was trying to "borrow" and post some pictures from LeMaster's book on the subject. I am not a relative of Bill Gates' so I failed miserably.
(For those of you who have the book, please look at Figures 9.1 through 9.17., chapter: Edging the Ski).
First about the picture from down under. I do not see how it can explain anything. (Thank you for posting it though). It does not show the reaction force from the snow onto the ski. What are the yellow lines supposed to represent? A lever to work must have a fulcrum to develop any leverage. Where is it? What forces are leveraged there? Can you answer those questions based on the picture posted? I can not.
Not having a sketch to help me state my point more specifically, I will have to use many words to describe rather than state what I think so please bear with me. I will ask you to make your own free-hand sketches:
Sketch a tipped skater (not a skier) on ice, as viewed from behind the skater. A horizontal line = ice surface. Another line at, say, 45 deg to the ice surface = line of action of the reaction force onto the skate and, at the same time, the axis of symmetry of the skate. This view is intended to show the real thickness if the blade. Let the blade thickness be represented by the thickness of the 45deg. line. Put a skate boot on the blade (a tilted rectangle centered on the 45deg. line sitting on top of the blade). For now let's assume that the skater is neither ankle-angulating nor knee-angulating. Somewhere within the boot and on the 45deg. line mark the location of the ankle.
Add the location of the knee on the same 45deg. line but higher, above the boot. For simplicity imagine that the skater turns on the one skate only (in a minute it will be the dominant outside ski).
There are three external forces acting on the skater: (1) the gravity force through the center of the skater's mass directed vertically down, (2) the centrifugal force through the same mass center directed horizontally and towards the outside of the turn, (3) the reaction force with which ice acts on the blade at the point where the blade contacts the ice. Draw these three arrows. For equilibrium to exist, the vector sum of forces (1) and (2) must be equal in value and opposite in sense to the reaction force, and must act along the same line as the reaction force (in this case, along the 45deg.line)- this is the condition of equilibrium. Therefore, if you want to mark the location of the center of mass, put a dot somewhere higher than the knee mark but on the same 45deg. line.
A few conclusions can be drawn from this picture:
A)The reaction force acting on the blade compresses the bones of the leg, compresses the ankle and the knee (We should have sketched the hip joint and the upper leg bone, but it would be somewhat painful for me to describe how to construct the picture; let's just say that the hip joint would end up very close to the 45deg line and not far away from the mass center. The upper leg bone, because of its shape, would undergo some bending, mostly its upper portion at the hip)
B)The ankle is not in bending
C)The knee is not in bending
D)The skater has no problem holding the edge
Now the skater decides to increase the blade angle from 45deg. to, say, 60deg. by angulating with his ankle and knee.
Sketch a construction arc with its center at the blade-ice point of contact and of a radius equal to the distance between the blade-ice contact point and the ankle joint. Mark the new ankle location on this arc and towards the center of the turn. From the new ankle joint location make a construction arc whose radius is equal to the distance between the ankle and the knee and mark the new knee location on this arc. Do not change, for simplicity only, the location of the mass center. Join with construction lines the blade-ice contact point (not changed) with the new ankle mark and the new ankle mark with the new knee mark. The reaction force is still on the 45deg. line. Sketch two parallel line segments perpendicular to the original 45deg. line through the new ankle mark and the new knee mark. The lengths of these segments if multiplied by the reaction force, represent the bending moment action that the ankle and the knee are exposed to. You could add a few parallel segments between the ankle and the knee to illustrate the bending of the lower leg bones, if you want to. The longer the segments, the greater the bending moment. Now the skater's muscles must hold the angles if he wants to hold the blade angles. If the segment lengths are short or approach 0, the bending moments are negligible or approach 0, and the skier's muscles do not have to hold the ankle and knee angles, as it was in the case of no angulation. I hope I explained what I mean by bone/joint bending. Usually the bones can take a lot of abuse, unlike the joints.
Let's get back to skiing. A skier turns on his outside leg, no angulation for now. Replace the blade with a ski whose width at the waist is much greater than the blade's width. Make a new sketch showing the ski cross-section (a rectangle) instead of the skate blade, with a boot (a rectangle) on it, at a 45deg. to the snow (a horizontal line) angle. First, the case of soft snow (an extreme case).
On deep snow the ski sinks so that the contact with the snow covers the whole width of the ski base and we can assume that the resultant reaction force is perpendicular to the ski base and on a line close to or exactly on the line of symmetry of the ski cross-section. Draw this line of symmetry and mark the location of the ankle and the knee (no angulation) on this line. Add an arrow representing the reaction force (on the same line at 45deg). Is there a difference between the ski on soft snow and the skate blade on ice if you do not angulate? No, and there is no bending action on the joints. And no problem with holding the ski edge. Your muscles do not have to hold the joint angles because there are no joint angles. If, in addition to inclination, you angulate, you move your joints to the inside of the turn, create the angles (that your muscles have to hold), load your joints with bending moments the same way as the skater does. Would a riser change anything here? Nothing, as long as the reaction force is close to or exactly on the line of symmetry of the ski cross-section. Your sketch can be helpful here; increase the thickness of your ski cross-section to simulate a riser. All you will see is the reaction force moving deeper into the snow, no angles changing.
Now the case of iced/hard-packed snow (opposite extreme case).
Sketch a ski at 45deg.to the horizontal (snow). This line represents now the line of symmetry of the ski cross-section. Put a boot on the ski and mark the ankle and knee locations. To make things simpler let's disregard the possibility of ankle angulation for now (let's say your boot will not allow any) and focus on the knee angulation. The assumption of no ankle angle allows us to mark both the ankle and the knee joints on the 45deg. line (line of symmetry). The new thing here is the reaction force. On hard slopes we are dealing with practically a point contact of the ski with the snow. If you do not angulate at all, the reaction force can be assumed to be perpendicular to the ski base while passing through the sharp corner of the ski edge. (And, of course, its line of action must pass through the CM point.) If you angulate, the reaction force line, while passing through the edge and CM, is not perpendicular to the ski base. This is a very important point.
Let's say you angulate. To sketch the reaction force draw its line of action. Make a straight line that passes through the corner of the ski edge, and misses both the ankle mark and the knee mark but passes between both marks. Put an arrow pointing to the ski edge. You have the reaction force. (This reaction force is, of course, equal to the vector sum of the gravity force and the centrifugal force). The distances of the joint marks from the line of the reaction force, if multiplied by the reaction force, would represent the bending moments on both joints. (The greater the distances, the greater the moments.)
Please note that the ankle bending moment wants to reduce the edge angle while the knee bending moment wants to increase the edge angle. In order to hold the edge angle your muscles must work. If you are on wide-waisted powder ski, (sketch a wider rectangle representing the powder ski cross-section) the reaction force line will pass further away from the ankle joint and closer to the knee joint, perhaps through the knee (this would mean no knee bending) or even would end up on the inside of the knee. The wider the ski, the more you have to work your muscles to hold the edge angle on hard snow. Will a riser help with the wide skis? Yes, it will. Again simulate it by increasing the thickness of the ski into the snow. The reaction force will move towards the outside of the turn. The more non-perpendicular to the base the reaction force is, the more help you will get. Depending on how high the riser is, (stilts?) you can move the reaction force so that it passes through the ankle joint or even further so that the reaction force is on the outside of the ankle. You may gain a lot as far as the ankle goes. But take a look at the knee bending moment that you will create. Fortunately, the skis for hard slopes are narrow at the waist. Theoretically, you can manipulate the location of the all-important reaction force by fooling around with the riser heights. But the knee will remind you of its existence.
If I had a problem of holding the edge on wide skis on a hard slope, I would try to support better the ankle joint with my boots rather than overdo the lifter heights.
This is how I think the lifters are supposed to work. I wish binding manufacturers were honest enough to tell us about the pros and cons before we decide what to buy (if we had a choice), rather than publish silly glossy brochures about enigmatic lever actions that are more confusing than explanatory.
I hope I haven't bored you to death.