How Do Figure Skates Enhance Skating Performance?

[HPsource]

Please help - How do figure skates work...?

How exactly do figure skates work? and how do the materials used help and how do figure skates improve a skaters ability to skate?

any links, commments and answers would be appreicated, thank you.

 

[Math Jeans]

Skates work because the blade has a low surface area, and when your weight is on those blades, and contact is made to the ice, it melts the ice at that area causing you to slide. Figure skates just have ridges at the tip of the skate so you can spin on it.

 

[rcgldr]

Skates work because the blade has a low surface area, and when your weight is on those blades, and contact is made to the ice, it melts the ice at that area causing you to slide. Figure skates just have ridges at the tip of the skate so you can spin on it.

The surface of the blade is concave with sharp ridges, for cornering. Without the sharp edges, it's difficult to turn without the skates sliding. Figure skates and hockey skates have curved blades, which allows spinning even at the middle of the blade. Speed skates are almost straight and can't be used for spinning.

 

[Cthugha]

Skates work because the blade has a low surface area, and when your weight is on those blades, and contact is made to the ice, it melts the ice at that area causing you to slide. Figure skates just have ridges at the tip of the skate so you can spin on it.

This is an old and wrong myth.
People are not heavy enough to melt ice significantly. The thin water film on top of the ice is a surface effect.

Although Wikipedia is usually a questionable source, in this case the article about skating is quite ok:

http://en.wikipedia.org/wiki/Ice_skating

 

[russ_watters]

It should be evident from trying to walk on ice that there is very little friction, so there is no need to melt ice for an ice skate to slide.

 

[rcgldr]

People are not heavy enough to melt ice significantly.

I updated my post. The sharp edges cut into the ice, increasing cornering "grip". The ridges at the front of figure skates is for "toe" jumps, they dig into the ice allowing a skater to "plant" the toe into the ice for a launch. The toe picks can't be used for spinning. Another type of jump uses an edge to "plant" into the ice (which is why they need to be sharp), and the final type of jumps are basically vaulting.

 

[Danger]

A new type of hockey skate has just been introduced to the market, and has been approved (at least for trial) by the NHL. There's an electric heating element in the blade that causes significant melting of the ice at the contact points. There have been claims of a 10% or more increase in skating speed, with improved manoeuvrability.

 

[FredGarvin]

The two edges created by the undercut in the blade (this feature is called the "hollow") is analogous to your edges on snow skis. When you get your skates sharpened, the sharpening accentuates these two edges. You can feel the two edges when you are skating and turning as your weight gets moved from one side to another. They are definitely needed for turning of any kind. A goalie skate has a pretty flat blade with little to no hollow. This allows a goalie to slide side to side in the net.

The curve of the blade you see if you look at the blade form the side is called the rocker. It varies depending on what the skater likes. Defensemen usually like a fair amount of rocker because it allows them to turn quickly to skate backwards. The drawback is that with the reduced surface area in contact with the ice, the pressure increases and it does slow you down a bit when compared to a flatter rocker.

 

[Dissonance in E]

This is somewha related so i didnt want to make a new thread. Could someone explain the mechanical principles and math behind the classic example of a figure skater pulling her arms close to her body and thus increasing her spin velocity?

 

[cepheid]

This is somewha related so i didnt want to make a new thread. Could someone explain the mechanical principles and math behind the classic example of a figure skater pulling her arms close to her body and thus increasing her spin velocity?

It has to do with conservation of angular momentum. Just as momentum "quantifies" translational motion, so does angular momentum quantify rotational motion. The angular momentum (L) of a particle is given by:

\mathbf{L} = \mathbf{r} \times \mathbf{p}​

where r is a vector from the centre of rotation to the particle, and p is its momentum. For the simple case of motion in a circle (edit: at a constant speed), r is always perpendicular to p, and so the magnitude of the cross product is simply given by

|L| = L = |r||p| = rp = rmv​

where |r| = r is the radial distance to the particle from the centre of rotation. You can see that as r decreases, v must increase for L to be conserved. For extended bodies such as figure skaters, the actual calculation of the angular momentum is more complicated, but the principle still applies, because we can consider the figure skater as an aggregate of a large number of such particles, each orbiting about the centre at a different distance (depending on its position in the figure skater's body).

If you're wondering intuitively why a large rotating body has MORE angular momentum than a small one rotating at the same speed, it has to do with the fact that it's harder (it takes more work) to get the larger body rotating at that speed in the first place. For translational motion, an object's inertia is quantified solely by its mass. But an object's rotational inertia (how much it resists a change in its state of rotational motion) depends not only on its mass, but also on how that mass is distributed in space. (It is quantified by the moment of inertia). This ties in with the idea of torque...that it's easier to make something spin if you push on it at a point farther away from the spin axis (e.g. it's easier to push open a door using the handle than it would be if you pushed on it at a point close to the hinges). In fact, just as force is the rate of change of momentum (dp/dt) so too is torque the rate of change of angular momentum (dL/dt). That's why, for a system like the figure skater, if no external torques are applied, angular momentum must be constant. There are many such analogies between dynamical quantities describing translational motion, and those describing rotational motion...if you spend some time looking at the area of classical mechanics known as rotational dynamics, you will see how everything ties together.

 

[Dissonance in E]

all right,cool. that makes sense. thanks!

 

[jachyra]

cepheid, that was a really awsome explanation!

There are many such analogies between dynamical quantities describing translational motion, and those describing rotational motion...

could you please list a few of your favourite sites that have these analogies?