ski impact 2

[andran]

hi all

I'm having a bit of trouble calculating impact forces on a 'ski'that I'm designing. the front of the ski is sloped, the slope has a length of 0.4m and a height of 0.2m. i need to calculate the forces on the front of the ski when it impacts a solid object as speeds ranging from 1 - 20 km/h. the ski will be attached to an object with a mass of 2131.53kg.

what i have done is used the conservation of energy. i said that the impact will raise the object 0.2m therefore increasing the potential energy. the kinetic energy will decrease by the same amount. i then calculated the change in horizontal velocity at various initial velocities. with this i calculated the impulse. i then used the average of the initial and final velocities and the 0.4m to calculate the time that the force would be applies. using this time and the impulse, i calculated the force.

i got a constant force for all speeds between 5 - 20 km/h and the impulse decreases with increasing velocity. this doesn't seem correct and i'm not sure where i went wrong.

any advice would be appreciated.

Thanks and regards
Andran

 

[rb1957]

i'm not so sure that the force you're calculating is a force applied to the vehicle by the surface. isn't it an inertial deceleration caused by the change in potential energy ?

the vehicle hasn't been slowed by an external force, because the delta KE has stayed with the vehicle as delta PE.

no?

 

[andran]

but wouldn't the PE be lost once the mass returns to the ground? it will not accelerate the mass in the horizontal direction. i've basically assumed that the change in PE is energy lost due to the deflection caused by the impact with the object.

i'm not sure if this is correct though. it seems more realistic that my initial model.

 

[rb1957]

before bump ... KE, and no PE
top of bump ... KE-deltaKE, deltaPE
after bump ... KE, and no PE

i think you've been assuming a perfect system, with no losses. strain energy would be one source of lost energy.

but again, i think the force is "only" an inertial deceleration, and not an externally applied force.

an externally applied force would appear something like ...
initial conditions KE
top of bump ... KE-deltaKE - impact force*time + deltaPE
and you've lost your relationship between deltaKE and deltaPE

 

[andran]

i agree.

but i've still kept the relationship between delta KE and PE. i've just changed the way i get the PE. there should be another term in there, but i'm hoping this at least gets me in the ball park.

 

[zekeman]

Andran,

What you are doing is not solving an impact problem but making a fictitious assumption that the moving ski with its attached mass is magically smoothly climbing up the incline and changing its linear momentum due to an increase in PE without thinking that
1) that change has to come from a horizontal force and
2) the change in PE is also an increase in vertical momentum which comes from a vertical force

I suggest that these small forces are the subsequent force you might get AFTER the impact force you seek occurs

You seem to be hung up with the time of impact which is NOT the time it takes to complete the vertical climb but in the millisecond or microsecond range where the maximum instantaneous rate of change of velocity is executed.

To answer your query
"thanks zekeman. but if i did calculated the force with the ski hitting a wall, there would be no way to determine the time interval."
That is a legitimate question but the model I suggested has a spring-mass-spring-mass, etc model which would allow 0 time for the first spring mass impact to occur.

For example, if you drop a mass from a height above a floor, at the moment of impact, the first molecules that hit the floor are stopped instantly and the distributed mass above those molecules compress and slow, etc causing a shock wave which settles down and the whole mass has a "impact time " equal to the settling time. Engineers are not going to solve every distributed type problem like this but use reasonable empirical impact time data for solutions which in practice are oftentimes way off the mark.

So far , I am not impressed with where this is going.

 

[chicopee]

What if you did some research with the steering mechanism (skis)of snowmobiles

 

[GregLocock]

I have had some success in predicting the impact force between a rubber tired wheel and a curb, at speeds up to 30 mph.

That success is limited to say a factor of two on peak forces, and is on the back of several actual instrumented experiments, and a very detailed knowledge of the suspension of the wheel, and the non linear stiffness of the tire.

With your proposed geometry and model there is a (fairly low) critical coefficient of friction at the impact point at which the 2 tonne mass will stop dead, whatever the initial speed. This is, frankly, absurd. If a model starts chucking out absurd results then it is not a good model. Your initial model also seems to ignore the case where the KE is less than the PE and so the ATV never surmounts the step. If on the other hand the projectile is driven at constant speeed then a simple energy balance won't work either.

In your case the model is too simple. Once you have made it complicated enough then you may be able to simplify it again.

I'll post the results of a model with some unspecified assumptions in it, and zero friction, after lunch.



Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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[GregLocock]

Vehicle weighs 2100 kg, obstruction is 0.2m diameter, leading edge of vehicle is a ramp 0.3m high by 0.6m long. Coefficient of restitution is 0.5, friction is 0. initial velocity is 4 m/s. Pitch inertia is 443 kg m^2

The target a long pole weighing 1 kg and is mounted on 1000 N/mm springs at each end restraining it horizontally and vertically. Each spring has a 2000 N/m/s damper in it. I used a round target to simplify the contact solution. Basically the target doesn't move much under the impact. In the real world targets with a local stiffness of 2 kN/mm are things like car bodies and so on.

Max force is about 70 kN vertically and 37 kN horizontally. The force profile is a half sine pulse 0.1 seconds long. Forward speed drops to 3 m/s, vertical speed rises to a maximum of 2 m/s as the ATV jumps over the log.



Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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[GregLocock]

That seems to have killed this topic.

The model needs validating

1) The impact time for a direct hit would be related to 2*pi*sqrt(m/k) seconds, 200 ms, so we are in the right sort of ballpark for the model as built.

2) a 2 ton vehicle hitting an 8 inch tree trunk at 10 mph would not generate 40g forces, therefore the model is wrong in some sense. A vehicle with a proper (soft) crash structure hitting a concrete wall at 30 mph generates 40g forces.




Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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