my experiments at Makaha, Sunset Beach and Waimea Bay using flattened bottom curves and minimal nose kick for very short boards. My fastest boards had wide tails, shallow concave bottoms and a single skeg for minimal drag.

Iggy had shaped a VERY fast 6' 2" short board in 1969 that he rode in 12-ft surf at Malibu, and he had the fastest board in the water. That design became the Weber "Ski", which my surfing buddy, Jimmy Blears used to win the 1972 World Surfing Campionships at San Diego, California.

My Conclusion: Flat is Fast! Wanna go really fast? Ride a paipo board! And stay High!

Not convinced? Take a look at this video of "Flyin' Fearless Phyllis" Dameron at Waimea Bay:

www.youtube.com/watch?v=fj3Y39gu1HY&NR=1

 

 

 

33

I've seen paipo's pass full guns, on the drop, at Waimea.   Early to mid 60's.    Was that you?     Looked like big guitar pick.

34

Howzit, Bill,

The guy you saw at Waimea Bay on his own designed 'Guitar Pick" style skegless wood paipo in the '60s was John Waidelich. He was a pioneer of big-wave surfing on paipo boards. There is video of him and his paipo buddy (forgot his name) surfing big Sunset and Waimea.

Those wide-tailed paipos are the fastest thing in the water, at least on waves that aren't too big and fast to be caught by a paipo rider. We can only paddle and kick maybe 3-4 MPH on those short boards (they're about 1/3rd as long as a big-wave gun). A surfer can paddle maybe twice as fast, so they are limited to about 35 ft waves (or 40 ft if you're Greg Noll!). Any bigger than that, and you need to be towed in. A 50-foot wave is travelling nearly 31 MPH, about 4 or 5 times as fast as a surfer can paddle his big gun, so he needs some help from that Yamaha WaveRunner.

I didn't surf Waimea Bay until the beginning of the '69/'70 winter season, not long after I came over here. I lived at Makaha for the first 15 years. In fact, I lived on Makau street for 4 years (from '73-'77). You must have seen me in the lineup if you went out early in the morning. I was usually the first guy out (at 'first light').

Gotta go! Time for the Evening News.                Aloha!

35

Holy Moly Batman!    I used to live at 84-246 Makau St.     Did you know the kid with the artificial leg, named Lucky.   Nice friendly kid.  Rell Sunn also lived down the street.    Small world, eh?

36

Hi Larry,

The thrust of this thread seems to be changing. In your
earlier posts to this thread, you were presenting your ‘Total Energy
Budget’ model as means of estimating the maximum speed that can be
achieved. One of the assumptions implicit in the model is that it is a
‘closed’ system and the total energy of the system is constant (and is
equal to the total energy at the time the surfer catches a wave). I
commented that one of the model predictions you listed was that the
fastest speeds occur low on the wave face (as all the potential energy
is converted to kinetic energy), and vice-versa high on the wave face (i.e. all the kinetic
energy has been converted into potential energy). From your subsequent postings we seem to be in agreement that
observations indicate that the fastest speeds are associated with surfing
high on the wave face instead of low. Hence there is a serious discrepancy between the Total Energy Budget model prediction and direct observation. This led me to point out that the model is
not a closed system and that there are both losses of energy associated with
drag, etc., and energy gains associated with the interaction of the
board with gravity and the sloping face of the wave.

With respect
to energy loss, you responded that since you were interested in
predicting upper bound for the maximum kinetic energy it was
appropriate to assume no energy loss. That’s fair enough. But you have
not addressed how to arrive at an upper bound for the rate of energy
input and which doesn’t lead to a total energy value that continues to
increase with the passage of time (normally the magnitude would be limited by the rate of loss of energy equalling the rate of input–but you’ve assumed that to be zero).

mtb

37

[quote="$1"]

It seems that ONLY the Breaking Wave Height has any effect on how fast you can go!

That makes perfect sense, since it's the Height of the wave that gave you all that Energy to play with. You obtained all that Energy Budget at the Drop-In! How fast you CAN go is dependent ONLY on the Energy you have available to you!

... 

Comments, anyone?

[/quote]

Great stuff here!  I think your above assumption only applies to big single drop waves like Waimea.  The way that I see it, the surfers muscles are effectively an engine that converts potential energy into kinetic energy.  The engergy put into doing a bottom turn allows the surfer to go back up to the top of the wave for another drop.  Like a forced pendulum, if you make several little pushes at the right times, then you can really get going fast.  The image that comes to my mind is of Laird riding a wave at Jaws.  He seems to be moving the fastest after making a few turns and drops.  He is rythmically increasing his velocity as he travels at an angle across the wave.

38

Howzit, guys!

Thanks to all of your input in this discussion! We are getting closer to resolving one of "mtb"'s questions about whether it is possible that a surfer on a wave is able to actually offset the loss of Total Energy due to Drag, and maybe even INCREASE his total energy somehow, enabling him to go faster. But...HOW?

Could it be that by "pumping" or climbing-and-dropping and powering through a series of turns, he IS actually able to increase his TOTAL Energy Budget? Fascinating question!

When I first theorized about the wave dynamics (as related to surfing) back in 1976, right after I bought my first scientific calculator, I assumed that the energy available to the surfer was obtained at the drop-in, and that it was thereafter a closed system. You take off on the highest part of the wave, and after turning, the wave gradually loses height in most cases.

In "Part 1" of my essay on "Surfer Speed vs. Wave Speed and Peel Angle", I used a right triangle, where the SHORT side represents the speed of a surfer moving WITH the wave, at the same speed as the wave moving toward the beach, so I called it "Vwave". It's the same as the speed of the wave over the bottom, so it is a GPS speed.

The component of the surfer speed that is occuring at a right angle to the wave motion toward the beach is represented by the LONGER side of the triangle (longer that is, if the Ride Angle, or Break Angle is greater than 45 degrees, or the Peel Angle is less than 45 degrees). I agonized over the question as to what I should call that speed. Should it be labled "Vsurfer,curl,boat" to describe the surfboard speed through the WATER (i.e., "boat speed"), or maybe just "Vsurfboard,curl"?

But because I wanted to reserve the use of "Vsurfer" to describe the resultant motion of the surfer moving 'across' AND 'with' the wave simultaneously (GPS speed), I decided to simplify the label for the speed of a surfboard that is JUST barely keeping up with the curl, without being passed up by the wave, (i.e., a close-out section). I decided to call it "Vcurl".

Note to "mtb": Vcurl is the maximum speed of the SURFBOARD, while staying just ahead of the curl. It's NOT the maximum speed of the curl itself. The maximum speed of the curl is INFINITY!

So, I ended up with a formula that said: (Vsurfer)^2 = (Vcurl)^2 + (Vwave)^2

Or, if you use the equivalent of the 'squared speeds' for the parts of the right triangle:

3.28gH = 2gH + 1.28gH

The ratio between Vcurl and Vwave is the Tangent of the Maximum Ride Angle (measured away from straight off), so

TAN B = Vcurl / Vwave, so, (TAN B)^2 = (Vcurl)^2 / (Vwave)^2

(Tan B)^2 = (2gH) / (1.28gH), = 2 / 1.28, = 1.5625

Tan B = SQRT(1.5625)  = 1.25

Angle B = 51.34019175 degrees ( = Maximum Ride Angle)

In other words, Vcurl is 1.25 times Vwave. That's the same as your drop-in speed, at the bottom turn. That's what you start out with.

 

If you prefer to use the Peel Angle, A (as measured away from the crest of the wave), then The Tangent of A is Vwave / Vcurl, so...

(TAN A)^2 = (Vwave)^2 / (Vcurl)^2,  =(1.28gH) / (2gH),  = 1.28 / 2,  = 0.64

TAN A = SQRT(0.64), = 0.8

Angle A = 38.65980825 degrees (= Minimum Peel Angle)

Note that Angle B is the complement of A:   B = 90 - A,     so, B = 90 - 38.65980825 = 51.34019175 degrees.

I ride a paipo board, where every takeoff is a late takeoff. We make our turn while we are still high up on the face of the wave, where the slope is steep, greater than 45-50 degrees. Sometimes we make angling takeoffs, beginning our high-speed run as soon as we get moving on the wave. I believe we are no more than 20-25% down from the lip of the wave when we catch the wave. Yet we can go fast from the get-go...no bottom turn required, and no climb-and-drop or 'pumping' needed to reach high speeds. Up high is where we can really fly!

My question for "mtb": Is the speed potential on any given wave, then, only a function of wave slope steepness, or only of wave height, or a combination of BOTH? Any idea how we could use your graphs to answer this vexing question?

I always believed that for a 'steady-state' condition, (surfer maintaining a high track, just ahead of the curl, not down low, IN the tube), the maximum makeable speed would be dependent ONLY on the wave height. I guess it might also depend on how Steep the wave is where he's riding.

So, guys: maybe we haven't yet discovered "How Fast Can a Surfer GO?"

But it's been fun trying to find the answer!

Any more ideas?

39

While that’s a valid (and interesting) question, that’s not my primary
interest, which is:

What is the maximum sustained speed that can be achieved when racing across
the face of a breaking wave?

By “sustained” I mean that the surfer maintains his position
relative to the curl (steady-state conditon). An example of a wave
approximately fulfilling this condition would be traversing across the wave
face at the maximum possible speed with the curl chasing (or leading, if in the
tube) at identically the same speed. The model I referenced in an earlier post
in this thread addresses this question in considerable detail and makes
predictions on speed, wetted area, location of the rider on the board, etc.
(and the effect of where one rides on the wave face and trims the board) that
are generally compatible with observation (see attached graphics) . In the
subsequent discussion. I’ll refer to this model as the equilibrium or steady-state model.

An alternative model is your ‘Total Energy Budget’ model. I have a problem
with this model in that it predicts that the fastest speeds occur when riding
low on the face of a wave, while direct observation reveals that the maximum
speeds are associated with riding high on the face of the wave. So another of
my questions is how does one  reconcile this discrepancy? 

{Note: the
maximum sustained speed predicted by the equilibrium model generally increases as the slope of the wave increases. However, there is a limit to this trend at very large slopes as the hydrodynamic (planing) efficiency of
the surfboard begins to diminishes as the wave slope becomes very steep.}

But getting back to your question about the effect of pumping…

The input of new energy into the system via pumping increases the energy of
the system such that the loss of energy via drag, etc. no longer balances the input of energy into the system via “pumping”. So the system will
respond with an increase in speed to bring the system back into balance. In my
opinion, estimating the magnitude of this increase is a more difficult question
to answer than calculating the maximum steady-state speed of a surfer racing
across the face of a peeling wave.

However, one might be able to obtain a rough estimate (and an upper bound) of
the increase in speed by assuming that the transfer of energy from the rider
to the board via the contraction and extension of his leg muscles occurs without loss of energy. As
best I can recall, the rate of energy loss from the equilibrium model simulation shown in the graphics system is about 3-4 horsepower. The
maximum rate of power generation by an Olympic class athlete is (briefly) a bit
over a horsepower. Power associated with kinetic energy varies as the cube of
the speed. Hence assuming that the rate of loss of energy (due to drag) is 3.5
horsepower, and the rate of power input from the rider pumping is 1.0 horsepower, one
might expect that the (maximum) increase in speed resulting from the pumping
action is approximately 9 percent:

Vpump / Vno_pump = cube root (POWERpump/POWERno_pump) = (4.5/3.5)^0.33 =
1.09  

**
**

Agreed…(and is the case in the equilibrium simulation model.

Again…the hydrodynamic efficiency of the surfboard decreases with increasing wave steepness (via the changes in shape and size of the wetted area) and so the maximum sustained speed occurs before the slope approaches vertical. So it would seem that for both practical (read: “survival”) and theoretical reasons one shouldn’t go the last little bit to where the slope becomes vertical.

The speed potential is a function of wave slope steepness and wave height – and more.

There’s lots going on in the interaction between the board and the breaking wave. The model only incorporates first order modifications to adapt empirical observations and relationships for planar surfaces to the significant spatial curvatures present in breaking waves. So I don’t put a lot of faith into the numerical accuracy of the model predictions. But I think the simulations at least help to indicate–and in some cases, quantify–how the surfboard responds to changes in the characteristics of the wave and changes in the board design. The effect of a sloping sea surface on the “lift” characteristics of the board is especially interesting.

One thing that I found interesting is that it is difficult to make changes in a board design that will result in really significant increases in board speed as nature fights back by decreasing the rate of energy input as the speed of the board across the face of the wave increases…

Yes, it directly depends on the slope of the wave face (along the pathline of the surfer). The latter, in turn, depends on the peel angle and the “straight-off” steepness of the wave at the location of the rider and board.




40

Hi, guys!

My original formula for finding Drop-In speed used the conversion of Potential Energy to Kinetic Energy to find that the surfer reaches a speed at the bottom that's about 1.25 times the Wave Speed, that's all.

But the surfer doesn't want to STAY there, obviously, so using his excess speed, he turns his board back up face of the wave, going up to where the wave is steeper, and the 'drive' of gravity is stronger, whereupon his board gains BACK the speed he lost climbing back up the wave face, and THEN SOME. There's more power up there, so he can pick up more speed, ending up maybe going 1.6 times as fast as the Wave Speed (GPS speed). That's what I called "Vsurfer" in the formula.

The equivalent speed through (or OVER) the water is what I called "Vcurl", and is the "boat speed". It is also 1.25 times the Wave Speed, same as the speed aquired at the Drop-In.

Bob Shepherd's speedometer experiments showed 27-28 MPH on "20-foot" surf at Waimea Bay, which breaks in at least 31 ft of water, so is actually at least 24 ft wave heights, including the Trough. The Wave Speed is at least 27.33 MPH. Pretty close...

Other guys shared their speed measurements in this forum, and they seemed to agree with my formula, so I was happy about that.

But, Paipo riders don't ever make bottom turns. We turn while still high on the wave, and we have no problem reaching high speed soon after catching the wave. So, it's evident that you really don't need to drive down very far on the takeoff to pick up the speed you need to make the wave. Stand-up surfers only do a bottom turn because they HAVE to. If they turn at the top, they'll probably spin out on the steep upper face of the wave. (Think Waimea or Pipeline: only bodyboarders can pull that off!)

Take another look at the "Fearless Phyliss" video clip. She makes a super-late take-off, turns high, then proceeds to cruise past the guys on their surfboards, who are trimming lower on the wave, and therefore going slower.

If you trim high on a 5 foot wave, you can't go as fast as a similar trim position on a 20 ft wave. So, it's obviously a combination of Wave Slope steepness and Wave Height that governs ultimate speed on the wave. That's assuming the SAME type of board design, of course.

So, now I have one more question for "mtb":

I love the information that is in your graphs, but I STILL don't know what Wave Height gives that Ideal speed of 22.7 MPH you quoted for one of your graphs, or the Maximum Tracking Angle of near 51 degrees for 11 degrees Angle of Attack and 48 degrees Wave Slope.

Also, I wanted to know if you are using the Logarithmic Curve, (plotted on a Polar Coordinate diagram), for the shape of the wave face. In other words, how far up the wave face is that optimum 48 degree slope to be found?

Three times in the last 40 years in Hawaii, I have had the thrilling experience of riding in front of a near-Vertical Wave Face, where the tradewinds were stiff enough to hold the wave (and ME) up and allow me to track just below the pitching lip of the wave. (Waimea Bay, 20 ft; Klausmeyers,15 ft, and "Papa Nui", half a mile out on the South Shore, about 15 ft, or triple overhead.) Only paipo riders and bodyboarders can get away with that.

"mtb"... which factor do you suppose might be the most important for going fast: Wave Height or Waveface Slope?

By the way, my center of gravity is positioned over the "hump" on my bellyboards, which is  thickest at about 18 inches from the tail. Your graph shows the surfer CG at about 3 ft from the tail. I'm laying down on my board, which is about 4.25 to 4.5 feet long, or about half as long as a surfboard suitable for 15-20 ft surf. Hmmm... Interesting...!

The wetted length of my boards is about 45 to 48 inches, and the straight part of the bottom is about 2/3rds of the total board length. Lots of planing area there...about 3.2 square feet, same as the wetted area.

Thanks for your fascinating graphs! I'd love to see the underlying functions, but I realize that is your propietary intellectual property, so I won't ask.

Gotta go for now. Time for dinner...