# Apolo Ohno Physics

It is winter Olympics time and time for physics. The event that I always gets me thinking about physics is short track speed skating. It is quite interesting to see these skaters turn and lean at such high angles. All it needs is a little sprinkling of physics for flavor.

Check out this image of Apolo (apparently, it is not Apollo).

I know what you are thinking…Fcent….what force is that? Yes, I am going to use the centrifugal force in this case – but remember that sometimes fake forces are awesome. In short, if I want to pretend like Apolo is not accelerating then I need to add the fake centrifugal force (which is in the opposite direction as the actual acceleration). Remind me later and I will re-visit this problem without using fake forces. Anyway, the centrifugal force will have the magnitude:

Here v is the speed that Apolo (or any skater) is moving and r is the radius of the circle he (or she) is moving in. I drew this vector for the centrifugal force as acting at the center of mass of the skater. This isn’t exactly true. The problem is that different parts of the skater are moving in circles of different radii. However, the difference probably (but I will look at it later) not that large that it matters.

For the other forces, notice that the ice exerts two forces (well, one force that I broke into two components). There is a component parallel to the ice. This is a static friction force where the skate blades cut into the ice. Also, the ice pushes up perpendicular to the ice. This is the normal force. I assumed that the kinetic friction force (which would be into the page opposite the direction of motion) is small enough to be ignored. Really, that is the cool thing about ice skating. Ice needs to be low friction in the direction the skate moves and high friction perpendicular to the blade.

Back to the force diagram, there are two things to consider. The forces must add up to the zero vector (because I am assuming the reference frame of the skater is not accelerating). Also, the torque must be zero about any point. For this case, I will choose the point where the skates touch the ice. This will give three scalar equations (two for the forces an one for the torque). Forgive me, but I am not going to go into all the torque details for now. Wait – I forgot one more parameter – the distance from the point where the skates contact the ice to the center of mass. I will call this distance s (for no particular reason).

Now I can make a substitution for both the centrifugal force (which I wrote above) and the frictional force. I will assume a coefficient of static friction of mus. I will also assume that the skater is just at the point of slipping. This means that the static frictional force is the greatest it can be (so there will be an equal sign and not a less-than-or-equal sign)

Substituting in for the friction and the centrifugal force in the x-direction force equation:

And again for the y-direction substituting for the centrifugal force:

There is another important relationship here. I am going to assume that the sum of the frictional and normal force must be directed towards the center of mass. This means that:

And now, using this in the x-direction force equation to eliminate mu. I get:

This gives me the speed of the skater in terms of the angle he (or she) is at and the radius of the circle the skater is moving in. It turns out I get the same thing if I solve the y-direction force equation (and that would have been a little simpler). Does this result seem reasonable?

• Do the units work? If g is in N/kg (same as m/s2), then g*r will be m2/s2. When I take the square root of this, I get units of m/s – that is good.
• If r is constant, what should happen as theta gets larger? This should be slower speed. It is sort of difficult to see from that function, so let me make a quick plot.

That plot looks pretty good. For an angle that approaches 90 degrees, the skater’s speed would be smaller. A skater wouldn’t have to lean at all if the skater was stopped. As the angle gets smaller (approaching zero), the skater would have to be going faster and faster. That is just what that graph shows.

So, let me see if this works. What is the radius of a short-track? According to the ISU (International Sk8ing Union) the inner radius must be 25-26 meters 8 – 8.5 meters (see correction in comments). What about the angle? From the picture of Apolo above, I get about 33 degrees (0.6 radians). Using these values, I get a speed of:

### note:

This is the part where I discuss why the speed was too great. I originally used the 25 meter radius and calculated a speed of around 42 mph. I will leave this part in here even though it is incorrect.

Clearly, this is way too fast. Apolo’s best time on the 500 meter race is 41.5 seconds. This gives an average speed of 12 m/s. Is the angle the problem? I don’t think so. Looking at the plot of the function above, the angle would have to be around 50-60 degrees for the speed to be 12 m/s. Is it because he is pushing on the ice with his hand? Again, I don’t think this is the case because sometimes they don’t touch the ground. What about the radius? Even moving the radius down to 23 meters doesn’t make that big of a difference.

The problem must be with one of my assumptions. I suspect the assumption that the “center of the centrifugal force” was at the same location as the center of mass. This would make a difference. Now I guess I will have to calculate that.

### Update

You should give me some credit for knowing something was wrong. According to commenter Milan, the radius is around 8.5 – 8 meters. You can take off some points in my internet looking-up scores. This gives a speed of 24 mph – that I am much happier with. I will fix the figures above.

1. #1 Dave X
February 17, 2010

Might it have something to do with part of the non-accelerating assumption or the kinetic friction force? He has to do an Olympian level of work climbing towards the center in order to maintain that 12m/s.

2. #2 Len Bonacci
February 17, 2010

The 25-m radius is for the regular track, I think — short track has a smaller radius.

3. #3 len Bonacci
February 17, 2010

Found it — the radius for the short track is around 8.5 m. That gives you a speed of about 11 m/s.

4. #4 bsci
February 17, 2010

Not related to your math here, but one of the talking sports heads mentioned that short track blades are actually slightly curved and not asymmetrical on the bottom of the skate.
Here’s a link I just found:
http://www.nbcolympics.com/short-track/insidethissport/equipment/newsid=261277.html

It says the asymmetry is to allow the skater to lean more without the boot touching the ice, but it must also move the blades slightly closer to the center of gravity during the leans.

The curved blade probably decreases friction in the desired direction of motion and adds more surface area and friction perpendicular to the blade.

There must be some fun calculations behind the blade locations and curvatures.

5. #5 Milan Merhar
February 18, 2010

The inner radius of the turns on a 111 meter short track course is 8.0 m although the “measured track” radius is 8.5 m. Top competitors try to hug the 8.0 m track edge, but that’s very difficult at their highest speeds, and in timed trials they often swing wider on turn entry and exit, approximating a wider turn radius.

Blades are offset on the boot to provide clearance on turns, and are both ground to a slight radius (tips higher than middle) and curved slightly longitudinally (center to right of tips) to maximize edge on ice on turns.

6. #6 Rhett Allain
February 18, 2010

@Milan,

I was thinking about the radius last night. After watching the olympics, I was sure I had the wrong value – thanks for your input.

7. #7 Claire
February 18, 2010

Wondered if the skater changing leg alters the amount of time friction is excerted on ice – on flat blade and forward. Look what happens when they change leg. You know when you see slow motion of animals running, at some point they are air born – less friction so more area covered over time. In this senese, is the curve of the skater also a result of the changes of friction by each change of leg on the ice surface?

8. #8 Claire C Smith
February 18, 2010

I meant that path of the curve the skater takes, not the angle of the skater.

9. #9 Ian Kemmish
February 19, 2010

Please do curling. The most I could find on the Internet was an explanation of why a curling stone curls in the opposite direction to, say, an upturned tumbler on a glass surface. But I don’t understand where the sideways force on the upturned tumbler comes from either….

10. #10 Sebastian
February 20, 2010

Concerning the discussion about why the skater makes a curve, doing Inline Speed Skating i’m experiencing, that the skate starts the curve as soon as it isn’t orthographic to the floor.
Even with only one foot on the ground for the whole corner (thus on air time) you can make it around the corner (see short track skaters especially).
I assume that cornering is caused by slight drifting.

11. #11 rubbertree
February 22, 2010

I love your work here, Rhett.

I have a suggestion for a future post, perhaps during the NBA finals: “Spudd Webb Physics.”

12. #12 Birger Johansson
February 23, 2010

On terminology: Simply call it “the centrifugal pseudoforce” and get on with it….the readers will understand what it means.

13. #13 Jen
February 25, 2010

I have nothing of importance to contribute here, but I just stumbled across this and had to say how awesome it is.

14. #14 Craig
February 25, 2010

As a former short track speedskater with a degree in Biomechanics I can confirm everything posted here.

For those who are still curious, I’ll throw in some more data.

Obviously they dont skate ON the layout. The actual radius that anyone skates ranges from as low as 8m to as high as 16m, all within the same turn. It is never consistent through the entire turn. So any scientific calculations could only be specific to the instantaneous sample. 12m/s is a very typical velocity for short track. The actual question to be asked with these calculations is “What radius is he skating at this exact point?”

The blades are both curved and offset towards the center of the corner. In most cases the heel will be centered. On the right foot the front of the blade passes approximately between the big toe and second toe. On the left foot, depending on the width of the skaters foot, it will pass +/- through the center of the second smallest toe.

The “rocker” of a blade ranges from 8m to about 12m radius. The “rocker” is the normal curve in the blade you would expect. A few skaters will have a consistent rocker of only one radius. Most elite skaters have a compound radius, flatter in the center and more round at the front and back.
Flatter glides better in the straights. Rounder allows tighter corners and pivots. Tweaking the ankle at different points of the track positions the force through a different part of the blade and thus a different rocker value.
The “Bend” of the blade is a curvature that allows for more blade contact while cornering. Both the right and left blades have a slight concave bend as referenced from the center of the track radius. As the skater leans over, more blade contacts the ice. This curve ranges from about 18m to around 26m radius.
These two parameters combined result in some very complex relationships with the ice. In real life the numbers are far to complex. Thus elite skaters spend most of their careers going through trial and error to find the best setup. Once they find something they like it becomes a hugely guarded secret. As an aside, if you look closely at any pictures that show the bottom of someones skate you will see tape next to where the blades attach to the boot. This tape has little pen marks to help the skater remember past positioning. In other words, all of these things are constantly being tweaked. EVERYONE has a wrench with them while on the ice during practice.

15. #15 Claire C Smith
February 25, 2010

To comment 14, Craig,

Hey thanks for that. All a bit clearer. I wasn’t suggesting that the skater becomes air born when changing leg – the animal analogy was to highlight what more can be seen when slowing down a process.

By the way, I posted my first comment on this page a few days ago here, thinking I was still on the New York Time Science section and commenting on that – I was reading an article on NYTS then clicked here thinking it was still the same site. So, now accidently quite glad I saw this kewl site for the fisrt time.

Thanks,

Claire

16. #16 Ethan
March 5, 2010

l like ice cream

17. #17 Shaner
September 13, 2010

I am a short tracker myself. I just wanted to note the hand touches the ice only when we are near top speed and need extra support

18. #18 Brett
September 28, 2010

I don’t know if you know this but Apolo has angled cups on his skates. the blade isn’t quite perpendicular to his foot which means that the angle your measuring might be off.

19. #19 Craig
September 30, 2010

The angle remains the same since what is being measured is the angle that is created from the line that starts at the point of contact on the ice and passes through the body’s center of mass.

Angled cups don’t effect this measurement. In other words, they don’t change the angle of lean. Just the contact angle of the blade to the ice.

20. #20 Barraldinho
December 6, 2010

Nice breakdown of the basic forces

But I think you overlook some things
1- The lean angle is not the same all the way thru the corner- you fall in to a maximum lean at the apex- the pop backout the exit- it’s an orbital tube – think of a swept volume tracing where all the body parts have been over time.

2. The COM is not fixed as the body moves – as you say the diff body libs have different orbital radii, and it really is a big difference – and all have different moments of inertia, especially the parts that swing- like legs and arms.

3. You ignore the corriolis force ( yes psudo forces are awesome, and are very real) especially on trailing limbs with reference to the first person frame of ref.

4. You ignore the Newtons law of conservation of angular momentum, Iw = Iw – this is a key equation for objects that change their orbit, it’s what gives good skaters that “slignshot ” effect when they tighten their track. It’s also why core stability is so important, without it you get no efficient conversion of angular momentum to radial spin/torque.
Newton says ” In a rigid system angular momentum is conserved” So rigidity is important,even when you are moving a lot, you have to keep some rigidity /core stability

Also there are many internal torques in the body that act to store and release rotational and compression energy (like plyo effects from muscles and tendons), when skaters fall they sometime spin out anti clockwise- as the rotations are contained by the centripedeal force of the blade path, then released when they fall

I’d say there is a bunch of Euler transforms involved and you would need a bunch of Quaternion math to solve it, espcially the blade dynamics.

The rotational dynamics are very complex, and are the hidden forces that can be quite non linear, as the blade behaves dynamically depending on angle of lean, blade profile and torque and vert loading applied to it.

this might be interesting to u

Great thread- I have some other writing on this I will dig out.

Cheers
Bar

21. #21 Barraldinho
December 6, 2010

Found this on the web- can’t remember where . .
Generating Angular Momentum
An object does not just typically have angular momentum. Recall Newton’s first law that an object in motion tends to stay in motion. Well, if a figure skater is just skating straight down the ice and then needs to perform a spin or jump with several rotations in the air, he or she needs to generate angular momentum. Angular momentum is generated by the skater applying a force against the ice. The ice then applies a ground reaction force on the skater. This ground reaction force causes gives the skater angular momentum.
The point of application and line of action of this force is critical. If the line of action of the force is directed through the skater’s axis of rotation, then he or she won’t spin. The force must cause a torque, or moment, which means it must be applied some distance from the axis of rotation AND have a line of action which does not go through the axis of rotation.

The larger the force or the farther the force is from the axis of rotation, the larger the torque. The larger the torque, the greater the angular momentum.
Another key consideration in generating angular momentum is the object’s moment of inertia. The larger an object’s moment of inertia, the more angular momentum the object can obtain. For example, if a figure skater wants to generate a lot of angular momentum, they should have their arms spread wide, which increases their moment of inertia. In this position, while the skater will have to have a large torque to start rotating, his or angular momentum:

will be larger due to the large I. A skater who starts spinning with his arms at his side, with the same angular velocity will have a smaller angular momentum. Moreover, this skater will not be able to increase his speed in the spin, because he will not be able to reduce his moment of inertia to increase his angular velocity. Two animated figures are provided to illustrate this idea.

The larger the moment of inertia the more torque it takes to start the object spinning. Thus, there is a trade-off between moment of inertia and angular velocity when generating angular momentum. In figure skating, the skaters do not usually have a problem with having producing large enough torque to start spinning. Accordingly, it is to their advantage to start every spin, or rotational trick, with a large moment of inertia. They accomplish this by having their arms and free leg held away from their body.

Some skaters reach rotation speeds of 7 rev/s during a jump. This corresponds to 420 rpm (revolutions per minute). This is as fast than the idling speed of the engine on some cars!!!!

Given the following moments of inertia and angular velocities of the skaters initiating spins, calculate their angular momentum and answer the questions that follow. Note that when calculating angular momentum, it is important to convert any angular velocity to readians/s before performing the calculation.

22. #22 Barraldinho
December 6, 2010

Also this is interesting, and often overlooked in ST technique.

—-

The Physics of Ice Skating – Angular Momentum
The angular component of linear momentum is angular momentum. When an object rotates around a fixed axis, the force acting on the object is called the centripital force. This force points inward, toward the center of the circle traced by the rotation. The velocity of the object points tangential to the circle traced. This is illustrated by swinging a ball on a string around your head (don’t hit any lamps though). If the ball becomes detached from the string, it goes flying in a straight line.
The vector for angular momentum points perpendicular to the velocity and force vectors. It goes according to the “right hand rule.” This is just a simple way of remembering where the angular momentum vector is pointing. Angular momentum is represented by the equation L=I where I equals the moment of inertia and is the angular rotation or the period of rotation divided by 2 . The moment of inertia depends on the mass of an object and also the distribution of that mass around the axis of rotation. So a skater can have a different moment of inertia based on whether their arms are extended or not. This can be compared to linear momentum where p=mv or linear momentum equals mass times velocity.
Angular momentum is conserved when no outside torques act on an object. As say, the moment of inertia decreases, the angular rotation has to increase to keep the same angular momentum. This is most evident when a figure skater spins. A skater starts the spin with arms outstretched (a large moment of inertia).  As the skater brings the arms in (decreasing the moment of inertia), the rotational speed increases. This is how those incredible spins skaters like Paul Wylie, Todd Eldridge and Kristi Yamaguchi are accomplished. Along with many long years of practice.
Most of the spins done by world class figure skaters are edge turns, meaning they are spinning while remaining on an edge. For beginners, often the first spin learned is the two-footed spin. A skater rides a large curve with most of their weight on an outside edge. As the curve spirals into the center, the skater rises up on the flats and begins to spin. One of the most important aspects of a spin is how to “center” a spin. This refers to the property that the spin should stay in one place and not travel all over the ice (which is quite hazardous). This requires converting all of the linear momentum into angular momentum. (Another conservation law)
Another example of conservation of angular momentum occurs when a massive star (meaning several times the mass of our sun) dies. As the star, which is already rotating, begins to collapse, it becomes a smaller sphere which decreases its moment of inertia. Since the star is an isolated system with no forces acting upon it, the angular momentum must be conserved and the rotational period of the star increases. If the star (known now as a neutron star) is emitting a beam of radiation, its rotational motion makes this beam appear to us like pulses. These stars are known as pulsars.
Back to the Physics of Ice Skating
by Karen Knierman and Jane Rigby

The Physics of Ice Skating – Torque
Torque is a rotational force, in fact the word itself comes from the latin for “to twist”. Torque, in a sense, causes rotation about an axis. Torque involves both the force applied to an object and the distance from the rotation axis you apply the force. Perhaps some examples will help. In order to open a heavy door, you need to apply a force. But force alone will not do the job. Where the force is applied and in what direction is also important. If you apply a force close to the hinge of the door rather than out by the doorknob, it is much harder to move the door. That’s why doorknobs are located at the opposite side of the door to the hinges; it’s much easier to move the door out there.
The definition of torque is the product of the distance from the axis of rotation (often called the lever arm) with the force that is perpendicular to the lever arm. (If you pull on a door parallel to the plane of the door, you do not rotate the door.) Another example of torque occurs when you turn a screw or bolt. Using a screwdriver (the non-electric kind) is often hard and time consuming since you must apply a large force in order to turn the screw (small lever arm). However, if you use a wrench for tightening bolts, you only need to apply a small force since you have a long lever arm. That’s why wrenches used to turn large bolts have much longer handles compared to those that turn small bolts. This enables the user to use less force since they have a long lever arm. Of course, the user must apply that force over a longer distance.  So there’s a tradeoff between force and distance.
So how do we get from tools to ice skating? A skater, in order to rotate, must exert a torque on his body by pushing against the ice. In edge spins, the skater pushes one foot against the ice to start the turn. You also see this in multiple rotation edge jumps. In these jumps, the skater takes off from the ice, turning the skate as he does so, which creates a torque.  Thus, the skater spins!

You can think about the physics of moving objects in two different and equally acceptable ways: in terms of forces, or in terms of energy. Which you use depends on which is more convenient; both will always work.  In this section, we’ll look at energy in ice skating.  It’s an energetic section!
Kinetic and Potential Energy Explained
Kinetic energy is the energy of motion (“kinetic” means motion). So an ice skater flying across the ice has lots of kinetic energy. When he slams into the boards, that is transferred abruptly into thermal energy, sound, and work done on the skater (ie compressing his chest, rearranging his face, etc.)  Seriously, to measure kinetic energy (KE), just measure the mass of the object and its velocity. KE = 1/2 mass*velocity*velocity or 1/2 mv^2.
If the movement is rotational instead of straight-line, then the equations are similar. (See Karen’s section on rotation.)  The idea is the same – how fast the skater is spinning is directly related to her kinetic energy.
It’s easy to “see” kinetic energy in the motion of an object. But what about potential energy? Potential energy is stored energy. Energy can be stored in chemicals, by compressing a spring, or by doing work against gravity (ex: by placing an object on a higher shelf.)  Muscles act like springs, so the chemical potential energy of muscles is converted, by the muscle applying a force, into kinetic energy of motion.
Energy Conversions in Jumping
In a jump, the skater uses chemical potential energy (muscle power) to gain speed across the ice. When she jumps, she’s also converting her chemical potential energy into kinetic energy. As she flies upward, her kinetic energy is converted into gravitational potential energy; as she slows, she gains height above the ice. At the top of her jump, she has no kinetic energy (for a moment!)  There, all of her former kinetic energy is now gravitational potential energy. As she falls back down, her potential energy is converted back into kinetic energy. At the moment she hits the ice, all the gravitational potential energy she had at the jump’s peak is again kinetic energy, so she hits pretty hard. If you measure the height of her jump, you can determine how hard she pushed off, which is also how hard she smacked down.

The Physics of Ice Skating – Isaac’s Third Law
Newton’s third law is one of the most-quoted in physics:   for every action, there is an equal and opposite reaction.  Lots of times, this is used out of context (to describe politicians, etc.)
The basic stroke in ice skating provides a good example of Isaac’s 3rd law in action. When you “stroke” (the basic push in ice skating), you apply a backwards force to the ice. The ice applies an equal, forward force on you, so you go forward. Skaters stroke at an angle, so part of the stroke is wasted. You’re pushing forward and to the side. The side push is resisted by the edge of your other blade. The forward push is resisted only by ice/blade friction, so you go forward.
Actually, this is exactly why sailboats go forward, not sideways. The wind pushes the boats forward and to the side. The sideways push is resisted by the long keel, but the forward push is relatively unresisted. Boats are designed to be aerodynamic (actually, “hydrodynamic) to forward motion and are intentionally unhydrodynamic to sideways motion.
Note that these forces apply only to direct pairs of objects. I push on the ice; it pushes on me. As I push in on the wall, it pushes me outward. Third law reactions never involve a third body.

23. #23 Barraldinho
December 6, 2010
24. #24 messerman
February 15, 2011

surprising that so much mental muscle has gone into accepting the fact that we lean round corners. Even my horse knows that (which can be daunting as he leans to go round a tree and near wipes you out). And the fact of maybe coming out of a turn faster than entry is clearly nothing to do with some imaginary sling-shot effect. Quite simply if your linear speed out > linear speed in then yo have inserted extra energy (more than reqired to ovecome resistive forces). which is what they do. QED

Of much greater interest – and perhaps not such risk of confounding by minutia – is the question of the “slalom” style of propulson of long track long distane (10km olympics for example) speed skaters. They constantly skate a curve path (to left and right) up their 100m straight. Putting aside the illusory sling-shot model, how do they insert forwards thrust into their bodies by swerving side to side. no other sport (motorcycle racing, even salom ski racers on their final straight swoosh to finish) – nobody believes that swerving increases speed.

25. #25 FirstLoser
July 20, 2011

Wow – thanks – this is great!

26. #26 FirstLoser
July 20, 2011

Wow – thanks – this is great!

27. #27 Mark
December 13, 2011

I stumbled across this photo looking for for images of skaters to put into a biomechanics exam that I am preparing. Funny that the photo I clicked on happened to be associated with the exact problem that I was formulating!