BUT even though at top of wave vert velocity is zero you can still have down the line velocity. force vectors can add. maybe think sculling a rowboat. the waater has zero velocity but you move right along, even accelerataing from a dead stop to whatever top speed is...
And, due to less air drag, terminal velocity for a tucked KBer is higher than for a squatting FBer--we should be weaaring those duck looking bicycling helmets for air flow
20 interesting statemente re quad v tri
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Bryan Jackson
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Your down the line speed is still a result of either the speed generated during your initial drop in to the wave (due to gravity) or the movement of the wave towards shore and/or the water as it rises up when the swell reacts to the pull of the ocean bottom (a speed which is always much less than terminal velocity).
I’ll briefly discuss each factor but with the caveat that it will be in general terms (that is, no fancy physics lingo or equations since those are always a turn-off
) and with slightly rounded figures 
. That way everyone can follow along 
. (If you want more precise calculations I can give you the links to several web sites with the relevant formulas and you can do the math yourself. 
)
I’m also going to use kilometers and meters. If you want to translate kilometers into miles divide by a factor of 1.6 and multiply kilometers by 1000 to get meters.
Swells are generated by wind blowing across open ocean water (known as fetches) I think a category 5 hurricane force wind is something like 265 kph (notice I said “something like”) and that is quite strong. A 320 kph wind force would be considered to be a 'superstorm'
.
Thus, at best, an open ocean swell could never travel faster than those speeds and, of course, they really don’t due to other factors and considerations. In fact, the average speed of an open ocean swell in the Pacific Ocean is about 56 kph and somewhat less in the Atlantic.
In addition, once the swell enters shallower water (a continental shelf for example), it is slowed down considerably as it “feels” the drag from the sea floor. That’s why the waves on oceanic islands such as Hawaii, Fiji, etc., are so much faster. There is no continental shelf to slow them down. They rise up directly out of deep water
.
But let’s be generous and say the breaking wave (which is the one you will be riding) is traveling along at 56 kph. That’s as fast as you’ll be able to move (in a line heading straight towards the shore, that is). Of course, you want to be able to turn and maneuver, so you will harness and use the force of gravity by dropping down the face of the wave to then “speed” down the line
.
Terminal velocity is reached when the driving force, that is, gravitational attraction (the force of gravitational attraction is about 10 meters per second squared), is balanced by the resistive force, that is, fluid friction (the air if you are in a freefall.) A human body in a normal, non-aerodynamic freefall (arms and legs outstretched), reaches a terminal velocity of about 195 kph. In an aerodynamic position, that is, arms and legs tucked in, you could get up to about 320 kph. It reaches this terminal velocity in about 7 seconds time and over a vertical distance of about 430 meters.
If you jump off a five story building you will have reached a speed of about 71 kph by the time you hit the ground. Of course, unless you’re towing in most of us don’t surf five story waves, so we could really never achieve that kind of speed. Tow-in surfers achieve such incredible speeds precisely because they have that much larger wave face to utilize and work with.
Terry Hendricks, Ph.D., (a fellow kneeboarder, BTW) discusses the hydrodynamics/physics of surfboards at www.rodndtube.com/surf/info/Hydrodynamics.html. Although written in the late 60’s, it is still quite relevant and the last time I checked the basic physics involved haven’t changed since then . He calculates the top potential speed of a surfer, in an ideal situation (e.g., 20’ clean vertical face ala Pipeline, board generating good lift, minimal drag, etc.) when dropping in as no more than about 32 kph, max! Remember that is under ideal conditions .
But for the sake of the argument let's just say you can totally eliminate all drag and more or less do a controlled freefall down the face of a wave 5 meters high (about 15'). You will have attained a speed of approximately 36 kph when you reach the bottom (and good luck pulling off your bottom turn ), only 4 kph faster than the max speed calculated by Terry Hendricks.
So you really think you can surf faster than that provided by gravitational attraction alone (much less terminal velocity) and/or the moving wave. Wow
Now that I gotta see .
But I propose a simple experiment, easily performed, to settle this controversy .
Get together with a buddy on say a 20’ day at a break with fairly vertical faces (Pipeline would do nicely). Both of you takeoff on the same wave. One of you drops in straight down the face, maintaining contact with the wave. The other bails at the moment of takeoff, jettisoning board and doing a freefall through the air down the face. See who makes it to the bottom first. Report back as soon as you recover from your thrashing!
I’ll briefly discuss each factor but with the caveat that it will be in general terms (that is, no fancy physics lingo or equations since those are always a turn-off
) and with slightly rounded figures . That way everyone can follow along . (If you want more precise calculations I can give you the links to several web sites with the relevant formulas and you can do the math yourself. )I’m also going to use kilometers and meters. If you want to translate kilometers into miles divide by a factor of 1.6 and multiply kilometers by 1000 to get meters.
Swells are generated by wind blowing across open ocean water (known as fetches) I think a category 5 hurricane force wind is something like 265 kph (notice I said “something like”) and that is quite strong. A 320 kph wind force would be considered to be a 'superstorm'
.Thus, at best, an open ocean swell could never travel faster than those speeds and, of course, they really don’t due to other factors and considerations. In fact, the average speed of an open ocean swell in the Pacific Ocean is about 56 kph and somewhat less in the Atlantic.
In addition, once the swell enters shallower water (a continental shelf for example), it is slowed down considerably as it “feels” the drag from the sea floor. That’s why the waves on oceanic islands such as Hawaii, Fiji, etc., are so much faster. There is no continental shelf to slow them down. They rise up directly out of deep water
.But let’s be generous and say the breaking wave (which is the one you will be riding) is traveling along at 56 kph. That’s as fast as you’ll be able to move (in a line heading straight towards the shore, that is). Of course, you want to be able to turn and maneuver, so you will harness and use the force of gravity by dropping down the face of the wave to then “speed” down the line
.Terminal velocity is reached when the driving force, that is, gravitational attraction (the force of gravitational attraction is about 10 meters per second squared), is balanced by the resistive force, that is, fluid friction (the air if you are in a freefall.) A human body in a normal, non-aerodynamic freefall (arms and legs outstretched), reaches a terminal velocity of about 195 kph. In an aerodynamic position, that is, arms and legs tucked in, you could get up to about 320 kph. It reaches this terminal velocity in about 7 seconds time and over a vertical distance of about 430 meters.
If you jump off a five story building you will have reached a speed of about 71 kph by the time you hit the ground. Of course, unless you’re towing in most of us don’t surf five story waves, so we could really never achieve that kind of speed. Tow-in surfers achieve such incredible speeds precisely because they have that much larger wave face to utilize and work with.
Terry Hendricks, Ph.D., (a fellow kneeboarder, BTW) discusses the hydrodynamics/physics of surfboards at www.rodndtube.com/surf/info/Hydrodynamics.html. Although written in the late 60’s, it is still quite relevant and the last time I checked the basic physics involved haven’t changed since then
. He calculates the top potential speed of a surfer, in an ideal situation (e.g., 20’ clean vertical face ala Pipeline, board generating good lift, minimal drag, etc.) when dropping in as no more than about 32 kph, max! Remember that is under ideal conditions .But for the sake of the argument let's just say you can totally eliminate all drag and more or less do a controlled freefall down the face of a wave 5 meters high (about 15'). You will have attained a speed of approximately 36 kph when you reach the bottom (and good luck pulling off your bottom turn
), only 4 kph faster than the max speed calculated by Terry Hendricks.So you really think you can surf faster than that provided by gravitational attraction alone (much less terminal velocity) and/or the moving wave. Wow
Now that I gotta see .But I propose a simple experiment, easily performed, to settle this controversy
.Get together with a buddy on say a 20’ day at a break with fairly vertical faces (Pipeline would do nicely). Both of you takeoff on the same wave. One of you drops in straight down the face, maintaining contact with the wave. The other bails at the moment of takeoff, jettisoning board and doing a freefall through the air down the face. See who makes it to the bottom first. Report back as soon as you recover from your thrashing!
Last edited by Bryan Jackson on Mon Jun 28, 2004 6:15 pm, edited 2 times in total.
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DrStrange
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So...a water skier who carves across the wake and goes from tow rope length behind the boat to nearly even with the tail of the boat, thereby catching up to said boat, is still only going at same or less speed than the boat? I would think that for short time the skier would have TOTAL speed (not just speed vector in line with boat travel) faster than boat speed? 
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Bryan Jackson
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First off, there were a few errors in my discussion regarding gravitational acceleration . None fatal to my argument but embarrassing nonetheless 
. The result of writing under a self-imposed deadline (I wanted to post it before going to work that day), thus not being able to thoroughly check my calculations and figures. (Hey what can I say. I wasn’t a physics major. It’s just something that interests me.)
Most embarrassing; I inadvertently gave the vertical distance and time needed to reach terminal velocity for, all of things, cats, not humans ! (The vertical distance and time needed for a human to reach terminal velocity is about 430 meters and about seven seconds. Not 5 stories and less than 2 seconds as stated in my posting).
An interesting phenomenon regarding cats is that they suffer far less serious injuries when they fall from heights of greater than 5 stories than, say, 2-3 stories ! It seems that as long as they sense they are accelerating they are frightened and tense up, holding their legs rigidly underneath them, which transmits the full force of hitting the ground to their body.
But once they reach terminal velocity (at about 5-6 stories distance), since they no longer experience the sensation of falling they relax. This is critical because being in a relaxed state helps them absorb the shock of their landing much better. They also then spread out their legs (like a flying squirrel), which helps slow them down a bit. (BTW, the terminal velocity of a cat is about 97 kph in an aerodynamic mode and with limbs outstretched only about 68 kph.)
Anyway I had the nagging feeling in the back of my mind that I made some mistake(s) in my posting , so I reread it at work today (on a break, of course) and cringed a bit. I then stopped off at the library on my way home to consult some physics texts. I have now made the appropriate corrections to that post . Corrections which, once again, do not materially affect my argument (that is, that a surfer cannot possibly go faster than that afforded by gravitational attraction).
To answer Dr. Strange’s objections, (i.e., a water skier catching back up to the boat on a hard arcing turn). Well the answer to that involves going into a discussion of centripetal force and angular momentum, which I don’t really feel like getting into at this time
. (Considering the comments of Kripchik, the expression “casting pearls before swine” somehow comes to mind ). Perhaps someone else more astute in physics and with too much “time” on their hands would like to explain these phenomenon 
.
Suffice it to say that the two situations are really not the same. If nothing else, just consider the simple fact that the water skier is “attached” to the boat by the rope and thus has an external source of propulsion, while the surfer on the wave has only gravity and/or the wave to propel him/her forward . (If you don’t believe me, then please consult a basic physics text and look up the two aforementioned subjects for a full explanation of what is going on with the water skier.)
Finally, there is an additional factor to be considered in calculating the top speed (on a wave, of course) attained by a surfer. That is, that the terminal velocity of a surfer riding a wave in a ‘normal’ manner is far, far less than that of a surfer in freefall through the air (i.e., down the face of a wave without only minimal contact with the wave itself).
The wave not only moves towards the shore but also rises up beneath you, and you seem to descend at a steady pace. In fact, when your movement down the wave‘s face is matched by the rising movement of the water you don’t actually descend as much as you ‘hover’ in place
. (To put it in fancy terms, you have reached a ‘dynamically stabilized equilibrium’! )
If you ascend the wave (turn upwards), you reach a higher and faster level. If you descend (turn downwards), you reach a lower and slower level. Regardless of these manuevers, you can never exceed the terminal velocity of a surfboard moving across/through the water .
And since even the fastest (paddle-in) board in existence must still move through/across the wave face (Laird Hamilton has that hydrofoil thing-a-ma-jiggy that rises up out of the water ), that means it will always be slower than a surfer in freefall down the wave’s face 
.
. None fatal to my argument but embarrassing nonetheless . The result of writing under a self-imposed deadline (I wanted to post it before going to work that day), thus not being able to thoroughly check my calculations and figures. (Hey what can I say. I wasn’t a physics major. It’s just something that interests me.)Most embarrassing; I inadvertently gave the vertical distance and time needed to reach terminal velocity for, all of things, cats, not humans
! (The vertical distance and time needed for a human to reach terminal velocity is about 430 meters and about seven seconds. Not 5 stories and less than 2 seconds as stated in my posting). An interesting phenomenon regarding cats is that they suffer far less serious injuries when they fall from heights of greater than 5 stories than, say, 2-3 stories
! It seems that as long as they sense they are accelerating they are frightened and tense up, holding their legs rigidly underneath them, which transmits the full force of hitting the ground to their body.But once they reach terminal velocity (at about 5-6 stories distance), since they no longer experience the sensation of falling they relax. This is critical because being in a relaxed state helps them absorb the shock of their landing much better. They also then spread out their legs (like a flying squirrel), which helps slow them down a bit. (BTW, the terminal velocity of a cat is about 97 kph in an aerodynamic mode and with limbs outstretched only about 68 kph.)
Anyway I had the nagging feeling in the back of my mind that I made some mistake(s) in my posting
, so I reread it at work today (on a break, of course) and cringed a bit. I then stopped off at the library on my way home to consult some physics texts. I have now made the appropriate corrections to that post . Corrections which, once again, do not materially affect my argument (that is, that a surfer cannot possibly go faster than that afforded by gravitational attraction).To answer Dr. Strange’s objections, (i.e., a water skier catching back up to the boat on a hard arcing turn). Well the answer to that involves going into a discussion of centripetal force and angular momentum, which I don’t really feel like getting into at this time
. (Considering the comments of Kripchik, the expression “casting pearls before swine” somehow comes to mind ). Perhaps someone else more astute in physics and with too much “time” on their hands would like to explain these phenomenon .Suffice it to say that the two situations are really not the same. If nothing else, just consider the simple fact that the water skier is “attached” to the boat by the rope and thus has an external source of propulsion, while the surfer on the wave has only gravity and/or the wave to propel him/her forward
. (If you don’t believe me, then please consult a basic physics text and look up the two aforementioned subjects for a full explanation of what is going on with the water skier.)Finally, there is an additional factor to be considered in calculating the top speed (on a wave, of course) attained by a surfer. That is, that the terminal velocity of a surfer riding a wave in a ‘normal’ manner is far, far less than that of a surfer in freefall through the air (i.e., down the face of a wave without only minimal contact with the wave itself).
The wave not only moves towards the shore but also rises up beneath you, and you seem to descend at a steady pace. In fact, when your movement down the wave‘s face is matched by the rising movement of the water you don’t actually descend as much as you ‘hover’ in place
. (To put it in fancy terms, you have reached a ‘dynamically stabilized equilibrium’! )If you ascend the wave (turn upwards), you reach a higher and faster level. If you descend (turn downwards), you reach a lower and slower level. Regardless of these manuevers, you can never exceed the terminal velocity of a surfboard moving across/through the water
.And since even the fastest (paddle-in) board in existence must still move through/across the wave face (Laird Hamilton has that hydrofoil thing-a-ma-jiggy that rises up out of the water
), that means it will always be slower than a surfer in freefall down the wave’s face .Group/phase speeds, and shallow-water wave speeds
Some clarification:
However, unless one is riding open ocean waves with one of Laird et.al.'s hydrofoils, you're riding "shallow water waves" and the speed becomes dependent on the depth of the water and the height of the wave. An approximate equation for the speed of progression then is:
V ~= square-root (g x (0.75H +h))
where H is the wave height (trough to crest), h is the depth of the water, and g = gravitational acceleration = 32.2 ft/sec^2. Many factors influence the ratio between the water depth and the wave height at the point of breaking, but typical values for the ratio H/h range from about 0.8 to 1.2.
Source: Oceanography and Seamanship. Wm. G. Van Dorn., Dodd, Mead, and Co., NY (highly recommended as a introduction to oceanography)
As an example, let's assume a ratio of 1.0. In that case, for a 8 foot high wave, the speed of progression of the wave at the point of breaking will be about 21 ft/sec (6.5 m/s), or 14.4 mph (24 kph).
Free fall speed from a height of 8 ft (starting with no vertical component of velocity and disregarding air friction) is given by the equation:
V= square-root (2*g*H)
For a free fall distance of 8 ft, that equals 22.7 ft/sec (remarkably close to the speed of translation of the wave...think there's a connection somewhere there with regard to the wave breaking?).
So if you're moving with the wave toward shore, but not laterally across the face of the wave, and are right at the top of the wave, and you get pitched out by the lip and free-fall to the bottom--then when you contact the sea surface at the bottom of the wave, your speed relative to the earth will be:
Speed (total) = square-root( V(horizontal)^2 + V(vertical)^2)
= square-root( (21^2 + 22.7^2)) = square-root(956)
= 31 ft/sec (9.4 m/sec) = 21 mph (~ 35 kph)
Actually, your speed will be somewhat higher since the water being pitched out undergoes a strong horizontal component of acceleration and increase in speed, relative to the crest of the wave, in the process of pitching out (e.g. check out the studies on the characteristics of wave-breaking by Grilli (at URI, I think)).
Your speed relative to the water you impact will be similar, but not identically the same, due to the motion of the sea surface relative to the bottom.
Now let's say you're racing across the face of the wave as well. Studies show (e.g. K. Black) that the angle the path of an expert surfer makes "over the bottom", relative to the alignment of the crest of the wave, ranges between 30 and 45 degrees. Assuming 30 degrees for an 8 ft wave (45 is more appropriate to bigger waves), the component of the surfer's speed parallel to the crest (and orthogonal to the velocity of progression of the wave toward shore) would be about 0.87 x 21 ft/sec = 18 ft/sec.
Now suppose that you're in trim and racing across the face of the wave, when the wave suddenly becomes steeper and you get carried up the face of the wave, your vertical component of velocity falls to zero just as you get to the crest, consequently you get pitched out, and you land at a sea surface elevation close to that of the trough of the wave. Now your total speed at the point of impact will be about:
Speed = square-root (21^2 + 18^2 + 22.7^2)
= 35.9 ft/sec (11 m/sec) = 24.4 mph (41 kph)
Some might wish to argue that you will lose some of your lateral speed while being carried up the face of the wave due to your increase in potential energy. However, that is not true since it is the increased vertical component of the water velocity (relative to the earth) that carries you upward, and the lateral component of velocity of the kneeboarder (or any surfer) if he maintains his path, is orthogonal to the force of gravity--i.e. the only force acting to retard that component of the surfer's velocity is the result of friction.
MT
That's the group speed for a swell with waves characterized by a dominant period of about 20 sec. The "group speed" is the collective speed of progression of the group of individual waves (i.e. it's the speed you should use when trying to predict the arrival of a swell).Bryan Jackson wrote:... the average speed of an open ocean swell in the Pacific Ocean is about 56 kph and somewhat less in the Atlantic.
A surfer rides a specific wave. That moves at the phase speed, not the group speed. In deep water (depth > 3 x wave-length of the wave) the phase speed is equal to twice the group speed--or in this case at 112 kph....But let’s be generous and say the breaking wave is traveling along at 56 kph.
However, unless one is riding open ocean waves with one of Laird et.al.'s hydrofoils, you're riding "shallow water waves" and the speed becomes dependent on the depth of the water and the height of the wave. An approximate equation for the speed of progression then is:
V ~= square-root (g x (0.75H +h))
where H is the wave height (trough to crest), h is the depth of the water, and g = gravitational acceleration = 32.2 ft/sec^2. Many factors influence the ratio between the water depth and the wave height at the point of breaking, but typical values for the ratio H/h range from about 0.8 to 1.2.
Source: Oceanography and Seamanship. Wm. G. Van Dorn., Dodd, Mead, and Co., NY (highly recommended as a introduction to oceanography)
As an example, let's assume a ratio of 1.0. In that case, for a 8 foot high wave, the speed of progression of the wave at the point of breaking will be about 21 ft/sec (6.5 m/s), or 14.4 mph (24 kph).
Free fall speed from a height of 8 ft (starting with no vertical component of velocity and disregarding air friction) is given by the equation:
V= square-root (2*g*H)
For a free fall distance of 8 ft, that equals 22.7 ft/sec (remarkably close to the speed of translation of the wave...think there's a connection somewhere there with regard to the wave breaking?).
So if you're moving with the wave toward shore, but not laterally across the face of the wave, and are right at the top of the wave, and you get pitched out by the lip and free-fall to the bottom--then when you contact the sea surface at the bottom of the wave, your speed relative to the earth will be:
Speed (total) = square-root( V(horizontal)^2 + V(vertical)^2)
= square-root( (21^2 + 22.7^2)) = square-root(956)
= 31 ft/sec (9.4 m/sec) = 21 mph (~ 35 kph)
Actually, your speed will be somewhat higher since the water being pitched out undergoes a strong horizontal component of acceleration and increase in speed, relative to the crest of the wave, in the process of pitching out (e.g. check out the studies on the characteristics of wave-breaking by Grilli (at URI, I think)).
Your speed relative to the water you impact will be similar, but not identically the same, due to the motion of the sea surface relative to the bottom.
Now let's say you're racing across the face of the wave as well. Studies show (e.g. K. Black) that the angle the path of an expert surfer makes "over the bottom", relative to the alignment of the crest of the wave, ranges between 30 and 45 degrees. Assuming 30 degrees for an 8 ft wave (45 is more appropriate to bigger waves), the component of the surfer's speed parallel to the crest (and orthogonal to the velocity of progression of the wave toward shore) would be about 0.87 x 21 ft/sec = 18 ft/sec.
Now suppose that you're in trim and racing across the face of the wave, when the wave suddenly becomes steeper and you get carried up the face of the wave, your vertical component of velocity falls to zero just as you get to the crest, consequently you get pitched out, and you land at a sea surface elevation close to that of the trough of the wave. Now your total speed at the point of impact will be about:
Speed = square-root (21^2 + 18^2 + 22.7^2)
= 35.9 ft/sec (11 m/sec) = 24.4 mph (41 kph)
Some might wish to argue that you will lose some of your lateral speed while being carried up the face of the wave due to your increase in potential energy. However, that is not true since it is the increased vertical component of the water velocity (relative to the earth) that carries you upward, and the lateral component of velocity of the kneeboarder (or any surfer) if he maintains his path, is orthogonal to the force of gravity--i.e. the only force acting to retard that component of the surfer's velocity is the result of friction.
MT
Experience gained is in proportion to equipment ruined.
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Beeline2.0
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willli
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DrStrange:
From personal experience having waterski'd since youth, you don't have to be directly behind the boat to realize its very easy to go faster than the tow boat, especially using say 50 feet of tow rope. with a slalom ski you can easily pass the boat out on the "whip" and when the line goes slack post a turn that gives you a moment to set your back and arms for the charge across the wake. viewed from above the skier will be making incredible speed going side to side while the tow boat makes constant speed in a straight line. I know. I still have the local record of eight cartwheels on top of the water falling while slalom skiing behind a tow boat going 40kts.
From personal experience having waterski'd since youth, you don't have to be directly behind the boat to realize its very easy to go faster than the tow boat, especially using say 50 feet of tow rope. with a slalom ski you can easily pass the boat out on the "whip" and when the line goes slack post a turn that gives you a moment to set your back and arms for the charge across the wake. viewed from above the skier will be making incredible speed going side to side while the tow boat makes constant speed in a straight line. I know. I still have the local record of eight cartwheels on top of the water falling while slalom skiing behind a tow boat going 40kts.
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DrStrange
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Willi--a little rhetorical action from me. The wave is the boat and the fin and rails are the tow rope??? I still think it is possible to go faster than wave speed. Saw Herbie Fletcher at Malaeaa (sp?) back in the early Brewer days. 6-8 foot faces. He was going unbelievably fast. Faster than anyone else. Only one making waves (about 1 in 10 at that). Only one way to really settle this one is to get a radar gun out there and shoot from different angles to get absolute surfer speed and wave speed.
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Bryan Jackson
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Good posts, all (no insults or rancor ).
Thanks, Willi, for the more involved intricate calculations. As I said, my discussion was only in general terms and using rounded figures (e.g., gravitational acceleration is actually 9.8m/s squared, not 10m/s) in an effort to make it of interest for those turned off by equations, technical talk, etc. . However, I'm not quite sure exactly what your conclusions were/are and, regardless of same, my theses still stand:
a) A surfer can never "go faster" (referring to his/her rate of acceleration) than in a freefall through the air (remember, if your board is in contact with the water your speed is less) down the face of the wave, and
b) A surfer's down the line speed (velocity) is entirely due to gravitational acceleration (and, of course, his/her ability to harness said gravitational acceleration) and/or the forward movement of the wave itself (which may or may not be at a speed greater than that afforded by the amount of gravitational acceleration available to the surfer).
Although it is a consideration, I ignored the forward movement of the surfer as he/she starts his/her freefall down the wave's face partly in order to simplify things and partly because it does not really factor in in a wave larger than 5 meters/15'. (NOTE: I am referring to the height of the wave's vertical face, since that is the part I surf on and occasionally freefall down ).
True, a surfer freefalling from the top of the wave would not normally drop in a vertically straight line (provided he/she was trying to catch it, that is). The freefall would actually describe an arc (sort of like taking a 'flying leap' off a building) because the surfer is initially moving forward with the wave. But the initial velocity is entirely horizontal, not vertical, and adds nothing to the downwards velocity if it is less than 9.8m/s and the freefall lasts more than one second .
Remember, at the end of one second of freefall you are falling at a rate of 9.8 meters per second (slightly over 22 miles per hour) and will travel a vertical distance of 4.9 meters (about 15 feet) in that first second. Once the surfer has caught and thus matched the forward speed of a wave traveling at or less than 22 mph, that component of his/her velocity (i.e., in a horizontal direction) will be of no further consideration if he/she should experience a freefall of at least 5 meters/15' distance and one second in duration.
Of course, if you are surfing a wave less than than 5 meters, you could not experience a full second of freefall and the wave's initial forward movement is an important consideration in determining the 'top' speed you can attain on that particular wave. In other words, you are limited to the forward speed of the wave being ridden and/or whatever vertical freefall time/distance you could possibly make (e.g., 1/2 second of freefall equals 4.9 m/s which is about 11 mph and a vertical distance of 1.3 meters which is about 4 feet) . As we all know oh so well, smaller waves mean slower speeds .
In regards to the speed of waves/swells, the longer the wavelength, the faster the swell; the shorter the wavelength, the slower the swell. The text I consulted gave the average speed of open ocean swells but the word "average" can have one of three meanings...mode, median, mean...and the text didn’t specify which . However, for the purposes of my argument that was/is not critical (nor is discerning between "phase" and "group". )
I did say I was being “generous” with a speed of 56 kph (35 mph) for the wave being ridden. In terms of waves ridden by us mortal human beings (paddle-in surfers), this speed would mean a wavelength such that the wave was not just large, but huge (of tow-in proportions) . A smaller wave (i.e., less than average) would be moving forward at a slower speed (such as Willi calculated ).
If you want to understand why a water skier can (temporarily) go faster than the tow boat, then you really need to discuss 'angular momentum' and 'centripetal force' . Although these phenomenon are also at work on a surfer, the situation is not quite the same because of the differing sources of propulsion .
Simply put, velocity is a combination of speed and direction. When you change direction while maintaining your speed you will experience acceleration . At first glance a somewhat strange concept to be sure but that's how it works in physics .
However, if you lose your source of propulsion (e.g., being pulled by the boat) after that initial 'burst' of acceleration you will quickly "decelerate" (i.e., slow down) due to the friction caused by both air and water .
While a water skier has the tow boat as a propulsive force, a surfer has only the (downwards) acceleration of gravity and the forward motion of the wave as propulsive forces. (Note: the water rising up as the swell enters shallower water does not propel the surfer forward. It can only lift him/her up to a higher level from which he/she can once again use gravity as a source of propulsion ) .
As a simple experiment, try letting go of the tow rope AT THE PRECISE MOMENT YOU BEGIN YOUR TURN and see if you catch back up to the boat . Although you will initially experience acceleration you will quickly slow down and definitely will not catch up to nor even come close to catching up to the boat .
Letting go of that tow rope is analogous to a surfer at the bottom of his/her turn. The surfer’s main source of propulsion for getting 'down the line' speed (that is, gravity) has now been lost
. Basically, he/she uses the speed generated by dropping in to race across the wave's face (at an angle) and/or turn back up the face of the wave in order to attain a higher and faster position on the wave (a position from whence he/she can once again make use of the force of gravity to accelerate) .
If he/she happens to get stuck at the bottom of the wave , the only thing that can be done is to (hopefully) somehow generate enough speed to get out in front of the whitewater, turn, and make it back up the wave’s face. And as we all know, the usual scenario in this situation is to have to staighten out as the breaking wave speeds on down the line without you.
And this is why surfing (like downhill skiing, snowboarding, and skydiving) is classified as a “gravity sport”, whereas water skiing is not .
I suspect that the terminal velocity of a surfboard moving across the face of the wave, even the fastest surfboard, is less than the terminal velocity of a water ski(s) . If you’ve ever towed on a surfboard behind a boat (I have) you will know that, although a blast , you really can’t attain the same speed as on water skis . In any event, a surfboard's terminal velocity would need to be known in order to calculate the top speed possibly attained by a surfer on any wave. I also suspect that, under normal conditions, most paddle-in boards are already being ridden at or near the terminal speeds a surfboard is theoretically capable of attaining .
Regarding Hamilton’s hydrofoil; believe it or not he actually has successfully ridden it at macking’ Jaws . Amazing but true (I saw the pictures) .
).Thanks, Willi, for the more involved intricate calculations
. As I said, my discussion was only in general terms and using rounded figures (e.g., gravitational acceleration is actually 9.8m/s squared, not 10m/s) in an effort to make it of interest for those turned off by equations, technical talk, etc. . However, I'm not quite sure exactly what your conclusions were/are and, regardless of same, my theses still stand:a) A surfer can never "go faster" (referring to his/her rate of acceleration) than in a freefall through the air (remember, if your board is in contact with the water your speed is less) down the face of the wave, and
b) A surfer's down the line speed (velocity) is entirely due to gravitational acceleration (and, of course, his/her ability to harness said gravitational acceleration) and/or the forward movement of the wave itself (which may or may not be at a speed greater than that afforded by the amount of gravitational acceleration available to the surfer).
Although it is a consideration, I ignored the forward movement of the surfer as he/she starts his/her freefall down the wave's face partly in order to simplify things and partly because it does not really factor in in a wave larger than 5 meters/15'. (NOTE: I am referring to the height of the wave's vertical face, since that is the part I surf on and occasionally freefall down
).True, a surfer freefalling from the top of the wave would not normally drop in a vertically straight line (provided he/she was trying to catch it, that is). The freefall would actually describe an arc (sort of like taking a 'flying leap' off a building) because the surfer is initially moving forward with the wave. But the initial velocity is entirely horizontal, not vertical, and adds nothing to the downwards velocity if it is less than 9.8m/s and the freefall lasts more than one second
.Remember, at the end of one second of freefall you are falling at a rate of 9.8 meters per second (slightly over 22 miles per hour) and will travel a vertical distance of 4.9 meters (about 15 feet) in that first second. Once the surfer has caught and thus matched the forward speed of a wave traveling at or less than 22 mph, that component of his/her velocity (i.e., in a horizontal direction) will be of no further consideration if he/she should experience a freefall of at least 5 meters/15' distance and one second in duration.
Of course, if you are surfing a wave less than than 5 meters, you could not experience a full second of freefall and the wave's initial forward movement is an important consideration in determining the 'top' speed you can attain on that particular wave. In other words, you are limited to the forward speed of the wave being ridden and/or whatever vertical freefall time/distance you could possibly make (e.g., 1/2 second of freefall equals 4.9 m/s which is about 11 mph and a vertical distance of 1.3 meters which is about 4 feet)
. As we all know oh so well, smaller waves mean slower speeds .In regards to the speed of waves/swells, the longer the wavelength, the faster the swell; the shorter the wavelength, the slower the swell. The text I consulted gave the average speed of open ocean swells but the word "average" can have one of three meanings...mode, median, mean...and the text didn’t specify which
. However, for the purposes of my argument that was/is not critical (nor is discerning between "phase" and "group". )I did say I was being “generous” with a speed of 56 kph (35 mph) for the wave being ridden. In terms of waves ridden by us mortal human beings (paddle-in surfers), this speed would mean a wavelength such that the wave was not just large, but huge (of tow-in proportions)
. A smaller wave (i.e., less than average) would be moving forward at a slower speed (such as Willi calculated ).If you want to understand why a water skier can (temporarily) go faster than the tow boat, then you really need to discuss 'angular momentum' and 'centripetal force'
. Although these phenomenon are also at work on a surfer, the situation is not quite the same because of the differing sources of propulsion .Simply put, velocity is a combination of speed and direction. When you change direction while maintaining your speed you will experience acceleration
. At first glance a somewhat strange concept to be sure but that's how it works in physics .However, if you lose your source of propulsion (e.g., being pulled by the boat) after that initial 'burst' of acceleration you will quickly "decelerate" (i.e., slow down) due to the friction caused by both air and water
.While a water skier has the tow boat as a propulsive force, a surfer has only the (downwards) acceleration of gravity and the forward motion of the wave as propulsive forces. (Note: the water rising up as the swell enters shallower water does not propel the surfer forward. It can only lift him/her up to a higher level from which he/she can once again use gravity as a source of propulsion
) .As a simple experiment, try letting go of the tow rope AT THE PRECISE MOMENT YOU BEGIN YOUR TURN and see if you catch back up to the boat
. Although you will initially experience acceleration you will quickly slow down and definitely will not catch up to nor even come close to catching up to the boat .Letting go of that tow rope is analogous to a surfer at the bottom of his/her turn. The surfer’s main source of propulsion for getting 'down the line' speed (that is, gravity) has now been lost
. Basically, he/she uses the speed generated by dropping in to race across the wave's face (at an angle) and/or turn back up the face of the wave in order to attain a higher and faster position on the wave (a position from whence he/she can once again make use of the force of gravity to accelerate) . If he/she happens to get stuck at the bottom of the wave
, the only thing that can be done is to (hopefully) somehow generate enough speed to get out in front of the whitewater, turn, and make it back up the wave’s face. And as we all know, the usual scenario in this situation is to have to staighten out as the breaking wave speeds on down the line without you. And this is why surfing (like downhill skiing, snowboarding, and skydiving) is classified as a “gravity sport”
, whereas water skiing is not .I suspect that the terminal velocity of a surfboard moving across the face of the wave, even the fastest surfboard, is less than the terminal velocity of a water ski(s)
. If you’ve ever towed on a surfboard behind a boat (I have) you will know that, although a blast , you really can’t attain the same speed as on water skis . In any event, a surfboard's terminal velocity would need to be known in order to calculate the top speed possibly attained by a surfer on any wave. I also suspect that, under normal conditions, most paddle-in boards are already being ridden at or near the terminal speeds a surfboard is theoretically capable of attaining . Regarding Hamilton’s hydrofoil; believe it or not he actually has successfully ridden it at macking’ Jaws
. Amazing but true (I saw the pictures) .
Last edited by Bryan Jackson on Wed Jun 30, 2004 9:44 am, edited 17 times in total.
Be a round rock, not a flat one. That way you can hit the reef without any help from the lip. In my case, three sounds more like the max number of times I have gone back over the falls on a single wipe-outstemple wrote:so if I hit the water at 24 mph. How many times will I skip before the lip pounds me into the reef? I personally have counted three down a single face.
MT
Experience gained is in proportion to equipment ruined.
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stemple
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oh yeah, I love that feeling of being picked up, then accelerated upward only then have a sudden feeling of freefall while you wait for crushing impact number 2. Been there baby!!
As you free fall through the lip are you following the classic parabolic trajectory or is there a more horizontal component to the trajectory. If your mass is included in the wave do you actually slow down the wave throwing?. When going over the falls, looking at it from the beach, it seems (when you can see someone) slower.
Also could you calculate (estimate) the force or mass of the water impacting on your body, say on 6 foot wave verses a 10 foot wave? What are the magnitude of forces that work you after missing that air drop at Mavs?
How much hydraulic punishment can the human body sustain without injury?
As you free fall through the lip are you following the classic parabolic trajectory or is there a more horizontal component to the trajectory. If your mass is included in the wave do you actually slow down the wave throwing?. When going over the falls, looking at it from the beach, it seems (when you can see someone) slower.
Also could you calculate (estimate) the force or mass of the water impacting on your body, say on 6 foot wave verses a 10 foot wave? What are the magnitude of forces that work you after missing that air drop at Mavs?
How much hydraulic punishment can the human body sustain without injury?
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DrStrange
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Back to flogging the top speed thingy:
from another thread, a GG quote
I saw some of this watching GG at Rincon and it was truely frickin' amazing. Literally faster than the eye/brain could follow...much faster than wave speed...
from another thread, a GG quote
What made Velo such a challenge to ride was that it had untold gears. Because of its deep displacement hull, it never peaked out. No matter how fast you were going, you could bury the forward rail into a bottom turn and break the fin out, and it would jump into the next gear.
I saw some of this watching GG at Rincon and it was truely frickin' amazing. Literally faster than the eye/brain could follow...much faster than wave speed...-
Bryan Jackson
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Once again I repeat : The speed you achieve on a wave can only be due to either the speed of the wave itself or gravitational acceleration (and, of course, the surfer's ability to make use of same) .
The less "wetted area" of your board in contact with the water, the less drag it will create, and the higher its terminal velocity/speed (which is the limiting factor as to just how fast you can go on the wave's face).
As this occurs your terminal velocity through the air, which is much higher than your terminal velocity through the water, starts to become the limiting factor . At the same time your rate of acceleration due to gravity also greatly increases .
Result: you literally "fly" down the face of the wave
Therefore, Doc, you are correct when you say Greenough was going "faster than wave speed". His flex board was able to both maximize gravitational acceleration and minimize drag such that no one in the surfing world had thought possible.
If it seems like he was going faster than these factors would allow , that is only because everyone else was going so much slower!
(I also believe that the spring/torsional characteristics of his flex boards had a role to play but that seems to be a somewhat controversial theoretical perspective , on this web site anyways, so I will not get into that) .
: The speed you achieve on a wave can only be due to either the speed of the wave itself or gravitational acceleration (and, of course, the surfer's ability to make use of same) .The less "wetted area" of your board in contact with the water, the less drag it will create, and the higher its terminal velocity/speed (which is the limiting factor as to just how fast you can go on the wave's face).
As this occurs your terminal velocity through the air, which is much higher than your terminal velocity through the water, starts to become the limiting factor
. At the same time your rate of acceleration due to gravity also greatly increases .Result: you literally "fly" down the face of the wave
Therefore, Doc, you are correct when you say Greenough was going "faster than wave speed". His flex board was able to both maximize gravitational acceleration and minimize drag such that no one in the surfing world had thought possible.
If it seems like he was going faster than these factors would allow
, that is only because everyone else was going so much slower! (I also believe that the spring/torsional characteristics of his flex boards had a role to play but that seems to be a somewhat controversial theoretical perspective
, on this web site anyways, so I will not get into that) .