Physics has a purpose

**TeleAl** · 12-09-2005, 07:38 AM

Originally Posted by YetiMan

What if the verticle component of skier's speed prior to takeoff and the air resistance cancel? Is that even possible?

Quick reply = yes it is possible.
It would differ with each skier (see density) and their vertical/downward velocity.
I find it unlikely that anyone could pull that off, but I think it is possible.

Here we would need to know the slope at the bottom of the cliff.
The above assumes a flat landing.
Without a flat landing, it gets complicated as launch velocity will determine vertical fall distance, which affects time.
Still possible, but less likely.

**shmerham** · 12-09-2005, 08:28 AM

Group of skiers pondering a huge cliff.

#1: "Can we get a quick number crunch on our survival rate"
#2: "uhh, let's see, should be about 22.1%"
#1 and #2 are talking about approach speed, whether to backslap, what to avoid, etc.
#3: just goes for it "LEEEEEEEEEEEEEEROY JENKINS!!!!!!!!!"

**runethechamp** · 12-09-2005, 08:34 AM

Why is irrelevant such a difficult word to understand?

Edit: Wow, 1300

**KANUTTEN** · 12-09-2005, 09:51 AM

Physics puzzle along the same nerdy lines as in this thread;

Assume no friction and no air resistance. Starting with a speed of 0 skiing on two slopes with angle at 25 degrees and 50 degrees respectively. After 10 seconds of skiing, at which angle do you ski the fastest?

**XtrPickels** · 12-09-2005, 10:06 AM

assuming no friction and air resistance.

The skier on the 50degree slope will accel at 7.5m/s^2. The Bunny Hiller will accel at 4.14m/s^s

I'll get back to you with velocity

**XtrPickels** · 12-09-2005, 10:14 AM

Its official,

The skier in the Chugach will be moving faster than the skier in Ohio

**Ireallyliketoski2** · 12-09-2005, 10:19 AM

I made a particle accelerator out of a blender and some surgical tubing. Although, I was unable to prove the existence of the 6-dimensional Calabi-Yau space, it does make awesome margaritas.

**Big Blue** · 12-09-2005, 10:21 AM

Originally Posted by TeleAl

To clarify, mass is irrelevant. Period.

Actually, you're wrong. It is the ratio of mass to drag coefficient that is important. Density plays a role in this, but so does the physical configuration of the object.

**cj001f** · 12-09-2005, 10:22 AM

Originally Posted by Ireallyliketoski

I made a particle accelerator out of a blender and some surgical tubing. Although, I was unable to prove the existence of the 6-dimensional Calabi-Yau space, it does make awesome margaritas.

Good thing you don't live in Alaska!

**shane-o** · 12-09-2005, 10:34 AM

Originally Posted by TeleAl

To clarify, mass is irrelevant. Period.

When air resistance applies, it is density, not mass, that matters.
Think of a ping pong ball versus a rock of same size.
Think of a basketball and rock of same mass.
Think density!

Al, I don't think I quite agree with you here. The two forces at play are gravity and air resistance. The gravity force on an object is propotional to mass, while the air resistance for a given velocity and object shape is approximately proportional to cross-sectional area. In the example of a sphere of a given average density, as the radius increases, mass increases cubicly while cross-sectional area increases quadratically. Thus for a given shape and density, a larger object will tend to fall faster in air than a smaller object. So yes, while density plays a role, ignoring the scale of the object is not correct. As an example, consider the terminal velocity of a particle of sand versus a boulder. A strong wind has no trouble blowing sand aloft, but I have never experienced wind that could hold a large boulder aloft.

In the case of a falling skier, I would expect the skis to provide the majority of relevant cross-sectional area, so a reasonable simplification would probably be to consider only the area of the skis and the mass of the skier.

Z · 12-09-2005, 11:25 AM

Originally Posted by shmerham

Group of skiers pondering a huge cliff.

#1: "Can we get a quick number crunch on our survival rate"
#2: "uhh, let's see, should be about 22.1%"
#1 and #2 are talking about approach speed, whether to backslap, what to avoid, etc.
#3: just goes for it "LEEEEEEEEEEEEEEROY JENKINS!!!!!!!!!"

hahahahhahahh. nice.

that'd be me.

**Mechmaster** · 12-09-2005, 03:19 PM

The velocity of the skier is a vector consisting of forward velocity and downward velocity.

Air resistance is directly proportional to the square of skier velocity and the coefficient of drag (Cd). Obviously, the Cd will vary according to the position the skier is in (bigger cross-section = larger Cd). The frontal area is required in this calculation (likely a rectangle approximatey the size of both skis side-by-side, although the skier's stance in the air will greatly affect this value).

During the freefall, the forces acting on the skier is a constant force caused by the interaction of the skier's mass with the earth's gravity, and a resistant force due to drag.

Let's assume a worst-case scenario; that is, no air resistance (Exact solution of the falling body problem with drag requires solution via partial differential equations). Then, elementary kinematics can calculate the final velocity prior to impact as well as the time for free fall.

Effects of air resistance only become important on large drops (since the effect of drag depends on the square of velocity). For typical hucks, the error introduced by neglecting air resistance is likely nominal.

To determine the impact force, solve for the acceleration required to slow or stop the skier. Multiply this by the skier's mass to yield the impact force.

Note that the skier is likely not landing on a flat surface. The decelerative force is thus a vector combination of a vertical load (absorbed by the skier) and a forward load (which pushes the skier down the hill).

**Vinman** · 12-09-2005, 03:22 PM

this all makes my brain hurt

**cj001f** · 12-09-2005, 03:32 PM

Originally Posted by Mechmaster

Air resistance is directly proportional to the square of skier velocity and the coefficient of drag (Cd). Obviously, the Cd will vary according to the position the skier is in (bigger cross-section = larger Cd). The frontal area is required in this calculation (likely a rectangle approximatey the size of both skis side-by-side, although the skier's stance in the air will greatly affect this value).

Air resistance = velocity ^2 * Cd * cross sectional area. A bigger cross-section != larger Cd, often times it can be smaller (why car manufacturers advertise Cd, not drag). Skiiers aim to reduce both cross sectional area and coefficent of drag

**TomK** · 12-09-2005, 03:45 PM

Originally Posted by Mechmaster

...Air resistance is directly proportional to the square of skier velocity and the coefficient of drag (Cd). Obviously, the Cd will vary according to the position the skier is in (bigger cross-section = larger Cd)...

And since skiers can get a smaller Cd by tucking, this is why a skier will always be able to go faster than a boarder. (all other factors being equal). Not "better", but faster.

**Mechmaster** · 12-09-2005, 03:46 PM

Originally Posted by cj001f

Air resistance = velocity ^2 * Cd * cross sectional area. A bigger cross-section != larger Cd, often times it can be smaller (why car manufacturers advertise Cd, not drag). Skiiers aim to reduce both cross sectional area and coefficent of drag

I'm sorry, what I should have typed was projected area NOT the cross sectional area. And yes, the Cd is a dimensionless parameter independant of the area term.

The implied point of the statement was that drag increases 1. As you go faster, 2. As you increase your projected area, and 3. As you vary your coefficient of drag. (2) and (3) are typically interconnected.

**alpinepronghorn** · 12-09-2005, 04:10 PM

It makes me very sad that 8 years ago this would have posed no problem, but today I don't even know where to begin...

Best not get airborne, I guess. But then, I always was a kinesthetic learner...

**surfsnowywaves** · 12-09-2005, 06:07 PM

Originally Posted by shmerham

Group of skiers pondering a huge cliff.

#1: "Can we get a quick number crunch on our survival rate"
#2: "uhh, let's see, should be about 22.1%"
#1 and #2 are talking about approach speed, whether to backslap, what to avoid, etc.
#3: just goes for it "LEEEEEEEEEEEEEEROY JENKINS!!!!!!!!!"

im going to have to go ahead and disagree with you. i do believe that its it THIRTY TWO POINT THREE PERCENT, REPEATING OF COURSE

**Big Blue** · 12-09-2005, 06:48 PM

Originally Posted by Mechmaster

Then, elementary kinematics can calculate the final velocity prior to impact

Again, much more easily done with kinetics.

Originally Posted by Mechmaster

To determine the impact force, solve for the acceleration required to slow or stop the skier. Multiply this by the skier's mass to yield the impact force.

Solving for the acceleration required to stop the skier is a complicated prospect. it's easier to use kinetics again here. the skiers kinetic energy is equal to the "impact force" times the distance (in this case the penetration distance into the snow + the distance that the skiers center of mass moves downward as his legs bend to absorb the shock (assuming he's landing on a flat surface on his feet)).

**powstash** · 12-09-2005, 06:58 PM

Originally Posted by XtrPickels

Its official,

The skier in the Chugach will be moving faster than the skier in Ohio

That took you 8 minutes to calculate? C'mon mang

edit - I'm actually enjoying this thread.

**Shepherd Wong** · 12-09-2005, 07:41 PM

Physics is Universal!

**AKBckntry** · 12-10-2005, 04:32 PM

No slope versus zero slope: during which condition would you be falling?

**TeleAl** · 12-10-2005, 06:57 PM

Originally Posted by Big Blue

Actually, you're wrong. It is the ratio of mass to drag coefficient that is important. Density plays a role in this, but so does the physical configuration of the object.

I stand corrected. I agree with you. And Shane-O too.
Remove the vacuum, and it get complicated quick.

Shall we explore altitude and air pressure as factors next?
Keep in mind, both are changing as the skier is in air. Will these have the same effect on two different objects?

**Daywalker** · 12-10-2005, 08:19 PM

Originally Posted by AKBckntry

No slope versus zero slope: during which condition would you be falling?

Slope is equal to vertical displacement over horizontal displacement. m=y/x. If y=0 then zero slope (flat ground). If x=0 then there is no answer (or however you phrase it) "no slope". You fall under the no slope condition.

This is the only subject I feel like addressing, all that drag stuff was last year and has been forgotten. If anyone gets on the subject of vibrations though....