First of all, I must thank Firehawk, who is oceans away,for helping me review, revise, and offering many instructive comments.
I may not ski well, but I sure can talk about it well enough. I’ve founded my own "Mouth School" to explore the physical mechanics of skiing. All content below is original — not necessarily all correct — and I welcome discussion, corrections, critiques, and even brickbats.
I really hate it when principles are made to seem as complicated as Olympic math problems (a hallmark of the Chinese education system), where half an hour of explanation leaves you more confused than ever, as if the goal is to make things as incomprehensible as possible. Why make knowledge so complex? Theory serves practice — you can ski just fine without it. But if you’re not skiing that well, then you’d better be good with words, because that can still impress people...
Please first take a look at this image:
[attach]2290125[/attach]
Image source: http://www.skiforum.it/forum/scuola-sci/62829-pmts-5-anni-e-possibile-4.html
(This is a very good diagram with many elements, but the theory I’m discussing isn’t much related to these elements, so please analyze everything in the picture on your own. The linked content is also rich in information.)
I want to ask: are points 4 and 12 the same? I remember someone previously posted a force analysis, summarized as follows: they pointed out that A is the turn entry and C is the turn exit, so A and C are different. But I think the only difference between points A and C is that one has the body facing left and the other facing right — everything else is exactly the same. If point A is the turn entry, then point C is also a turn entry... I don’t think it’s possible to instantly go from point C to the next point A. A and C are essentially the same thing. With this view, the following analysis becomes much easier. We only need to complete the analysis from A→C to understand the entire turning process in skiing. No problem with that, right? Although most people are asymmetrical on both sides, mechanical analysis isn’t affected — it’s purely idealized.
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Below, points A, B, and C will be used to represent the apex, midpoint, and base of a turn. So what happens at point A? In reality, the skis at point A are basically not horizontal (relative to the horizontal plane) — they almost certainly have an angle facing downhill [very few people draw the semicircle as shown in the diagram; usually it’s just an S shape. Reality is definitely like this, unless you deliberately turn your skis uphill. It’s impossible to ski such a perfect semicircle, which I’ll explain later — this is also the result of my dynamic mechanics analysis]. At this point, there may be three scenarios:
1. The center of gravity is on the uphill, outer side of the ski.
2. The center of gravity is on the ski or between the two skis.
3. The center of gravity is on the downhill, outer side of the ski.
Almost all diagrams explain this based on scenario 2, because advanced skiers can achieve the edge change just as the center of gravity crosses over the ski (here we consider it as having crossed the ski as long as it’s in the middle, since this moment is very brief). This is also one of the goals we strive for. But even achieving scenario 1 is fine, because it’s a sign of becoming an advanced skier — you’ll soon cross over. As for scenario 3, I think athletes possess this quality. Athletes change edges as needed, and can do it at any point during the turn (like in mid-air). But let’s not get into that — the skill gap is just too big. If it’s scenario 1, the center of gravity will also cross the ski surface at some point between A→B, so for the majority of the turn, the center of gravity should be on the inside of the ski. Little tip: keep practicing to bring the edge-change point closer and closer to point A, until they coincide.
A rant about “center of gravity”: I actually don’t like using the term “center of gravity,” but since everyone keeps using it, I have to explain my view. First, the center of gravity is the mass center of an object. For a person, it’s roughly located somewhere behind the navel. Of course, this is when standing upright. If you bend over or twist your legs, the center of gravity might even be outside your body. In popular skiing understanding, “center of gravity” is mostly interpreted as the point where the body exerts force on the snow surface. Saying the center of gravity is inside the ski is understood as the equivalent pressure point being on the inside of the ski. Actually, this force is a combined effect of two forces: not just gravity, but also the reaction force from the ski. The resultant of these two forces forms a combined force — complex and elegant. Here, I’ve coined the term “center of gravity equivalent action line” to understand the line connecting the center of gravity and this action point. When standing upright with feet together, the center of gravity equivalent action line goes straight down the middle of the body. During single-outrigger skiing, it’s equivalent to the line from the center of gravity to the contact point of the outer ski with the ground. This line can be seen in many diagrams explaining counteracting, for example:
[attach]2290127[/attach]
Image source: http://www.yourskicoach.com/glossary/SkiGlossary/Angulation.html
The pose in the picture is quite exaggerated — in reality, there’s no need to go that far. Just having the foot, knee, and shoulder roughly in a line is enough. The green line in the diagram is what I call the “center of gravity equivalent action line.” What people refer to as the “center of gravity” is actually the intersection of this line with the ground.
We often think of the center of gravity as the equivalent point of the gravitational force on the snow surface, believing that skiing is caused solely by gravity. But it should be said “not entirely,” because there’s also the reaction force from the ski on the snow surface. These two forces are the two main forces in my dynamic mechanics analysis today.
High-speed air resistance and wind force — can these be ignored? Based on experience, a cyclist at 50 km/h experiences about 4–5 kg of wind resistance. Ski clothing isn’t as form-fitting as a cyclist’s skinsuit, so I estimate the resistance at around 10–15 kg (cycling wind resistance data: http://bbs.8264.com/thread-1232166-1-1.html). The wind resistance formula is:
[attach]2290128[/attach]
Where
Cd is the drag coefficient
is the air density
S is the frontal cross-sectional area
v is the speed.
I assume the ski suit area increases threefold, estimating this 10–15 kg of force — not insignificant, and it can indeed affect skiing. I’ll reflect this in the dynamic calculations below, but because the calculations are too complex, it’s just a reference force for now.
Alright, the three forces for doing physical mechanics analysis in skiing have emerged: gravity, ski reaction force, and resistance (air resistance and ground resistance). I’ve gone on about a bunch of dry theory, but please bear with me, because the formulas and calculations that follow are even drier — if you’re not interested, just give a “like” or “thumbs up”!
Newtonian mechanics (a manifestation of classical mechanics) analyzes the magnitude, direction, and point of application of forces (luckily I haven’t forgotten it all — see http://zh.wikipedia.org/wiki/%E7%AE%80%E5%8D%95%E6%9C%BA%E6%A2%B0). The forces we need to analyze today total three. What are their three elements respectively?
1. Gravity:Magnitudeis always equal to your body weight,Directionis always vertically downward, as long as you haven’t flown off the Earth;Point of applicationrequires careful analysis—depending on how your body moves, whether leaning, arching backward, adjusting front to back, etc., the point of application shifts slightly, which also corresponds to the shift in your body’s center of mass.
2. Snowboard reaction force:Magnitudeis always equal to the pressure exerted on the snowboard. This pressure may come from the component of gravity or the force you apply by pushing down with your thighs. Some might wonder, “Doesn’t this pressure always equal the component of gravity in that direction, whether or not I’m pushing down?” I thought about this for a long time and finally got it. First of all, that understanding is incorrect—it’s not the component of gravity in the direction of the reaction force, but rather the component of the reaction force in the vertical direction that happens to equal gravity. Logically, this pressure can be split into two components: one contributes to the centripetal force needed for turning on the slope, and the other changes as your body moves up and down, altering the pressure through pressing down on or relaxing the snowboard, and shifting your center of gravity up or down. Why do I say “logically”? Because in reality, there is only one force; I’m just dividing it into two parts for easier understanding. So, to reduce pressure on the snowboard, certain adjustments are needed;Direction—if we consider the snowboard as equivalent to a single point, then the direction of this force points directly along what I mentioned earlier as the “center of gravity equivalent action line”;Point of applicationis always at the contact point between the snowboard and the ground. Well, that’s a given.
3. Air resistance:Magnitudeis related to the square of velocity (there’s a formula above),Directionis always opposite to the direction of motion,Point of applicationcan be considered as acting on the center of mass.
[attach]2290129[/attach]
Image source: http://www.tudou.com/programs/view/PPGNZhOrQnM/?FR=LIAN
Actually, analyzing resistance involves very complex mechanics, with an entire branch of physics dedicated to it—I, being a math person, simply can’t grasp it all. Here, I boldly make some assumptions (math folks love making assumptions and then deriving a bunch of theories they barely understand themselves):
1. Air resistance is only related to speed. Let’s assume: at 30 km/h, air resistance is 5 kg; at 40 km/h, 10 kg; at 50 km/h, 15 kg; at 60 km/h, 25 kg. I’ll go with these assumptions for now—if they’re wrong, we’ll adjust later...
2. The direction of air resistance always points toward the rear of motion. Whether turning or adopting various postures, resistance always pushes backward.
Next, we have to make a few more assumptions—no way around it, otherwise we just can’t proceed:
1. Assume the human body can be equated to a cylinder, with the cylinder’s axis being the aforementioned “center of gravity equivalent action line”;
2. Assume the two snowboards, in the direction of force, are equivalent to a single point. Typically the outside ski does the sliding, so this equivalence holds. Even if the forces are split 50-50 between both skis, the assumption still works—the only difference is that the cylinder would be slightly thicker.
3. The snowboard, along its length, can be equated to a single point. Although the board is long and wide, for mechanical analysis purposes it can be simplified to a point.
4. The pressure exerted by the person on the snowboard is singular. Earlier I said that this force can logically be divided into two parts, but when analyzing, we treat it as one. Here, factors like adding or reducing pressure or edging techniques aren’t considered—we’re only looking at the net effect of the pressure.
One more thing: all my analyses don’t distinguish between Skidding (sliding turns) and Carving (carving turns). Many assume Carving makes it easier to ski half-circles and often use Carving as an example to analyze such turns. But according to my calculations below, it’s actually only Skidding that can produce a true half-circle; Carving can’t produce a perfect half-circle. Surprising, right...?
Based on these assumptions, I can draw a simplified diagram of the forces acting on the slope:
[attach]2290130[/attach]
The image shows an inclined surface, where the orange line segment at point A represents a person standing vertically on the snow (on the slope). In reality, people rarely stand perfectly perpendicular to the snow surface—there’s usually some angle—but here we ignore that and use it only as an analytical assumption. The angle between the person and the direction of gravity equals the slope angle θ. At this moment, the person is only affected by gravity (we temporarily ignore resistance because its direction does not affect centripetal force). Actually, at point A, besides having speed in the horizontal direction (defined as the direction perpendicular to the fall line), there is also speed along the fall line. This speed is used to quickly shift the center of gravity, but for ease of calculation, we temporarily ignore this directional speed and set it to zero. Therefore, the mechanical analysis at point A is as follows:
[attach]2290131[/attach]
Where:
F_centripetal is the centripetal force at point A, directed toward the center of the circle;
F_perpendicular is the magnitude of the pressure exerted perpendicularly on the snow surface. This force is the component of gravity acting on the snow;
F_downhill is the resultant force along the fall line. Since all of gravity acts as centripetal force, the resultant force downhill is exactly zero at this moment;
Note: This analysis takes the snow surface as the reference frame, and all subsequent analyses also use the snow surface as the reference frame [very important!] The snow surface is an inertial reference frame, and we can still analyze it using the method of components and resultant forces. The three component forces analyzed here are:
1. Centripetal force, which always points toward the center of the circle, changing direction but not speed;
2. Pressure perpendicular to the snow surface, which does not affect direction or speed;
3. Resultant force along the fall line, which is the force that accelerates, decelerates, and also changes both direction and speed;
These three component forces are not perpendicular to each other. I’m only providing the mechanical analysis method—specific calculations are beyond my derivation due to insufficient math skills. Clearly, at point A, only the components of gravity in various directions are present, but at point B, things are different. At point B, both gravity and the reaction force from the snowboard jointly influence the glide (resistance is still ignored for now because resistance is always perpendicular to the centripetal force and does not affect direction).
[attach]2290132[/attach]
The analysis at point B is as follows: Point B is at the same horizontal level as the center of the circle O. At point B, the person leans sideways toward the snow surface. Assume the angle between the body and the snow surface is α. At this moment, the snowboard aligns with the fall line, and the body is facing directly toward the fall line. The centripetal force is entirely horizontal, and gravity does not affect the direction of the glide (because the component of gravity along the fall line direction of the snow surface is perpendicular to the centripetal force—this component can only contribute to acceleration along the fall line without affecting the centripetal force). Therefore, the three component forces at this moment are:
[attach]2290133[/attach]
Explanation—if incorrect, please correct me. At point B, the body is leaning sideways, making an angle α with the snow surface. Due to the slope of the surface, the component of gravity perpendicular to the snow surface is Gcosθ, which is also the normal force exerted by the body on the snow;
The centripetal force is the horizontal component of the reaction force from the snowboard. Since the vertical component of the snowboard’s reaction force equals the component of gravity perpendicular to the snow surface, the magnitude of the snowboard’s reaction force equals Gcosθ/sinα (where α is the angle between the body and the snow surface). The centripetal force thus equals Gcosθ/sinα * cosα = Gcosθ/tanα;
The force toward the downhill direction is the component of gravity along the fall line of the snow surface, which at this point only contributes to acceleration.
Due to the uniqueness of point B, at this position, the body's gravity and the centripetal force are perpendicular, and the downhill force gives the fastest acceleration. Therefore, overcoming the fear at point B as quickly as possible is essential in skiing. Point B has the fastest acceleration and provides the most noticeable feeling of weightlessness, but the point of highest speed is actually midway between B and C—at point B′, which will be discussed below.
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First, let’s talk about point C. Analysis shows that it is exactly opposite to point A. I previously mentioned that points A and C are entirely the same, but for half a turn, A and C are symmetrical:
[attach]2290135[/attach]
At point C, the centripetal force and the fall line direction are exactly opposite, so the signs are reversed, and the resultant downhill force remains zero.
We’ve now analyzed points A, B, and C. Next, we’ll further analyze three additional points—one between A and B, and two between B and C. The point A′ between A and B is chosen arbitrarily. The points B′ and B″ between B and C are selected more specifically: B′ is the point of highest speed, largest lean angle, and best photographic position; B″ is the starting point for pressure relief and edge change.
[attach]2290136[/attach]
Let’s first analyze point A′. Assume the angle between the body and the snow surface is α, and the angle between the arc and the fall line is β. Then the analysis of the three forces yields the following formulas:
[attach]2290137[/attach]
At this moment, the centripetal force consists of two components: one from the snowboard reaction force, and the other from the component of gravity perpendicular to the fall line (pointing toward the center of the circle)—a bit complex, but this is the actual situation. The centripetal force from the snowboard reaction force still satisfies the previous analysis, equaling Gcosθ/tanα; the component of gravity in this direction is the second part of the above equation.
As with points A and B, the force perpendicular to the snow surface always equals the component of gravity on the slope—this remains constant at any point;
At this moment, the resultant force along the fall line consists of three parts, all derived from sub-components. The main two are: the component of the snowboard reaction force toward the center of the circle, then its component along the fall line (with an angle β between the centripetal force and the fall line), and the component of gravity along the fall line. The difference between these two is the resultant force along the fall line; then subtract the component of resistance along the fall line.
Since point A′ is chosen arbitrarily, the trend of the transformation of the three forces from A to B at point A′ follows the above formula: α becomes smaller (lean angle increases), β becomes larger (until 90 degrees, which is point B). This formula also applies to points A, B, and C. In fact, it holds true at any point on the entire circle—try proving it yourself!
[attach]2290138[/attach]
The mechanical analysis at points B′ and B″ is exactly the same as at point A′, but since β is greater than 90 degrees, cosβ becomes negative.
[attach]2290139[/attach]
Points B′ and B″ are very special. From B to C, there are two main phases: first acceleration, then deceleration. The resultant force along the fall line gradually decreases (positive) to zero [at B′]; then gradually increases (negative) to its maximum [at B″]; then gradually decreases (negative) to zero [at C]. That is, you accelerate up to B′, then decelerate to B″, and then decelerate again to C. Point B′ is where you reach maximum speed, and point B″ is where you begin to release board pressure.
The analysis shows that the lower half of the turn is much more complex than the upper half. If you truly want to ski well, mastering the lower bend is essential!
My understanding of the general learning sequence is as follows:
1. First, master basic parallel turns, keeping the snowboards roughly parallel;
2. Try to initiate inward lean early, setting the edge from point A and entering the turn as soon as possible;
3. Carefully distribute your strength and balance the three phases from point B to C—this way, you can achieve a refined steered turn.
If no one actively participates and lets the body fall freely, then the speed from point B to point C will definitely keep increasing, eventually causing an uncontrolled launch off point C. However, due to the active control of leaning by the skier, as can also be seen from the above formula, the analysis of all forces includes the lean angle α. The most crucial aspect of skiing is fully controlling this angle — too large, and you fall over; too small, and you tip sideways. This brings us to the “lean angle theory” mentioned in my title. In skiing, the lean angle is the sole factor that determines the state of the glide, speed, and power. Below, I’ll provide the calculation for this angle, though it’s bound to get even more tedious.
If the speed at point B’ reaches its maximum, then the resultant force in the direction of the fall line should be precisely at the moment when it changes direction — that is, the point where the resultant force is exactly zero (the point of zero resultant force is also the point of maximum tangential speed, which you can prove on your own). In the above formula, the magnitude of F_downhill equals zero, that is:
Therefore we have
[attach]2290140[/attach]
Explaining the formula is far too complicated. To give everyone an intuitive sense first, let’s temporarily ignore the resistance component. After simplification, the lean angle α only relates to β. I made a table in Excel to give you a rough idea.