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I had posted with tips on how to successfully skid down the hill on the heel edge, as shown in the photos. It was a physics-based explanation focusing on the importance of one's center of mass being vertically over the engaged edge of the snowboard.

A commenter said this:

Your center of mass NEEDS to be outside the heel edge to counteract the force of sliding down the hill forwards, that's why it's so tricky, because you've got to lean a little further back than you feel comfortable with. Part of the battle is getting used to knowing that you need to shift your center of mass outside of where your body knows it can balance, coupled with the slipping feeling.

To which I replied:

There is no "force of sliding down the hill." A force is a push or a pull from another object. In this case, there are three significant forces on the rider/board object: gravity, friction with the snow, and the normal force (the push of the snow 90 deg to its surface). Skidding at constant velocity, these forces must be balanced (the vector sum must be zero). Torques about the pivot point (the heel edge of the snowboard) must also sum to zero. When they don't, you fall. You can see from the diagram that friction and the normal force act through the pivot point, the edge of the snowboard. Hence, no torque from them. That's why weight must be stacked over the edge. If it's not, they'll be a torque from the rider's weight and they will fall.

Admittedly, I did not provide a comprehensive answer likely to convince anyone unfamiliar with physics. Instead, I attempted to thread the needle between a comprehensive response and a concise response. In particular, I did not define torque, nor explain why positioning one's center of mass over the heel edge would eliminate it. So maybe this experiment in outreach was doomed to fail. Someone closed an extended, acrimonious exchange by calling me a "dork." ☺️ (For the record, I identify as "geek," though "nerd," sadly, is often more descriptive of my state of being.)

After some reflection, I began to appreciate that a big part of the confusion was a failure to differentiate between skidding down a hill on the heel edge and making a turn on the heel edge. While making a turn, it is absolutely normal and desirable to position one's center of mass to the inside of the turn. Much advice about body position in turns relates to getting the correct angle between the contact point of the engaged edge and center of mass. What might that angle be?

Let's simplify things a bit by imagining a turn along a circular arc, at constant speed, on level ground, and with a negligible difference between the radius of the turn and the (slightly smaller) radius traversed by the center of mass. We apply rotational dynamics to this scenario. (Original photo: Snowboard Addiction)

Figure 1A. Applying the right-hand rule about a point where the snowboard touches the snow, the torque in the turn is directed out of the page.

Figure 1B. With velocity into the page, and applying the right-hand rule once again, the angular momentum about the same point is directed down and to the left.

Figure 1C. The vertical component of angular momentum is not changing in either direction or magnitude. But the horizontal component is changing direction as the boarder traverses the turn.

Looking from the top, we examine the radius of the turn R and the projection of angular momentum onto the horizontal plane.

Rotating the orange triangle back into the side view, we see that dL points out of the page, the same direction as torque. Writing Newton's second law in angular form:

We recognize this as the same formula derived from force considerations for a banked turn without friction. Indeed, the two scenarios are analogous, as both objects lean into the turn, and both exist in a non-inertial reference frame as a result of their circular motion about a stationary point with radius R.

In a typical scenario, with a rider cruising at 12 m/s (27 mph) and turning with a radius of 9 m (30 ft), the necessary angle would be about 59 degrees.

Googling around, it's possible to find many different workable postures for effective carving: some folks with deep bends at the knees, others standing more erect. Are these differences related to the fact that lean angle appears in the formula, but not distance from the engaged edge to the center of mass (r)?

Every snowboarder has had the experience of leaning in too far on a turn. If the rider retains grip, they don't simply fall over. Instead, the board tracks with a smaller and smaller radius of turn until there is a kind of spin out. The larger the angle of lean θ, the smaller must be the radius of the turn. Unless of course, the rider balances out some of the torque by putting their inside hand or forearm on the snow.

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Link: https://shelterphysics.com/blog/Body-Position-While-Skidding-and-Carving-on-a-Snowboard-a2/

May 3, 2025
Quarter Pipe Heights

Here's an article prompted by an interesting post in one of my snowboarding groups.

The OP

Is it possible to pull off a 30 meters high air in a quarter pipe?

Ever since 1996, when Ingmar Backman did his legendary backside air up in Riksgränsen, I’ve been wondering: how big can a human actually go in a quarter — purely in theory?

Today, approximately 29 years later, I finally chat:ed my way to the answer (feel free to double-check the math and send feedback).

With the right height and transition on the quarterpipe, and a carefully calculated speed that a pair of well-trained thighs can withstand. I’ve come up with this:

I crunched the numbers using the laws of physics and assumed the rider can handle 4 g in the transition. Here are the formulas and calculations:

1. Total g-force in a quarterpipe:

g_total = (v² / (r × g)) + 1

Where:

• `v` = speed in m/s

• `r` = radius of the quarterpipe (in meters)

• `g` = 9.82 m/s² (gravitational acceleration)

2. Maximum speed without exceeding 4 g:

v² = (4 − 1) × r × g = 3 × r × g

3. Converting speed to height (all kinetic energy → height):

h = v² / (2g)

Insert v² from above:

h = (3 × r × g) / (2g) = (3/2) × r = 1.5 × r

So the maximum height above the lip becomes:

h = 1.5 × r

4. Example with a 20-meter radius quarterpipe:

h = 1.5 × 20 = 30 meters (above the lip)

Total height from ground = 20 + 30 = 50 meters

5. Speed required:

v² = 3 × 20 × 9.82 = 589.2

v = √589.2 ≈ 24.26 m/s ≈ 87.4 km/h

```

Conclusion:

If you have a quarterpipe with a 20-meter radius and can handle 4 g in the transition — then, theoretically, you can pull off a 30-meter-high air.

So... what are you all waiting for out there?

Ingmar Backman at the apex of his remarkable jump

Ingmar Backman's backside air, Riksgränsen Sweden, May 1996

My Reaction

My first response built on the assumptions in the OP. Given a landing in a 20-m radius quarter pipe, the maximum g's occur at the bottom of the quarter pipe, where normal force is opposite weight, not at the lip, where normal force is perpendicular to weight. So the calculation shows maximum height from the bottom of 30 m, and maximum height above the lip of 10 m, ignoring friction and air resistance.

If the person center frame is 1.8 m (71") tall, the height of this jump is 1.8•4.6 = 8.3 m above the lip, a good fit for a theoretical maximum of 10 m.

Then I Googled for and watched video of the jump. I realized some of the assumptions in the OP needed revision. The launch speed was limited by the terrain of the lead-in to the quarter pipe, and the ability of the rider to max out on speed during the lead-in. The landing zone could definitely not be modeled as a circular arc. Better to model it as a steep inclined plane, with landing impact on the slope. And finally, the parabolic trajectory of the jump is too large to ignore. The take-off ramp was not vertical at the lip.

To achieve a jump height of 8.3 m, the vertical component of Ingmar's speed at the lip would have to be 13 m/s, or 29 mph. With the angle of the ramp taken into account, the resultant speed would be no more than 15 m/s, or 34 mph. The long, uphill lead-in to the ramp scrubbed off a lot of speed.

Here is a good analysis of g's during snowboard landings, and the impact of a sloped landing zone. While in theory snowboarders might experience 14 g's at landing, the actual number is usually quite a bit less. What matters is the impact velocity perpendicular to the slope, and the time it takes for that velocity to zero out.

Frame-by-frame analysis of the landing shows that Ingmar bends his knees some, but also bends hard over at the waist, allowing his lower body to continue falling, and almost allowing his head to graze the snow. The sequence takes 5 frames or 0.2 seconds.

If the speed at impact is 15 m/s, and the angle between the slope and that velocity is 40 degrees, the component of velocity perpendicular to the slope would be no more than 10 m/s. With a stopping time of at 0.2 seconds, the stopping force would be 5 g's, what a roller coaster rider might experience and close to the 4-g maximum assumed in the OP.

For comparison, a Google search yields some compelling analysis to suggest that g's during carved turns are usually less than 2 g's and more often around 1 g or less. For quick reference, a turn at 15 m/s with a radius of 20 m yields a little over 1 g. But these are sustained g's, and they likely wobble quite a bit around the time-averaged value due to bumps and corrections to form.

Estimating the height of the jump

A still frame showing Ingmar's landing posture

Conclusion

Ultimately, the limitation on height of a snowboard jump is speed and angle of approach going into the jump, and the ability of the human body to absorb the forces needed to land. All in all, what a fascinating opportunity to explore the physics of snowboarding.