B
Bubba
Guest
Well that's different. Get him a physics book and make a math genius out of him. 
360 on a skateboard ?
- Thread starter Leslie
- Start date
unclehobart
New Member
Yeah, yeah. I just havent had to think about it in years. Physics and geometry and the like didnt gel in my head until my Air Force JROTC instructors placed it in terms that just made it all come together. I'm an air nav dead recoking mofo now. A few visual aids does what pages and pages of printed word cannot.
I just havent had to think about it in years. Physics and geometry and the like didnt gel in my head until my Air Force JROTC instructors placed it in terms that just made it all come together. I'm an air nav dead recoking mofo now. A few visual aids does what pages and pages of printed word cannot.
unclehobart
New Member
http://skateboard.about.com/gi/dynamic/offsite.htm?site=http%3A%2F%2Fwww.drskateboard.com%2Fcurriculum%2Fskateboard.htm
frontside 180 physics overview. 'westside!'
frontside 180 physics overview. 'westside!'
unclehobart
New Member
Something for you sickos that get a stiffy from math
This is an example of a "real-life" problem, where we have to decide which pieces of information are relevant to the answers desired.
Following our class discussions so far, and the discussions in the text, we can model this as a 1-dimensional, constant acceleration problem. We know that such problems involve
displacement, x - x0, over some time period, t,
initial velocity, v0,
velocity, v, after some time period, t, and
the (constant) acceleration.
They involve nothing else; therefore, we should not need the various weights, nor the angle of the board to the horizontal. Hopefully we have "made sufficiently many measurements" to determine the desired answer.
That desired answer, from the problem statement,
is the rate that the skateboard slows down as it coasts up.
This is just the acceleration (negative) in the one-dimensional problem, which is assumed constant. We do know the displacement, x - x0, and the elapsed time. We of course also know, although not explicitly given, the velocity at the end of that elapsed time, namely 0, since it has coasted to the top, and is about to start rolling back down.
We do NOT know either the initial velocity, v0, or the acceleration, a. [Notice that we cannot simply say that acceleration is that due to gravity, i.e., -g, since we are coasting upward at some angle, AND we know that there will be some sort of friction, unspecified, between any real skateboard and any real "ramp."]
Although we do not want the initial velocity, nonetheless it enters our equations, AND we need two equations to determine the two unknowns. The "obvious" two equations are ones we have become quite familiar with by now, namely
v = v0 + a t , and
x - x0 = v0t + ½a t² .
We may enter the known numbers into these two equations to obtain
0 = v0 + (6 s)a , and
95 feet = (6 s)v0 + (18 s²)a .
To determine a from these equations, we resolve the first one for v0, which it says is equal - 6a, and insert that into the other one, giving us
95 feet = - 36 a + 18 a , which implies that a = -5.278 ft/s².
One should note that it is possible to go ahead and eliminate v0 from the two equations, before entering in with the numbers. If one does this, the result will be the last of the set of 5 equations in the little table on p. 21 of the text; therefore, one could have simply chosen that one equation. I chose not to do that, because I don't want to have to deal with too many different equations, when I can choose just a few, and manipulate them with simple algebra to obtain the desired results.
This is an example of a "real-life" problem, where we have to decide which pieces of information are relevant to the answers desired.
Following our class discussions so far, and the discussions in the text, we can model this as a 1-dimensional, constant acceleration problem. We know that such problems involve
displacement, x - x0, over some time period, t,
initial velocity, v0,
velocity, v, after some time period, t, and
the (constant) acceleration.
They involve nothing else; therefore, we should not need the various weights, nor the angle of the board to the horizontal. Hopefully we have "made sufficiently many measurements" to determine the desired answer.
That desired answer, from the problem statement,
is the rate that the skateboard slows down as it coasts up.
This is just the acceleration (negative) in the one-dimensional problem, which is assumed constant. We do know the displacement, x - x0, and the elapsed time. We of course also know, although not explicitly given, the velocity at the end of that elapsed time, namely 0, since it has coasted to the top, and is about to start rolling back down.
We do NOT know either the initial velocity, v0, or the acceleration, a. [Notice that we cannot simply say that acceleration is that due to gravity, i.e., -g, since we are coasting upward at some angle, AND we know that there will be some sort of friction, unspecified, between any real skateboard and any real "ramp."]
Although we do not want the initial velocity, nonetheless it enters our equations, AND we need two equations to determine the two unknowns. The "obvious" two equations are ones we have become quite familiar with by now, namely
v = v0 + a t , and
x - x0 = v0t + ½a t² .
We may enter the known numbers into these two equations to obtain
0 = v0 + (6 s)a , and
95 feet = (6 s)v0 + (18 s²)a .
To determine a from these equations, we resolve the first one for v0, which it says is equal - 6a, and insert that into the other one, giving us
95 feet = - 36 a + 18 a , which implies that a = -5.278 ft/s².
One should note that it is possible to go ahead and eliminate v0 from the two equations, before entering in with the numbers. If one does this, the result will be the last of the set of 5 equations in the little table on p. 21 of the text; therefore, one could have simply chosen that one equation. I chose not to do that, because I don't want to have to deal with too many different equations, when I can choose just a few, and manipulate them with simple algebra to obtain the desired results.
unclehobart
New Member
The trick is performed by slamming the tail of the board to the ground before takeoff. If a skateboarder doesn't kick hard enough, the board won't get enough vertical rise. If we throw a baseball into the ground as hard as we can, it will bounce up. The same principle is at work here. If a skateboarder exerts enough force, the board will bounce up beneath his or her feet.
When the board is completely in the air, the skater slides his front foot forward, using the friction between his foot and the board's surface to drag the board higher.
Eventually, gravity takes over, and the board and skater together will begin to fall until they land.
When the board is completely in the air, the skater slides his front foot forward, using the friction between his foot and the board's surface to drag the board higher.
Eventually, gravity takes over, and the board and skater together will begin to fall until they land.
unclehobart
New Member
Efinal = Einitial
Kfinal + Ufinal = Kinitial2 + Uinitial
½ mv2 + mgh = ½ mv2 + mgh + Wapp
The work performed/energy added can be evaluated using 5 variables
the height of the ramp (h)
the radius of the ramp (r)
the height of the air (a)
the mass of the skater (m)
the height of the skater(s)
( we made friction and air resistance negligible)
The terms all vary with respect to their role in determining how the energy of the system changes
1) the height of the ramp is 3.66m and this term determines potential energy (u)
2) the radius of the ramp is less than its height (height – radius = amount of true vertical), and is usually around 3.05m. This term influences conservation of angular momentum, and this is where the work is done.
3) The height the skateboard reaches above the top of the ramp is the direct result of the work performed. This can range anywhere from 0m to 4.42m (the current world record was performed by Danny Way)
4) the mass of the skater, if using any conservation laws would drop out the equation because it is constant. However, when determining the work done, work is dependent upon force applied. For this project we made the mass to be 80 kg.
5)the height of the skater is, curiously, the most important variable. What drives the energy of the system is the work performed by the skater, which utilizes the principle of conservation of angular momentum. This will be expanded upon
later in this document. The variable that move the skater up the ramp is the change is the radius as the skater’s center of gravity changes. We calculated the center of gravity of a 1.75m tall skater on his skateboard. While standing upright, it is 1.067m and while crouching it is 0.813m.
The radius of the ramp is where the work is done. This is the area where gravity is used to add energy…
At any point on the curve,
Radial component
S Fr mg cosq - n + Wapp =mar
tangential component
S Ft mg sin q + Fr = mat
The resultant force (Fr) in the tangential direction is due to the work done by the skater in the radial direction. This causes a tangential acceleration.
The gravitational force is conservative. The work varies over the distance through which it is applied, so
Wapp = D k +D V
W=ò xixfFxdx
(D k + D u) =ò xixfFxdx
What exactly is the work done? The Wapp is a function of angular momentum to add work going down the ramp, the skater crouches into the vertical and stands up near the bottom. To add work going up the ramp, the skater crouches into the bottom and stands up near the vertical.
The net effect is to travel in a tighter radius than the curvature of the ramp.
Conservation of angular momentum states that to counter the decreased radius, the linear momentum must increase. Since the mass of the skater is constant, his linear velocity must increase.
Angular momentum L = mvr sin q
q is the angle between the linear momentum (mv) and the radius, so for a ramp skater,q = 90° and sin q = 1
angular momentum for the skater becomes
L = mvD r
= mv(rf – ri)
The amount that the skater compresses/crouches is inversely proportional to the change in angular velocity. As L and M are constant
As r1< r2, then w 1> w
This change in angular velocity is due to the energy added by the skater. This energy is equal in magnitude to the work done, and thus velocities can be determined by using the principle of conservation of energy.
From point 2 to point 3 no work is done. The only force acting on the skater is gravity. This in conjunction with the fact that the skater’s velocity at his highest point (here 2m) is zero, simplifies the process of finding the velocity at point 2.
E3 = E2
K3 + U3 = K2 + U2
½ mv32 + mgh3 = ½ mv22 + mgh2
(80kg)(9.8 m/s2)(5.66m) = (.5)(80kg)(v2)2 + (80kg)(9.8 m/s2)(3.05m)
4.437 kg m2/s2 = 40 kg (v2)2 + 2391 kg m2/s2
40 kg(v2)2 = 2046 kg m2/s2
v2 = 7.15 m/s
From point 1 to point 2, the curvature of the ramp enables the skater to exploit conservation of angular momentum to add energy. This is where the work is done, and the force component of this work is the centripetal acceleration.
Remember that
M=mass of the skater
V=velocity at point 2
D=the distance from the skater’s crouched center of gravity to his upright center of gravity
r=the distance from the center of the ramp’s radius to the skater’s average center of gravity (between crouching and upright)
W=Fd
= mad
=m(v2/r)d
This equation makes two assumptions
that v1 is roughly equal to v2
2)that the radius used to determine centripetal acceleration is constant
d dupright - dcrouched
(1.067m) – (.813m)
0.254m
r = rramp - { (dup – dcr) } + dcr
{ 2 }
=(3.05m) - { (.254) } + .813
{ 2 }
=2.11m
E2 = E1 + W
K2 + U2 = K1 + U1 + W
½ mv32 + mgh3 = ½ mv22 + mgh2
(.5)(80kg)(7.15 m/s)2 +(80kg)((3.05m)=(.5)(80kg)(v1)2 + (80kg)(9.8 m/s2)(3.05m) + (80kg)[ 7.15 m/s)2 /2.11](.254)
2045 kg m2/s2 + 2391 kg m2/s2 = 40 kg (v1)2 + 492 kg m2/s2
40 kg(v1)2 = 3944 kg m2/s2
v1 = 9.93 m/s
Kfinal + Ufinal = Kinitial2 + Uinitial
½ mv2 + mgh = ½ mv2 + mgh + Wapp
The work performed/energy added can be evaluated using 5 variables
the height of the ramp (h)
the radius of the ramp (r)
the height of the air (a)
the mass of the skater (m)
the height of the skater(s)
( we made friction and air resistance negligible)
The terms all vary with respect to their role in determining how the energy of the system changes
1) the height of the ramp is 3.66m and this term determines potential energy (u)
2) the radius of the ramp is less than its height (height – radius = amount of true vertical), and is usually around 3.05m. This term influences conservation of angular momentum, and this is where the work is done.
3) The height the skateboard reaches above the top of the ramp is the direct result of the work performed. This can range anywhere from 0m to 4.42m (the current world record was performed by Danny Way)
4) the mass of the skater, if using any conservation laws would drop out the equation because it is constant. However, when determining the work done, work is dependent upon force applied. For this project we made the mass to be 80 kg.
5)the height of the skater is, curiously, the most important variable. What drives the energy of the system is the work performed by the skater, which utilizes the principle of conservation of angular momentum. This will be expanded upon
later in this document. The variable that move the skater up the ramp is the change is the radius as the skater’s center of gravity changes. We calculated the center of gravity of a 1.75m tall skater on his skateboard. While standing upright, it is 1.067m and while crouching it is 0.813m.
The radius of the ramp is where the work is done. This is the area where gravity is used to add energy…
At any point on the curve,
Radial component
S Fr mg cosq - n + Wapp =mar
tangential component
S Ft mg sin q + Fr = mat
The resultant force (Fr) in the tangential direction is due to the work done by the skater in the radial direction. This causes a tangential acceleration.
The gravitational force is conservative. The work varies over the distance through which it is applied, so
Wapp = D k +D V
W=ò xixfFxdx
(D k + D u) =ò xixfFxdx
What exactly is the work done? The Wapp is a function of angular momentum to add work going down the ramp, the skater crouches into the vertical and stands up near the bottom. To add work going up the ramp, the skater crouches into the bottom and stands up near the vertical.
The net effect is to travel in a tighter radius than the curvature of the ramp.
Conservation of angular momentum states that to counter the decreased radius, the linear momentum must increase. Since the mass of the skater is constant, his linear velocity must increase.
Angular momentum L = mvr sin q
q is the angle between the linear momentum (mv) and the radius, so for a ramp skater,q = 90° and sin q = 1
angular momentum for the skater becomes
L = mvD r
= mv(rf – ri)
The amount that the skater compresses/crouches is inversely proportional to the change in angular velocity. As L and M are constant
As r1< r2, then w 1> w
This change in angular velocity is due to the energy added by the skater. This energy is equal in magnitude to the work done, and thus velocities can be determined by using the principle of conservation of energy.
From point 2 to point 3 no work is done. The only force acting on the skater is gravity. This in conjunction with the fact that the skater’s velocity at his highest point (here 2m) is zero, simplifies the process of finding the velocity at point 2.
E3 = E2
K3 + U3 = K2 + U2
½ mv32 + mgh3 = ½ mv22 + mgh2
(80kg)(9.8 m/s2)(5.66m) = (.5)(80kg)(v2)2 + (80kg)(9.8 m/s2)(3.05m)
4.437 kg m2/s2 = 40 kg (v2)2 + 2391 kg m2/s2
40 kg(v2)2 = 2046 kg m2/s2
v2 = 7.15 m/s
From point 1 to point 2, the curvature of the ramp enables the skater to exploit conservation of angular momentum to add energy. This is where the work is done, and the force component of this work is the centripetal acceleration.
Remember that
M=mass of the skater
V=velocity at point 2
D=the distance from the skater’s crouched center of gravity to his upright center of gravity
r=the distance from the center of the ramp’s radius to the skater’s average center of gravity (between crouching and upright)
W=Fd
= mad
=m(v2/r)d
This equation makes two assumptions
that v1 is roughly equal to v2
2)that the radius used to determine centripetal acceleration is constant
d dupright - dcrouched
(1.067m) – (.813m)
0.254m
r = rramp - { (dup – dcr) } + dcr
{ 2 }
=(3.05m) - { (.254) } + .813
{ 2 }
=2.11m
E2 = E1 + W
K2 + U2 = K1 + U1 + W
½ mv32 + mgh3 = ½ mv22 + mgh2
(.5)(80kg)(7.15 m/s)2 +(80kg)((3.05m)=(.5)(80kg)(v1)2 + (80kg)(9.8 m/s2)(3.05m) + (80kg)[ 7.15 m/s)2 /2.11](.254)
2045 kg m2/s2 + 2391 kg m2/s2 = 40 kg (v1)2 + 492 kg m2/s2
40 kg(v1)2 = 3944 kg m2/s2
v1 = 9.93 m/s
B
Bubba
Guest
Uncle scares me sometimes.
B
Bubba
Guest
What book did ya get that from Unc? I used Serway & Faughn when I took two years of college physics more than a decade ago.
unclehobart
New Member
I got that online. If you put 'skateboarding physics' into Google it spits out near 7000 replies. I beezed through the top 100 and found that... of course that monolith formulae up there is for doing a half pipe jump.. not a 360.
B
Bubba
Guest
Oh well, that explains everything. That throws the whole equation off.
flavio
Banned
I can do a Frontside 180 and can tell you that the above is easily the most important part of the physics. Grip tape is what it is called. Also the Ollie to start the trick was well explained by Unc, but in my own words you basically kick the tail against the ground with your back foot and then quickly lift that foot to give the board clearance to bounce up in the air.
The same idea if your were to dribble a basketball a few inches from the ground. You push the ball to the ground but must also lift you hand immediately to give it room to bounce up.
Now the board would be rising at an angle so you take your front foot and kick forward to level it out. This is where friction, grip tape, and soft rubber soles start to come into play. This would be the basic Ollie. To add the 180 part spin your body and the the board will come with you thanks to more friction.
unclehobart
New Member
no. he would have been like 5 at the time.