# Why does a curling rock curl?

In the game of curling, players "curl" a granite "rock" (of precise size and roughly a flattened cylinder) down a "sheet" of ice towards a target; the "rock" will curve in its path in the direction of the motion of the leading edge. The question here is, what is the mechanism for the curling of the rock?

Standard frictional physics indicates that angular velocity should not have any effect on translational vector; while moving, the bottom surface of the rock is subject to kinetic friction. Due to frictional deceleration, the leading edge of the rock exerts (slightly) more force on the ice than the trailing edge; this would seem to indicate that the rock would curve in the opposite direction from what it does.

Intuition would indicate that the reason for the curling motion would be due to kinetic friction being modified by velocity; that is, along the "left" side of the rock (assuming a rock moving away from the observer) for a rock rotating counterclockwise, the velocity along the ground is slower than for the "right" side; if kinetic friction is decreased as velocity increases, this would be a good explanation for the motion. However, everything I've seen seems to indicate that the coefficient of kinetic friction is constant; is this correct? If so, what is the reason for the (well-observed) phenomenon?

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It seems you must have watched a curling event, so I direct your attention to the sweeping done by the players ahead of the stone. Perhaps I may come back with an answer, but I have some other things to finish right now. – Mark C Nov 25 '10 at 4:25
@MarcC: the sweeping is relevant to the magnitude of the effect, as it's been proven to have a melting effect on the ice, but it doesn't explain the directionality of the effect, which occurs in the absence of sweeping. – Paul Sonier Apr 5 '11 at 4:10

The coefficient of friction for a curling stone is not constant - motion on ice is different from motion on an unchanging solid surface, thanks to melting, which in turn is proportional to most of the factors that affect friction (e.g. contact area, velocity). So, exactly as you suspect, friction decreases as velocity increases.

Because of the changing CofF, the stone curls more strongly at the end of its trajectory (as it gets slower) than at the beginning.

Additionally, as Mark C says, the team can affect the trajectory after delivery by sweeping - the entire purpose of the sweepers is to melt or at least smooth the ice ahead of the stone (or not, as required), which tends to straighten and lengthen the trajectory by reducing the effect of friction on the leading edge of the stone.

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""...solid surface, thanks to melting, which in turn..."" Are the stones heated? – Georg Apr 4 '11 at 14:49
@georg: no, friction (and pressure) causes the heating which causes the melting. It's the same melting that occurs with ice skates. – Paul Sonier Apr 5 '11 at 4:09
@McWafflestix, friction might melt some ice, the rotation would add to this. But pressure melting is a unban legend for skates, in case of those curling stones even more. – Georg Apr 5 '11 at 8:49
This answer is no good. If the x-y plane is the ice, and x the direction of motion, the motion in y is symmetric to reflections, so that for every spot going down there is a spot going up in y somewhere else So a varying coefficient with speed cannot cause an upward force not the answer. – Ron Maimon Dec 9 '11 at 9:23
Why do you think the front exerts more pressure due to slowing? There is no reason to think this, it is incorrect. The pressure is even for a symmetric rock. – Ron Maimon Dec 9 '11 at 9:26

I have considered the wet friction idea too. Ice when compressed will form a liquid layer where the pressure is applied. Yet one thing many seem to overlook is that the surface of the ice that the curling stones move over is not smooth. to prepare the ice for curling we first use a scraper to smooth it and level it. the scraper is a very sharp knife with a 5 feet long cutting edge that is ground to a 0.001 inch tolerance of flatness. Tension is used to pull the ends up a few thousandth of an inch to prevent the edges form cutting a ridge. This knife is run down the ice sheet shaving off the top of the ice. This is done till the all the high spots are removed and the ice is level. Then we take a sprinkler and sprinkle water drops over the ice in an even uniform pattern. This leaves drops about 1/8 to 3/16 inch across and about 1 mm high on the ice surface the spacing of the drops (we call them pebbles) is such that there is about an 1/8 inch spacing between drops. the rocks are not smooth on the bottom the running surface of the rock is a ring about 5.25 inches in diameter and about 5 to 7 mm wide this ring is smooth. The rock surface away form the running surface is a smooth arc on each side (inside and outside the ring) such that the bottom of the rock is cupped inside the ring and oval away from the ice surface outside the ring.

When the rock is running down the surface it does not slide smoothly, it bounces up and down the pebbles as it slides over them. This makes the rumble or roaring sound the stones make as they travel. This bouncing is what I feel is key to the stone's curl. The running surface of the stone forms a plane, so the stone typically would rest on three of those pebbles when stationary. When running down the ice the three pebble resting is not the case, the stone as it slides on to and off any given pebble will move up and down and in some cases the stone may bounce up and momentarily be in contact with the ice on 0, 1 or 2 pebbles. Each time it is in contact with one pebble it will be pinned in place and will rotate about the pinned axis. Statistically the pinned locations will be uniformly spread about the entire running surface and the lateral motion imparted by the pinning would tend to cancel out to a zero mean. but the stones forward motion combined with the spin means that the ring surfaces on either side will be moving at different speeds relative to the ice. On the side that is advancing relative to the stone forward motion, the average time the pebble will pin the stone will be less than that of the pebbles pinning on the slower side of the ring. this on average longer time rotating one way vs the other will bias the lateral motion of the stone towards the direction the rotation is moving the leading edge of the stone. So a stone traveling north with a clock wise rotation viewed from above (leading edge moving east) will be biased to move laterally to the east. The pinning idea is that when a freely spinning disk (think a frisbee) is pinned near the edge of the disk the angular momentum of the disk will cause it to spin about the pinned location as the new axis. When released from the pin the disk will now travel along a new path in the direction the center of mass was going when the pin was released. Thus some of the angular momentum is converted to lateral motion.

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The stone creates a contact patch with the ice. In this case it is roughly circular in shape with approximate radius of $a=0.721\,\sqrt[3]{W\,D\left(\dfrac{1-\nu_{1}}{E_{1}}+\dfrac{1-\nu_{2}}{E_{2}}\right)}$ where $D$ is the diameter of the spherical contact surface (the crown diameter of the stone), $W$ is the weight of the stone, and $\nu$, $E$ are the Poisson's ratio and Youngs modulus for the two materials contacting (ice and stone).

Now because the surface is uneven, the pressure on the contact is not necessarily symmetric, and may have an pressure center of axis from the spin axis of the stone. At most the eccentricity is equal to $a$, yielding the maximum possible torque applied to the stone to be $T=\mu\,W\,a$ where $\mu$ is the coefficient of friction.

The torque the is responsible for altering the spin the stone. Once the stone has spin, then the hydrodynamic contact properties alter the path of the stone.The friction is not constant as it depends on the pressure applied on each spot. Think of the top layer of ice as a lubrication layer, and the stone has to plow through it. As far as the mechanism that this complex interplay creates a side force (tracking force) I am not sure I can explain it. Maybe someone else can.

Ref: "Formulas for Stress & Strain", Roark, 3rd ed. 1954

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I was aware that the reduction in the spin on the stone is due to the torque of the applied coefficient of friction; the real concern is why the vector of the motion changes as it does; yes, the friction is not constant, since it depends on the pressure applied at each spot, but statistically, it evens out, and a pressure-based approach would (as I mention in the original question) indicate a change in vector opposite of the actual effect (due to the front of the rock exerting more pressure due to slowing). – Paul Sonier Apr 5 '11 at 4:19
I found this in Wikipedia: ""The only part of the stone in contact with the ice is the running surface, a narrow, flat annulus or ring, 0.25 to 0.50 inch (6.3 to 13 mm) wide and about 5 inches (130 mm) in diameter; the sides of the stone bulge convex down to the ring and the inside of the ring is hollowed concave to clear the ice."" – Georg Apr 5 '11 at 9:00

In order to move up, the back part of the rock must have slightly more friction on it than the front part.

Let x be the direction of motion, and y span the plane of the disk. The easy case to consider is the rolling stone limit, where the stone is moving forward and rotating with angular velocity $R\omega=v$ counterclockwise as you normally draw the x-y plane. In this case, the top point on the edge isn't moving at any point, the back part is going down in y, and the front part is going up. The friction on the back must be slightly higher to give a net up force.

The asymmetry in the friction must be due to the fact that the front edge encounters new ice, while the back edge is going through ice that has been contacted by the front edge already. This must mean that new ice has less friction than ice that has exerted some friction, so that a smooth sheet of ice has a coefficient of friction that increases with the amount of sliding done on it. This is strange, I would have expected the opposite, but it is the only way to explain the sliding.

The leading pressure is the same as the trailing pressure for a symmetric stone, and asymmetries in the stone do not explain the effect, since they average out over a full turn.

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Research in 2013 produced a paper (here:http://www.sciencedirect.com/science/article/pii/S0043164813000732)

In the abstract, they agree that frictional forces are not the cause:

the curling motion has been attributed to an asymmetrical distribution of the friction force acting on the sliding stone, such that the friction on the rear of the stone (as seen in the direction of motion) is higher than that on the front. In a recent paper, we could show that no such redistribution of the friction, no matter how extreme, can explain the magnitude of the observed motion of a real curling stone.

While the paper is not available for free, a university release points to the mechanism the paper attributes to the curl (from https://www.uu.se/en/news/news-document/?id=2582&area=2)

...the curved path is due to the microscopic roughness of the stone producing microscopic scratches in the ice sheet. As the stone slides over the ice the roughness on its leading half will produce small scratches in the ice. The rotation of the stone will give the scratches a slight deviation from the sliding direction. When the rough protrusions on the trailing half shortly pass the same area, they will cross the scratches from the front in a small angle. When crossing these scratches they will have a tendency to follow them. It is this scratch-guiding or track steering mechanism that generate the sideway force necessary to cause the curl.

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why so complicated? Turning the steering wheel of a vehicle clockwise miraculously makes the vehicle rotate to the right. A clockwise rotating stone has its running surface on the right side more closely matching the speed of the pebble below it especially while decelerating than the left side which is sliding faster over the pebble. If the one foot diameter stone was a wheel traveling down the ice it would be rotating about fifty turns on a standard 130 to 140 foot sheet. The right side surface would be just about stationary relative to the pebble while the left side would be sliding frictionless across the pebble. My experience is based on uncommon sense regarding rotational motion and maintaining curling ice for many years.

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What does a car's steering wheel have to do with a curling rock? how a car turns is completely different from how a curling rock curls – Jim Feb 24 '15 at 15:27

## protected by Qmechanic♦Feb 24 '15 at 15:32

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