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- Oct 24, 2009
One thing that should be mentioned is that it might take less time to reach the max height than the time it takes to come down back to the ice. So maybe it take 0.10 second to reach the max height and 0.37 seconds to come down?
Some scientific analysis of Mao's 3A
- Thread starter Bennett
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Yikes.
1. Except for Mathman, everybody who wants to expound on the physics of jumps go read a textbook on classical mechanics first!
2. Impressions looking at a photograph, or memories of competitions past, are not measurements. They may call into question whether the 0.45 seconds is correct, but they are not measurements.
So some of you question the 0.45. So do I. But that is just the collective OPINION. None of you (or me for that matter) have a measurement that says 0.45 is the wrong answer for Mao, and the correct anwer is 0.xxx instead.
And speaking of measurements, I dropped in at El Segundo at lunch time, hoping to time Evan, but he wasn't there. To illustrate again what skaters do (and what real data is) ...
The Australian Junior Men's champion (at least that's who I was told he was) was there working triples. His jumps were from 0.6 to 0.8 seconds. Most were around 0.7 to 0.75. Some Nov-Sen ladies working on double Axel were typically at 0.5 seconds. The Senior lady was also working triples, and her toe jumps were 0.6 to 0.7 seconds. Her best was 0.75 (a lot better than when I timed her at Southwest Pacific). Some Novices were mostly around 0.5 secs for their jumps. Carolina was working triples and was also 0.6 to 0.8, with 0.65 to 0.75 the most common. Her best that I timed was 0.81. The double Axels at 0.5 seconds work out to 12 inches high and 5 rotations per second (average). Carolina's triple at 0.81 second works out to 2.6 feet and 3.7 rotations per second. The Junior man was executing around 2 feet high and 4.2 rotations per second.
Carolina and the Junior man are examples of how height relieves the need for super fast rotations, while the 0.5 sec double Axels are a little on the low side and requires faster than average rotation to get around on the jump.
Notice that with these times, most of the skaters in this sample were consistently jumping 1-2 feet in height.
1. Except for Mathman, everybody who wants to expound on the physics of jumps go read a textbook on classical mechanics first!
2. Impressions looking at a photograph, or memories of competitions past, are not measurements. They may call into question whether the 0.45 seconds is correct, but they are not measurements.
So some of you question the 0.45. So do I. But that is just the collective OPINION. None of you (or me for that matter) have a measurement that says 0.45 is the wrong answer for Mao, and the correct anwer is 0.xxx instead.
And speaking of measurements, I dropped in at El Segundo at lunch time, hoping to time Evan, but he wasn't there. To illustrate again what skaters do (and what real data is) ...
The Australian Junior Men's champion (at least that's who I was told he was) was there working triples. His jumps were from 0.6 to 0.8 seconds. Most were around 0.7 to 0.75. Some Nov-Sen ladies working on double Axel were typically at 0.5 seconds. The Senior lady was also working triples, and her toe jumps were 0.6 to 0.7 seconds. Her best was 0.75 (a lot better than when I timed her at Southwest Pacific). Some Novices were mostly around 0.5 secs for their jumps. Carolina was working triples and was also 0.6 to 0.8, with 0.65 to 0.75 the most common. Her best that I timed was 0.81. The double Axels at 0.5 seconds work out to 12 inches high and 5 rotations per second (average). Carolina's triple at 0.81 second works out to 2.6 feet and 3.7 rotations per second. The Junior man was executing around 2 feet high and 4.2 rotations per second.
Carolina and the Junior man are examples of how height relieves the need for super fast rotations, while the 0.5 sec double Axels are a little on the low side and requires faster than average rotation to get around on the jump.
Notice that with these times, most of the skaters in this sample were consistently jumping 1-2 feet in height.
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The actual height is still 9.6 inches regardless of the horizonal distance.
The "v0" in your formula is the verticle component of the initial velocity. If v1 is the horizontal component, then the distance is simply d = v1t. If you want to maximize your horizontal distance, aim for a 45 degree angle of take-off.
Exactly. In this case, Vx = Vo * cos(45), Vy = Vo * sin(45)
So if she's jumping at 60 degree angel or 90 degree angle, her Vy would increase, thus increase the height and decrease the horizontal distance, no?
I'll probably regret getting ito this, but,
Vo is not the independant parameter. Vx and Vy are controlled seperately by the skater. Vo = Vx at the instant before takeoff. The skater generates Vy in the takeoff and you have a new Vo. The arc of the jump is controlled by the amount of Vy the skater generates in the takeoff relative to the Vx at the instant of takeoff. Also, in the takeoff most skaters lose some Vx, so the Vx at the instant they leave the ice is not the Vx they had approaching the jump, also affecting the magnitude and direction of the velocity vector at the instant the skater leaves the ice. The skater does not have a fixed Vo and adjust the takeoff angle. They adjust Vy and Vx from which you can determine the velocity vector at the instant they leave the ice (Vo), and thus the trajectory of the jump.
Vo is not the independant parameter. Vx and Vy are controlled seperately by the skater. Vo = Vx at the instant before takeoff. The skater generates Vy in the takeoff and you have a new Vo. The arc of the jump is controlled by the amount of Vy the skater generates in the takeoff relative to the Vx at the instant of takeoff. Also, in the takeoff most skaters lose some Vx, so the Vx at the instant they leave the ice is not the Vx they had approaching the jump, also affecting the magnitude and direction of the velocity vector at the instant the skater leaves the ice. The skater does not have a fixed Vo and adjust the takeoff angle. They adjust Vy and Vx from which you can determine the velocity vector at the instant they leave the ice (Vo), and thus the trajectory of the jump.
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The skater can control the angle, the take off force. Both directly control Vx and Vy.
http://library.thinkquest.org/2779/Balloon.html
See this simulation. You can adjust the angle and see yourself that the height and the travel distance directly correlate with take off angle. Mao does not travel far, her jumps are high.
According to the fuzzy math, she would be like 9.6 in jumping up and down and probably 5 in off the ground in reality with the distance she travels, which is totally bogus.
I think the timing is really off. Even my jump lasts longer than half a sec.
http://library.thinkquest.org/2779/Balloon.html
See this simulation. You can adjust the angle and see yourself that the height and the travel distance directly correlate with take off angle. Mao does not travel far, her jumps are high.
According to the fuzzy math, she would be like 9.6 in jumping up and down and probably 5 in off the ground in reality with the distance she travels, which is totally bogus.
I think the timing is really off. Even my jump lasts longer than half a sec.
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- May 15, 2009
Can you explain the sporting term "explosive jumper" that we hear in basketball and track? Even in skating if Scott is announcing
Is it possible for one person to not only jump higher but also jump faster, to accelerate into the air at a faster speed than another?
I saw Charles Barkely do this many, many times......atleast it appeared he was doing it.
If this was true, isn't it possible for a figure skater to have an explosive jump - and to jump up at a faster rate, than the speed they will descend back to the ice with? I don't mean the DrJ move you accidentily gave Jordan credit for
Is it really impossible ?
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Exact measurements are not needed. It is easy to see by using your brain. Stop ignoring the examples that have been given.
Do you need to measure the height on Brian Joubert's Triple Flip to tell that it is higher than Caroline Zhang's? No, you do not. It is obvious.
Again, if I am 6 feet tall and someone jumps on top of my head, they have jumped 6 feet.
I have been right next to Mao when she has done jumps and they went higher than 10 inches.
It is plain to see even on video that Mao's 3Axel goes higher than 10 inches.
I knew I would regret it!
Your example does not apply to what the skater is doing. The skater controls Vx and Vy seperately. The takeoff angle is given by ArcTan(Vy / Vx).
Starting with Vx and Vy at the instant of takeoff:
The height of the jump will be 0.5 * Vy ^ 2 / g, where g is the acceleration due to gravity.
The distance traveled will be 2 * Vx * Vy / g
To get maximum distance for a given Vx, the skater must generate a Vy = Vx, but Vx is an independant parameter controlled by the skater.
The Vx is determined by taking the integral of the force applied to the ice with respect to time divided by the mass of the skater, taking into account the fact the impulse to the ice is an inelastic process and some energy is lost compressing anf fraturing the ice, etc.
You fail the physics lab. You know she jumps higher than 10 inches because you know she jumps higher than 10 inches!
(I will know how high she jumps when I measure it for myself in Vancouver.)
Your example does not apply to what the skater is doing. The skater controls Vx and Vy seperately. The takeoff angle is given by ArcTan(Vy / Vx).
Starting with Vx and Vy at the instant of takeoff:
The height of the jump will be 0.5 * Vy ^ 2 / g, where g is the acceleration due to gravity.
The distance traveled will be 2 * Vx * Vy / g
To get maximum distance for a given Vx, the skater must generate a Vy = Vx, but Vx is an independant parameter controlled by the skater.
The Vx is determined by taking the integral of the force applied to the ice with respect to time divided by the mass of the skater, taking into account the fact the impulse to the ice is an inelastic process and some energy is lost compressing anf fraturing the ice, etc.
You fail the physics lab. You know she jumps higher than 10 inches because you know she jumps higher than 10 inches!
(I will know how high she jumps when I measure it for myself in Vancouver.)
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This should be a simple question.
Do you believe Mao is only 9.6 in off the ice?
I don't.
Btw, I was responding to Mathman who asserted that the height is the same regardless of distance travel. It is not.
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- Jun 21, 2003
Flatt Fan said:
Yes...but...in this problem the independently measured variable is the total time,"T," not V0 = |V|. "T" is determined solely by the vertical component (T = 2Vy/a). For fixed "T" (say, t = 0.45), the maximum height H is the same no matter what the angle is: H = h(T/2) = h(Vy/a) = Vy^2/2a.
(Something like that. 
)The 45 degree angle comes in when you want to maximize the distance for fixed V0 (but you knew that.
)Edited to add: I see Dr. R. beat me to it.
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You fail the physics lab. You know she jumps higher than 10 inches because you know she jumps higher than 10 inches!
I have failed nothing. YOU have failed at reading, however.
I know she jumps more than 10 inches because I know how tall I am. 10 inches is below my knees. When a skater jumps right in front of me and the jump is definitely higher than my knees, they are getting quite a bit more than 10 inches of height in the jump.
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Yes...but...in this problem the independently measured variable is the total time,"T," not V0 = |V|. "T" is determined solely by the vertical component (T = 2Vy/a). For fixed "T" (say, t = 0.45), the maximum height H is the same no matter what the angle is: H = h(T/2) = h(Vy/a) = Vy^2/2a.
The 45 degree angle comes in when you want to maximize the distance for fixed V0 (but you knew that.
Something is lost in translation with the "professor" whose "serious" analysis on a skater totally disregard the take off speed, the distance cover, and only gives the total time. Yeah, color me skeptical.
Wasn't there an article a while back showing a maximum revolution humanly possible is 7rev*Hz?
Anymore than that, quintuple axel would have been done in competition.
This should be a simple question.
Do you believe Mao is only 9.6 in off the ice?
I don't.
I have no idea how high Mao is off the ice and wouldn't presume to say without making the measurement myself. Is 9.6 inches wrong? I will find out for myself in Vancouver. BoP seems to know for sure, but I don't. I don't possess his highly calibrated knees, nor his ability to exactly measure angles and distances by eyeball only.
Depends on exactly what situation you are describing for the jump. What Mathman says is true if Vy is constant and you only vary Vx. However, if you have Vo constant and vary the takeoff angle, the height and distance both vary for the same Vo. Also all jumps with the same time in the air have the same height.
If a skater generates 5 ft/sec Vy on every jump, the height of each jump will be the same for all those jumps. If Vx is different for each of those jumps the distance travelled will be different while the height stays the same. But there are also combinations of Vx and Vy where the distance will be the same and the heights will be different.
But again, the skater does not chose a Vo and a takeoff angle. They don't think on the takeoff "I want to jump at an angle of 30 degrees at 20 ft/sec." They control the Vx speed at which they approach the jump and the force they apply to the ice to get Vy.
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- Jun 21, 2003
Can you explain the sporting term "explosive jumper" that we hear in basketball and track? Even in skating if Scott is announcing
Yes. it means that the jumper has superior take-off velocity. Like the muzzle velocity of a high powered rifle compared to that of a pop-gun.
By the way, if a jumper were so explosive that he took off with an initial velocity of 11.2 kilometers per second, he would escape the earth's gavitational pull altogether and never stop until he reached the next galaxy.
If a jumper jumps "faster" -- that is, with a higher initial velocity -- then he will automatically jump higher, no question.
Once a jumper's feet leave the ground, however, he cannot accelerate (increase his speed while in the air). Our friend Isaac tells us why: Force = mass times acceleration. Once the jumper leaves the floor, no other force acts on him except gravity (ignoring minor effects like wind resistance.)
I think the main reason for that effect is Charles' size and fierceness. Like a charging water buffalo, it just seems like he's oming on like a freight train.
Not really.
(But I have to admit that Dorothy Hamill looks like she defies the laws of physics on her delayed Axel jump.
)I don't mean the DrJ move you accidentily gave Jordan credit for
Actually, I remember reading a study once in which Elgin Baylor turned out to be the hang time king.Yes.
By the way, this is why the CoP gives extra bonus points on spins when a skater can accelerate when going from one position to another. It is really hard. Skaters can accomplish it by manipulating their angular momentum.
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- May 15, 2009
Yes. it means that the jumper has superior take-off velocity. Like the muzzle velocity of a high powered rifle compared to that of a pop-gun.
If a jumper jumps "faster" -- that is, with a higher initial velocity -- then he will automatically jump higher, no question.
Once a jumper's feet leaves the ground, however, he cannot accelerate (increase his speed while in the air). Our friend Isaac tells us why: Force = mass times acceleration. Once the jumper leaves the floor, no other force acts on him except gravity (neglecting minor effects like wind resistane.)
I think the main reason for that effect is Charles' size and fierceness. Like a charging water buffalo, it just seems like he's coming on like a freight train.
(But I have to admit that Dorothy Hamill looks like she defies the laws of physics on her delayed Axel jump.
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Thanks for the answers - I believe you, and it was the Dorothy delayed axle example that got me
BTW, I have never been next to Mao to see her jumps- but I have met Barkely and you would be surprised how "untall" he actually is. Not much more than 6'4
Geez, he was fierce, especially under the boards or going to the basket.
Actually the rule reads "For camel, sit and layback positions, once the position has been established a clear increasing of speed will be considered a difficult variation."
I.e., it's really hard to accelerate while staying in the same position, so that's what gets the extra feature.
Some changes going from one position to another accelerate by their nature, e.g., going from camel to sit would be the most obvious example. It's not really hard to accelerate in that situation.
I have a question for grossano. What method or equipment do you use to measure the time skaters are in the air to hundredths of seconds?
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- Jun 21, 2003
Well, the actual article (I think) is a brief blurb written by a newspaper reporter, mentionaing Ikagami's study.
If you Google Yasuo Ikegami, 131,000 references come up -- as far as I can tell, almost all of them referring to his scholarly work. Here is one:
http://www.apissc2009.org/biomechanics symposium.htm
Scroll down to the third presenter at this symposium. The title of Dr. Ikegami's talk was "Biomechanical Study on Figure Skate Techniques: Focusing on Jump Motion." Here is the abstract:
By the way, if you do Google this guy, the third reference that comes up is -- this thread.
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9.6 inches is definitely wrong because no human is capable of spinning faster than 7 revolution per sec.
The fastest scratch spin in the Guinness World Record is 308 rpm, which is 5.2rev per second. Do you think Mao Asada, as mighty as she is, is capable of spinning faster than the current world record? Granted, it was a scratch spin, but at that level, it's pretty damn fast.
I asked a simple question if you believe Mao is only 9.6 inches off the ice. Presume away.