Wednesday, August 22, 2018

Wave and Surfing Speeds


William A. Wurts


Below is the progression of thought and calculations that led to these two equations.



A day or two after calculating wave speeds for a hydrofoil thread at Swaylocks, it occurred to me that the physics of maximum surfer speed are incredibly simple.  They are directly related to the speed of an object sliding down a frictionless slope or free-falling from the same initial height.  After the object has dropped or slid to the bottom, the final speed is identical for both free-fall and slope.  So, I compared final slide/drop speeds with wave speeds for various wave heights and realized that final sliding/falling speed is the maximum speed a surfer can reach on a big wave.

Final "speed" at the bottom of a frictionless slope is identical to final free-fall speed when initial height is the same.  But, the “velocity” is vertical for a free-fall and horizontal at the bottom of a slope.  Since speed is the same, the equation for calculating free-fall speed is simpler.  This speed assumes (actual) initial velocity (Vo) of zero
  
After that, ride angle (A) and transverse velocity calculations are simple.

Yep.  There are some difficult parameters to quantify, e.g. the classic debate over "pumping."

To begin sliding down a wave face, surfer shoreward velocity must first equal wave speed (C), net velocity = 0.  Essentially, a surfer's (actual) Vo at the top of a wave for take-offs and/or re-entries must = C.  (However, Relative Vo = 0; surfer shoreward speed minus wave speed equals zero).

To move back up the face of a wave after a drop, surfer shoreward speed has to be less than wave speed C.  By the time the surfer is back at the top, he/she must achieve wave speed C again to begin sliding down the face.  So, this roller coaster approach does not add additional starting speed for each new drop.  Again, maximum velocity is related to wave height.

The Vmax velocity vector is the average maximum sustainable velocity.  This is pretty much the straight-track drop and ride on a big wave.

Finally, my values are very close to those of LarryG's at Swaylocks -- independent results substantiating one another.  Our results were based on similar and/or related equations.

The form of surfable waves will range from plunging to spilling.

I posted this at Swaylocks today:



The upper limit speed of a surfer/surfboard on a wave is a simple calculation.  This upper limit speed assumes no drag/friction or upward movement of water in the wave face.  Actual maximum surfing speed would be slower.

The calculation is simple high school physics and algebra.  Maximum sustainable surfing speed cannot be greater than the free-fall speed (Vmax) of a surfer falling from a height equal to the height of the wave.

Vmax = square root of 2gh
where
g = gravity (32 ft/sec^2 or 9.8 m/sec^2)
h = height (in this case wave height)


This first table shows the upper limit Vmax speeds for several wave heights (Vmax, assumes no friction and no upward movement of water in the wave face.)


The minimum surfing speed cannot be less than the wave speed (C).  Surfable waves are shallow water waves.  Approximate wave speed can be calculated with the following equation.
C = square root of dg
where
d = depth
g = gravity (32 ft/sec^2 or 9.8 m/sec^2)

Water depth for breaking waves depends on wave types (Galvin, 1968; J. Geophys. Res.)
For plunging waves, d = 0.9h
For spilling waves, d = 1.2h
Where
d = water depth
h = wave height
Speeds for several heights of plunging and spilling waves are shown in this table.



The maximum angle of surfboard travel (A) relative the direction of wave travel is determined by the maximum surfing speed.
CosA = wave speed (C) divided by maximum surfing speed (Vmax)
CosA = C/Vmax
Assuming no friction or upward water movement in the wave face, (using high school trigonometry) approximate maximum values of ride angle A would be;
48 degrees for plunging waves,
and
39 degrees for spilling waves.

The following figure is a view of the velocity vectors -- from above the wave -- for Wave Speed (C), Transverse Velocity (Vtrans) and Maximum Velocity (Vmax).  Ride angle A is 48 degrees.


Vmax is the long velocity vector that allows the surfer to travel forward at wave speed (C) while traveling perpendicular to the direction of wave motion at transverse velocity (Vtrans).
Trigonometry is based on the geometry of right triangles.  Vmax is the hypotenuse of that right triangle and C is the side adjacent to ride angle A (39-48 degrees).  

Vtrans is the side opposite ride angle A.  Vtrans can be calculated using SinA (39-48 degrees).

SinA = Vtrans/Vmax

Sin 39 degrees = 0.629
Vtrans = 0.629 x Vmax

Sin 48 degrees = 0.743
Vtrans = 0.743 x Vmax

For a 25-foot wave, plunging or spilling, Vmax = 27.2 mph.
For a 25-ft plunging wave,
Vtrans = 20.2 mph
For a 25-ft spilling wave,
Vtrans = 17.1 mph
_____

I suspect pumping has greatest impact when surfer speed has dropped below Vmax.  On big waves, there is little or no pumping.  It is pretty much a straight-track, single drop. 

Certainly, there is a terminal velocity due to water drag.

If you measure the rider angle relative to the direction of wave motion, you can readily determine surfer velocity based on wave speed (C).
This is what mtb said in an earlier post.  (But I am thinking drag may negate this):
mtb wrote:
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  
Also I think LarryG's points regarding GPS measuring of pro-surfer velocities are good ones.  It is likely the GPS measurements are artificially high due centripetal/centrifugal acceleration being inaccurately measured by the sensor as velocity increases.
_____
27 August 2018
This was originally posted at the following link (Swaylocks):

https://www.swaylocks.com/comment/551632#comment-551632


Vo = initial velocity
Voh = initial horizontal takeoff velocity
Vor = relative initial takeoff velocity (Vor = Voh - C = 0)
C = wave-form speed
Vmax = maximum velocity when Vo = 0
Vult = ultimate upper limit horizontal surfing speed
The following will make the surfing speed discussion significantly more abstract, especially since we have already crossed into Relativity with initial take-off velocity Vo. We know that a surfer's initial horizontal takeoff velocity (Voh) equals wave speed C.

This new abstract concept took hold after showing "relative" initial takeoff velocity (Voris zero (Voh – C = 0).

While Vor = 0 at takeoff, Voh = C.  This opens the possibility of an additive horizontal speed resulting from the take-off drop, Vult = C + VmaxDrop velocity starts out vertical at the top but changes to horizontal by the bottom/trough of the wave.

To achieve Vult = C + Vmax, there can be no friction/drag or water movement toward and up the wave's face.

[I debated this concept with my brother over the phone today while he was driving out of town for a camping trip.  He is better at physics and math than I am.  My brother and I have concluded it is very likely that Vult = C + Vmax.]

Using the table below, Vult for a 10-ft (3.048-m) plunging wave is 28.7 mph (46.2 kph).  Ride angle would increase also.

Vult = 11.5 + 17.2 = 28.7 mph

_____
30 August 2018

My brother and I both agree that ultimate surfer speed on a wave is accurately described by by Vult = C + Vmax.
While not a perfect example, the following is a good example demonstrating the additive speed of a surfboard dropping down the face of a wave traveling at speed C.
A train is moving down the ttrack at 8.2 mph.  There is a skateboarder on top of a 5-ft ramp inside a car pulled by the train.  The stationary skateboarder and the ramp are moving along with the train at 8.2 mph.

Now the skateboarder leans forward and drops down the ramp.  At the bottom of the ramp, the skateboarder shoots forward at 12.2 mph (Vmax) towards the front of the train -- 12.2 mph faster than the train is moving.  Relative to the ground outside the train, the skateboarder is traveling at a speed of 20.4 mph (8.2 mph + 12.2 mph = 20.4 mph).
_____
31 August 2018

I am thinking an effective technology for measuring surfer/surfboard speed might be a hot-wire anemometer.  It could be placed in the leading edge of a center fin or just in front of a center fin (minimal sensor drag).
_____
1 September 2018


My calculations were to determine the absolute maximum (ultimate) instantaneous sufer speed that is possible while obeying the laws of physics.  My calculations also assume frictionless acceleration and velocity.  Wave height is assumed to be measured from trough to crest/peak -- a liquid skateboarder's ramp.  Also, I used Cyril J. Galvin's research (1968) for wave speed estimates.

I think most of us realize that a frictionless (ideal) wave, without water movement in the face, does not exist in the real world (i.e. the line-up).  Drag, as well as water movement toward and up the wave face, prevent this "upper limit" speed from ever occurring.

(Galvin, 1968; J. Geophys. Res.)
_____

For additional discussion about surfable wave speed, click the following link:
http://bgsurfphysics.blogspot.com/2016/09/surfable-wave-speed.html

Reference
Galvin, C.J.  1968.  Breaker Type Classification on Three Laboratory Beaches.  Journal of Geophysical Research, 73 (12), 3655



Sunday, September 4, 2016

Surfable Wave Speed

Using the equation below and Table 8-4 at the bottom of this post, the estimated shoreward speed of a good, surfable 6-foot wave would be approximately 11 mph.

Surfing waves would be shallow water waves.  You can calculate approximate shallow water wave velocity with the second equation below.


Inline image 1

Where v = velocity, g = gravitational acceleration, d = depth and λ is wavelength.

Based on the table below, I would say good surfing waves are "plunging" waves.  The table provides ratios of water depth to wave heights for different types of breaker waves.  A simple calculation where d = water depth and h = wave height can be used to estimate water depth for the different the types of wave described in the table.  For plunging waves, a 6-foot wave would break in 

d/h = 0.9
d/6 = 0.9
d = 0.9 x 6 = 5.4 ft of water.

Use 5.4 feet for d (depth) in the shallow water wave velocity equation above.  Velocity of a 6-foot wave moving toward shore would be aproximately 13.1 mph.

For an expanded discussion about wave and surfing speeds, click the following link:
http://bgsurfphysics.blogspot.com/2018/08/wave-surfing-speeds.html

Wave Velocities (C) for Selected Wave Heights and Types



Inline image 2

Saturday, May 2, 2015

Circular Velocity/Acceleration in Ocean Waves & Surfing Physics


Below are some figures/diagrams of circular velocity and acceleration. Circular/elliptical motion is observed in the water movement of deep and shallow water ocean waves.

Related links:

http://highered.mheducation.com/olcweb/cgi/pluginpop.cgi?it=swf::640::480::/sites/dl/free/0072826967/30425/14_04.swf::Fig.%2020.4%20-%20Orbital%20Motion%20in%20Shallow%20Water

http://www.acs.psu.edu/drussell/Demos/waves/Water-v8.gif


http://www.acs.psu.edu/drussell/demos/waves/wavemotion.html


The green line is the relatively constant speed of the ocean water moving along its circular or elliptical path.  The left and right walls of the square/rectangle describe the vertical motion at the front and back of the wave, respectively.  The top and bottom 
of the square/rectangle describe the horizontal motion at the top and bottom of the wave, respectively.


Circular Acceleration & Velocity





Elliptical Acceleration & Velocity





Red and Blue lines show how "speed" changes in the vertical and horizontal dimensions.



Surfboard Trim Dynamics


My latest grapics:

In order to take off, the surfboard must achieve wave(form) speed (C) before the wave passes under the surfer.  Wave slope affects the time (t) required to achieve C.  The surfboard's shoreward speed (Vs) must be Vs = C to remain in trim.  The surfboard's transverse velocity (Vt) along the wave's face must be greater than or equal to the transverse velocity of the breaking crest (Vbc).  After the wave has been caught, if shoreward surfboard velocity (Vs) drops below wave speed (C), the wave will pass under the surfer/surfboard leaving them both behind.

I think many mistake the angle of the surfboard relative to the X axis (transverse movement viewed from the shore) as the angle that affects the continued forward velocity.  The angle of the wave face slope relative to the Z axis (direction of forward/shoreward movement) provides the acceleration needed to maintain continuous critical velocity (gravity).  The slope of the face is the curved ramp (inclined plane).



I didn't add labels to the axes because I would have had to reduce figure size to fit them in.  The figure would get a bit busy to look at with labels added.

The colored curved lines represent the slopes of three waves, red being steepest and blue gentlest.  The colored vertical lines, tangential to the top of each curve, represent the wave height.  The long black arrow represents the point on the wave face where the surfer achieves a shoreward velocity equal to wave speed (C) on take-off.  For an individual/independent wave, it could also represent the point where wave slope provides the angle necessary to create the acceleration needed to maintain a shoreward speed (Vs) equal to wave speed C (necessary for trim).


The point where the velocity arrow intersects the curve on each wave is also representative of the horizontal distance traveled, and therefore, the time required to achieve waveform speed C, dropping down from the crest.  The length of the slope curve decreases as wave slope increases.  Net acceleration increases as slope increases (curved ramps/inclined planes).  Position on the wave-face slope determines where the shoreward velocity reaches C.