Some scientific analysis of Mao's 3A | Page 14 | Golden Skate

Some scientific analysis of Mao's 3A

antmanb said:
Lu chen was the master of delayed rotation (almost sometimes tot eh detriment of landing them!) but she always had thefirst rotation go slowly and the second and third really sped up.

Ant
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I was also gonna say Lu Chen!
 
silverlake22 said:
Yes, I believe that this is considered not the best technique for a 2a, and that's why some have issues with Yuna's 2a. In reality, I think someone thought I was a Yuna-bot (which I'm not - I like other skater's too) and wanted to point out a flaw in her skating, which I can understand now, but I still think her 2a is fine. I like it when she does it on the ground :agree:
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Yuna's 2a is very quick, but does not get great height.
 
☆Genie;448989 said:
Yuna's 2a is very quick, but does not get great height.
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This seems to have pretty good height to me. It also make it look too easy:
2A

This was pretty good comparison (thanks to silverlake):
3a vs 2a

In short, they both have very high quality jumps.
 
☆Genie;448989 said:
Yuna's 2a is very quick, but does not get great height.
Click to expand...

From what gfskater measured in this thread, it seems that airtime of the Yuna's 2A is greater than Mao's 3A, which actually directly translated to greater height if I remember my physics correctly.
 
sleepymom said:
From what gfskater measured in this thread, it seems that airtime of the Yuna's 2A is greater than Mao's 3A, which actually directly translated to greater height if I remember my physics correctly.
Click to expand...

From what my eyes tell me, the height between Mao's triple-axel and Yuna's double-axel is comparable, but Yuna has more width, which would also explain why Yuna has greater airtime.
 
n_halifax said:
In some strange way, do you who Yu-Na's double axel reminds me of? Katarina Witt's. Not in the take-off, but in the air position and landing. And that's a compliment! Katarina had (and I'm sure still has) a beautiful double axel.
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Yes, Two beautiful actresses on the ice.
 
hurrah said:
From what my eyes tell me, the height between Mao's triple-axel and Yuna's double-axel is comparable, but Yuna has more width, which would also explain why Yuna has greater airtime.
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Hm.. it's been more than decade I've studied physics, but let me try this way. Consider the vertical path they would move during their jump.. On half-time, they would be top of the jump, correct? from there to the landing the only acceleration is gravity , so the distance ( or height in this case ) can be calculated by

height = g * t ^ 2 / 2

g = gravitational acceleration
t = half of jump air time in this case.

So in this case, from gfskaters air time measurement we can get height for both of them, without considering their jump distance. Urgh.. I don't know whether I'm making this clearly.
 
sleepymom said:
Hm.. it's been more than decade I've studied physics, but let me try this way. Consider the vertical path they would move during their jump.. On half-time, they would be top of the jump, correct? from there to the landing the only acceleration is gravity , so the distance ( or height in this case ) can be calculated by

height = g * t ^ 2 / 2

g = gravitational acceleration
t = half of jump air time in this case.

So in this case, from gfskaters air time measurement we can get height for both of them, without considering their jump distance. Urgh.. I don't know whether I'm making this clearly.
Click to expand...

You are right.

Distance has nothing to do with air time(hang time).
Only height makes the difference.
 
SkatingAnalyst said:
You are right.

Distance has nothing to do with air time(hang time).
Only height makes the difference.
Click to expand...

Oh man this brings me back to my physics days. :biggrin:

In a theoretically ideal physics world, the x and y components of movement (horizontal and vertical) are independent of each other. Ugh, it's been let out that I'm a huge skating loving science nerd. :rofl:
 
Wooowwww. This is really difficult for me, but:


Y=VyT+0.5GT^2

Y=vertical height, Vy=initial vertical velocity, T=hang time, G=acceleration due to gravity

If we assume that Mao's triple-axel and Yuna's double-axel height is approximately the same (which is how it seems to me), and Yuna's hang time is greater than Mao's, according to the above formula, is Mao's initial vertical velocity going into the jump faster than Yuna's? That sort of correlates with what my eyes tell me.
 
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Caculating the Height

If Mao's hang time is 0.45 seconds...
It'll take 0.225 to go up and...0.225 sec to come down.

The Earth's G is 980cm / sec...

So...we can caculate her height of the jump
980cm X 0.225 X 0.225 = 49.61 cm ... about 19.5 inches.

That is slightly less than a third of her body height.

You must know the speed of the skater and trajectory to caculate the distance.
You cannot calculate it only with hang time.
 
That's not the correct formula. This thread has gotten really long and I haven't read it all but Mao is in the air for more than .45 seconds. The original "study" was incorrect with its numbers.
 
Nevermind, I saw Blade of Passions' answer and decided to delete. I'd love it though if someone with full knowledge of physics could explain.
 
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Formula

Blades of Passion said:
That's not the correct formula. This thread has gotten really long and I haven't read it all but Mao is in the air for more than .45 seconds. The original "study" was incorrect with its numbers.
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If mine is wrong, what is right the formula?
You must know something Newton does not.

I'm more than willing to learn from you.
 
SkatingAnalyst said:
If mine is wrong, what is right the formula?
You must know something Newton does not.

I'm more than willing to learn from you.
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Well, what about the one I got from http://library.thinkquest.org/15384/physics/index.htm?

You said that:

SkatingAnalyst said:
You must know the speed of the skater and trajectory to caculate the distance.
You cannot calculate it only with hang time.
Click to expand...

But you know, our eyes do give us an approximation of the trajectory. I can see that Mao's triple is of approximately equal height to Yuna's double. I can also see that Yuna's double has more horizontal distance compared to Mao's triple. And so, the only factor that we can make no assumption whatsoever within the formula:

Y=VyT+0.5GT^2

Y=vertical height, Vy=initial vertical velocity, T=hang time, G=acceleration due to gravity

is the the speed, i.e., velocity. If we assume that vertical height between Mao's triple and Yuna's double is the same, and if we assume that Yuna's hang time is longer than Mao's, both of which assumptions might be wrong of course, but assuming they are correct, then Mao must have more speed (in the upward direction, that is) going into her jump as compared to Yuna.

Furthermore, if we assume that Yuna's hang time is longer than Mao's and Yuna's jump covers more horizontal distance as our eyes tell us, then, according to the logic of:

X=VxT

X=horizontal distance, Vx=initial horizontal velocity, T=hang time

Yuna must have more horizontal speed across the ice when she does her double as compared to Mao, which would indicate that she is more likely to telegraph her double axel.

Conclusions which can be drawn according to these laws of physics correlates with what I see. I have see Mao's triple popping quickly up into the air. I have seen Yuna telegraphing her double axel.
 
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hurrah said:
Well, what about the one I got from http://library.thinkquest.org/15384/physics/index.htm?

You said that:



But you know, our eyes do give us an approximation of the trajectory. I can see that Mao's triple is of approximately equal height to Yuna's double. I can also see that Yuna's double has more horizontal distance compared to Mao's triple. And so, the only factor that we can make no assumption whatsoever within the formula:

Y=VyT+0.5GT^2

Y=vertical height, Vy=initial vertical velocity, T=hang time, G=acceleration due to gravity

is the the speed, i.e., velocity. If we assume that vertical height between Mao's triple and Yuna's double is the same, and if we assume that Yuna's hang time is longer than Mao's, both of which assumptions might be wrong of course, but assuming they are correct, then Mao must have more speed (in the upward direction, that is) going into her jump as compared to Yuna.
Click to expand...

You have the RIGHT FORMULA. But...

We do not know skater's vertical speed or any speed as a matter of fact.
All that given to us is hang time which is more than enough to caculate
the hight from where the skater is falling from... by using your formula

Y=VyT + 0.5GTxT

At the peak of the jump Vy=0, G=980 and T is hang time which is 0.45
Now we are caculating the hight from the peak to bottom which is

Y=0x0.45 + 0.5x980x0.45x0.225=49.61 cm

Exactly same results as mine...

Still have any questions?
 
No, no question. Thank you.

However, we should note that as Blades of Passion said, actual time measurements of Mao's hang time exceeded 0.45, so she might actually be jumping higher than that.

This discussion was interesting because now, I can say with some confidence that if Yuna's hang time is longer, and if Yuna's jump is covering more horizontal distance, then Mao's vertical speed going into her triple axel is probably faster than Yuna's.
 
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