creasing the ski stance on the total force required .
As one pushes laterally on the handlebars at a spec height ( h ) from the ground , if one overcomes the force that the opposite ski can support spaced a distance apart equal to . 5x of the total stance ( S ), the sled will begin to tip due to the moment about the far ski contact point . This assumes infinite friction coefficient ( or that the sled won ’ t slip on the floor at all means : We only care about tipping – not slipping across the floor ). It would read like this : The force of the push times the height equals the force of gravity time times half of the stance , or Fpush * h = Fg *(. 5 * S ). And then , solving the equation for the force of the push , it would be Fpush = ( Fg *(. 5 * S )/ h )
This indicates that the force required to tip the sled varies linearly at a slope of . 5x the stance and height from the ground .
What if we ’ re not pushing completely parallel with the floor , and instead in an upward direction ?
Remember how statics requires the summation of forces to equal 0 in both vertical and horizontal directions ? We ’ ll effectively decrease the weight of the sled pushing downward into the floor by the magnitude of applied force relative to the sine of the angle to the horizontal , while also decreasing the effective lateral tipping force by the factor relative to the cosine of the angle . Yes , you may have gladly left all of the sine / cosine / tangent mumbo-jumbo behind years ago in your trigonometry class , but engineers still deal with it every day !
Putting all of that into an equation , summation M provides about the same point ( opposite ski ) shows that Fg *(. 5 * S ) = Fpush * cos ( θ )*( h ) + Fpush * sin ( θ )*( S ), or solving against for the force of the push , Fpush = (. 5 * Fg * S ) / [ h * cos ( θ ) + S * sin ( θ )]. Easy , right ?
If holding the height of application of force constant between both cases ( pushing laterally against an upward angle ), the magnitude of the force required to cause tipping is still intimately linked to the stance , or “ S ” in the equation , albeit through their respective trigonometric multipliers .
What does this mean for you in riding land ? In theory , its possible to calculate how much effort it takes you in every lean or turn to cause that ski to lift . ( Kind of … just watch your units . Slugs and kilograms and Newtons and inches and meters and feet per second , oh my !)
In reality , these types of equations are only the base of sorting through how a sled will interact with its driver on a given ride due to changes in width , weight , length , height , etc ., as well as the surfaces being crossed . If we free up variables such as changing the center of gravity , how shocks will actuate or even how the sloshing of gasoline in the tank provides a ‘ surge ’ of acceleration to one side which aids in lift … There are countless iterations and simulations factory engineers can and do run on each machine to guarantee a layout they deem fit for their customer base .
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