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Help with video game problem (acceleration on an incline)

  1. Jan 24, 2006 #1
    I'm trying to write a 2-D snowboarding game. I've forgotten my simple physics from college. Anyway, the problem is...

    Say you have a snowboarder falling straight down at velocity V and being accelerated by Gravity G. Say snowboarder then lands on a slope of Angle 45 degrees. I want to now now what the snowboarder's new X and Y components for Velocity are.

    I know that the acceleration down the slope(parallel to the friction) due to gravity when on the slope is:

    A = G sin(Angle)

    And also I know that I have to account for friction so really this becomes:

    A = G (sin(Angle) - Friction(cos(Angle)))

    And, so I can calculate Ax and Ay. But, what is confusing me is I would think that when the snowboarder hits the ramp, the snowboarders Velocity in the Y direction would *decrease*, not increase as the above formula would suggest. So, what am I missing? I guess the incline pushes back somehow to slow down the rider so they don't go thru the ground. What's that equation? I feel that the snowboarders Velocity should somehow be in that equation? ie, if the snowboarder is dropping at a Vy rate of 10 meters/sec, the ground would push back harder in the -Vy direction, no?

  2. jcsd
  3. Jan 24, 2006 #2
    Why would the velocity decrease when you hit the incline? The acceleration would decrease, but the velocity would still increase, only not as fast.
  4. Jan 24, 2006 #3
    The ground push is the normal force provided ny the surface. But as the suraface slant change, or the motion was vertical before and then it is not fully horizontal and so there is some vertical component.
    The force acting on the particle is due to earth and if it has mass M and then the force acting on it is Mg(don't use G, it conventionally means constant of gravity). The component perpendicular to the surface is Mgcos(angle) and so there is a normal force of same magnitude in the opposite direction. So force on Y-direaction due to normal force is Mgcos^2(angle), draw a diagram and clear what I am talking about.
    Similarly you agreed about the slanting force, Mgsin(angle) - KMgcos(angle). And the vertical component of this is Mgsin^2(angle) - KMg cos(angle)sin(angle) but in the downward direction.
  5. Jan 24, 2006 #4
    Civil Dude is correct anyway. But there is a probalbility that the friction is so large as well as the angle of slanting is less and then the velocity might decrease. I think Yoshele wanted to make sure when the velocity increase and when it decreases from his formula and he was confused.
  6. Jan 24, 2006 #5
    will upload graphic shortly

    I still don't see why the velocity would increase. I will upload graphic shortly to show what I mean. Say in the extreme example, the rider lands on a 1 degree slope (basically flat ground). Vy is going to quickly go to a really small value, right?
  7. Jan 24, 2006 #6

    http://learnworkplay.com/temp/sb_diag1.jpg [Broken]
    Last edited by a moderator: May 2, 2017
  8. Jan 24, 2006 #7
    You please read my post clearly. I made it clear that if you include firction at some cases the velocity would decrease and at some case the velocity would increase depending on ex-factors.
  9. Jan 24, 2006 #8

    sorry, i was replying to Civil_Dude's first post. thanks, I will try these new equations out!
  10. Jan 24, 2006 #9
    still a problem

    I still think that the velocity of the object should come into play here. To illustrate, say object is falling at 1 million meters per second straight down. Vy = -1e6 m/s. And, then it hits *flat* ground. Doesn't the earth push back with a much greater force than if the object was just placed on the ground with no initial velocity? Maybe your equations still account for that and that somehow it cancels out?? But, I am having trouble seeing it.

  11. Jan 24, 2006 #10
    friction dependent on velocity?

    Maybe the friction component is velocity-dependent?
  12. Jan 24, 2006 #11


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    With no friction my analysis of the motion down the incline comes to
    since the acceleration is constant in the y-direction you can use the constant acceleration equations, same for the x acceleration
    the relationship between the two velocity components should satisfy
    so you need to calculate only one of the two - this requirement will ensure that the guy is moving along the incline at all times. What happens as he hits the incline - only God ... The snow will compact up to a point (and decrease his velocity or decelerate him), then it will become rock hard and he will bounce off it, but normally a snowboarder absorbs the impact by bending his knees... so what to do? A hard ball will bounce off a hard incline - like reflection with incident angle is equal to the reflection angle that is equal to [itex]\theta[/itex] - resulting in projectile motion with some initial x velocity and a decreased initial y-velocity?
    Last edited: Jan 24, 2006
  13. Jan 24, 2006 #12

    aha! i think this is what i'm looking for. i want to make sure the rider stays on the line. And, yes, for now i would assume that the snowboarder absorbs exactly enough of the impact to not bounce off the slope. i'm sure i will add some code later that if the rider hits with a large enough normal force they will die. :)
  14. Jan 25, 2006 #13


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    With friction it seems the accelerations are:
    downwards, and the horizontal acceleration away from the incline
    [tex]a_x=g(\cos(\theta)\sin(\theta)-\mu \cos^2(\theta))[/tex]

    The velocity of impact with the slope is altered via the coefficient of restitution, [itex]e[/itex], which is a positive fraction depending on the two types of materials involved in the impact. If [itex]V[/itex] is the velocity of the snowboarder vertically downwards before impact then the snowboarder will bounce off the slope with a new velocity [itex]V'[/itex] with horizontal and vertical components given by

    [tex]V'_y = V(\sin^2(\theta) - e\cos^2(\theta))[/tex]
    downwards and an x-component away from the slope
    [tex]V'_x = V(\sin(\theta)\cos(\theta) + e\sin(\theta)\cos(\theta))[/tex].
    Last edited: Jan 25, 2006
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