Score Discussion

[builder83]

Curious if those can adapt for a 50+ time? Scoring is fun this year. Still 80-90 cm is crazy good gaps!

[SOPomo]

Curious if those can adapt for a 50+ time? Scoring is fun this year. Still 80-90 cm is crazy good gaps!

Almost everyone at NorCal's State could have hit most target times. Rankings are now basically just, 'who has the longest gap' for anything under 50 seconds in target time.

[awscioly]

At Solon invitation, the winner had 85.7 gap total

I just started sci oly this year, but I'm wondering how in the world you can get a gap score that large with just three gaps. Are there any pictures of the gaps from nationals last year or just descriptions? I'm not sure how it's even possible even though it obviously is.

[sciolyperson1]

20 seconds required but she missed 3 seconds. Another one get 91.6 gaps. it was pity. They missed last gaps. If the 91.6 succeed, they would be the winner. I think they can make it with more practice.

I'm confused - did the roller have 85.7cm or 91.6cm gap? Also, to restate one of the questions stated earlier, was this Springhouse's parents' roller?

[jander14indoor]

At Solon invitation, the winner had 85.7 gap total

I just started sci oly this year, but I'm wondering how in the world you can get a gap score that large with just three gaps. Are there any pictures of the gaps from nationals last year or just descriptions? I'm not sure how it's even possible even though it obviously is.

I can't speak to exact details of how people are doing it this year, but based on nationals last year:
- First, you have to jump across the diagonals.
-- Diagonals are sqrt(50**2+50**2)=70.7 cm
-- You lose a couple of cm because the ball can't go all the way into the corner, so keep the ball small. For a 1 cm ball you lose 1.4 cm, best case, this gives no room for the track to guide the ball, so call it 2 cm.
-- You need the 5 cm over-run at end of gap and some amount of run up to accelerate the ball, say 10 cm minimum.
-- That takes you to a theoretical max gap of 70.7-2-5-10=53.7 cm. Not saying you can make that, but it gives you more room than along the 50 cm sides.
- Second, you have to maximize the distance you can fly for a given launch speed. You must kick up at the beginning of the gap to approx 45 degrees launch angle. I say approx, because that is the theoretical max range angle. Air drag modifies that, so you'll have to experiment to get true ideal. Typically a little lower than 45 degrees. If you haven't learned the physics of this yet, find someone who has and ask for some help.
- Third, you have to maximize your speed. That means starting with the max height possible (which costs you height score, so here you have a trade off. If you lower the height 1 cm which costs three points, you can't lose even 1 cm of gap distance which costs five points). And it means dropping far enough to cross that theoretical maximum gap of 54 cm. My quick, rough calculations say you have JUST barely enough energy to make that jump if you start right at the top and use ALL the devices height. But, that leaves you no room for a timing area and no room for the second gap!
- Fourth, you have to MAINTAIN that speed/energy from jump to jump! Here's where some of the really clever designs came in. And where I'm going to stop detail. All of the above you can figure out from the basic physics of the problem.
-- The first part of maintaining the energy is to minimize drag/rolling resistance, one reason paper coasters aren't great. You want a track with hard surfaces that provides accurate guidance.
-- The second part takes clever reasoning and design, it has been hinted at or explained in previous strings either this or last year. I'll outline it though. The END of the gap needs a clever shape to reverse the direction of the ball, launch it back upward, with as little loss of speed as possible.

Summary. You have room for single jump maximum of 54 cm. You have barely enough energy to jump that distance. With drag you probably won't. And to have room for a timing section you'll have to give up some of that height and reduce the jump maximum. You now have to maintain that energy for the second and third jump, but even best designs will lose some.

Based on all that, I see a first jump of 40 cm second of 30 cm and third of 20 cm as reasonable. Note easy, but reasonable.

Jeff Anderson
Livonia, MI

[builderguy135]

At Solon invitation, the winner had 85.7 gap total

I just started sci oly this year, but I'm wondering how in the world you can get a gap score that large with just three gaps. Are there any pictures of the gaps from nationals last year or just descriptions? I'm not sure how it's even possible even though it obviously is.

I can't speak to exact details of how people are doing it this year, but based on nationals last year:
- First, you have to jump across the diagonals.
-- Diagonals are sqrt(50**2+50**2)=70.7 cm
-- You lose a couple of cm because the ball can't go all the way into the corner, so keep the ball small. For a 1 cm ball you lose 1.4 cm, best case, this gives no room for the track to guide the ball, so call it 2 cm.
-- You need the 5 cm over-run at end of gap and some amount of run up to accelerate the ball, say 10 cm minimum.
-- That takes you to a theoretical max gap of 70.7-2-5-10=53.7 cm. Not saying you can make that, but it gives you more room than along the 50 cm sides.
- Second, you have to maximize the distance you can fly for a given launch speed. You must kick up at the beginning of the gap to approx 45 degrees launch angle. I say approx, because that is the theoretical max range angle. Air drag modifies that, so you'll have to experiment to get true ideal. Typically a little lower than 45 degrees. If you haven't learned the physics of this yet, find someone who has and ask for some help.
- Third, you have to maximize your speed. That means starting with the max height possible (which costs you height score, so here you have a trade off. If you lower the height 1 cm which costs three points, you can't lose even 1 cm of gap distance which costs five points). And it means dropping far enough to cross that theoretical maximum gap of 54 cm. My quick, rough calculations say you have JUST barely enough energy to make that jump if you start right at the top and use ALL the devices height. But, that leaves you no room for a timing area and no room for the second gap!
- Fourth, you have to MAINTAIN that speed/energy from jump to jump! Here's where some of the really clever designs came in. And where I'm going to stop detail. All of the above you can figure out from the basic physics of the problem.
-- The first part of maintaining the energy is to minimize drag/rolling resistance, one reason paper coasters aren't great. You want a track with hard surfaces that provides accurate guidance.
-- The second part takes clever reasoning and design, it has been hinted at or explained in previous strings either this or last year. I'll outline it though. The END of the gap needs a clever shape to reverse the direction of the ball, launch it back upward, with as little loss of speed as possible.

Summary. You have room for single jump maximum of 54 cm. You have barely enough energy to jump that distance. With drag you probably won't. And to have room for a timing section you'll have to give up some of that height and reduce the jump maximum. You now have to maintain that energy for the second and third jump, but even best designs will lose some.

Based on all that, I see a first jump of 40 cm second of 30 cm and third of 20 cm as reasonable. Note easy, but reasonable.

Jeff Anderson
Livonia, MI

The maximum distance possible to jump a ball starting from height h is 2h. This is disregarding any kind of friction though. Also, since the ball has friction, a lot of velocity is being translated to angular momentum.

[jander14indoor]

Dang, forgot angular momentum, slows initial launch, but might help relaunch...
Hmm, h to 2h doesn't sound right, back to the algebra...
...unless you can point me to the derivation and save me some time?

Jeff Anderson
Livonia, MI

[knightmoves]

Dang, forgot angular momentum, slows initial launch, but might help relaunch...
Hmm, h to 2h doesn't sound right, back to the algebra...
...unless you can point me to the derivation and save me some time?

Balls are spheres, so they roll. Moment of inertia of sphere = 0.4 m r^2
Rolling condition: v = rw (that's an omega, = angular velocity)
Gravitational potential energy = m g h
Kinetic energy of rolling ball = 0.5 m v^2 + 0.5 I w^2 = 0.5 m v^2 + 0.2 m r^2 w^2 = 0.7 m v^2

So launch velocity v = sqrt(g h/0.7)
Launch at 45 degrees, so vertical velocity = sqrt (g h/1.4)
Ball reaches peak of parabola when vertical velocity is zero, => sqrt(g h/1.4) - g t = 0
So t = sqrt(h/1.4 g)
Double that to get time to land again, then multiply by horizontal velocity to get
gap distance = 2*sqrt(h / 1.4 g) * sqrt (g h/1.4)
= 1.42 h

This assumes no energy loss, of course. Real world numbers will be smaller than this.

If I take your 54cm theoretical gap, you could jump it with 38cm of height loss in magic no-energy-loss land.

[builderguy135]

Dang, forgot angular momentum, slows initial launch, but might help relaunch...
Hmm, h to 2h doesn't sound right, back to the algebra...
...unless you can point me to the derivation and save me some time?

Balls are spheres, so they roll. Moment of inertia of sphere = 0.4 m r^2
Rolling condition: v = rw (that's an omega, = angular velocity)
Gravitational potential energy = m g h
Kinetic energy of rolling ball = 0.5 m v^2 + 0.5 I w^2 = 0.5 m v^2 + 0.2 m r^2 w^2 = 0.7 m v^2

So launch velocity v = sqrt(g h/0.7)
Launch at 45 degrees, so vertical velocity = sqrt (g h/1.4)
Ball reaches peak of parabola when vertical velocity is zero, => sqrt(g h/1.4) - g t = 0
So t = sqrt(h/1.4 g)
Double that to get time to land again, then multiply by horizontal velocity to get
gap distance = 2*sqrt(h / 1.4 g) * sqrt (g h/1.4)
= 1.42 h

This assumes no energy loss, of course. Real world numbers will be smaller than this.

If I take your 54cm theoretical gap, you could jump it with 38cm of height loss in magic no-energy-loss land.

v=sqrt(2gh)
v stays constant throughout, and this v is at the bottom of the theoretical slope/jump w/o starting the gap yet.

v_x and v_y are both equal to v sin 45, or (v sqrt 2)/2.

t = 2*v_y/g = 2*(v sqrt 2)/2g = (v sqrt 2)/g.

obviously, the ball can only travel with constant velocity v_x while in the air for time t.

d = ((v sqrt 2)/2) * ((v sqrt 2)/g)
=(2v^2)/(2g)
=v^2/g

v=sqrt(2gh)
d=v^2/g
d=2gh/g
d=2h

sorry for the bad explanation :\

[knightmoves]

v=sqrt(2gh)

This assumes you have a frictionless ball sliding down the ramp, and so all the gravitational potential energy is available for the kinetic energy of the center of mass. You don't, and it isn't. The ball rolls, and so some of the energy is converted to rotational energy instead.

d = v^2/g is fine, but setting 0.5mv^2 = mgh is not. When you include the rotational energy that the ball must have, your expression is 0.7mv^2 = mgh,
which gets you d = (1/0.7)h