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ski impact
2

ski impact

ski impact

(OP)
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

RE: ski impact

Purely out of interest, what are you trying to ski that is 2 tons at 20mph?

RE: ski impact

(OP)
it will be fitted to the bottom of an acv.

RE: ski impact

"force due to impact" ... all impact problems revolve around the time duration of the impact. there are multiple threads on this topic.

the movement of the ski is obviously tied to the suspension of the ski. the shape of the object struck, is there a ramp or a bluff face ? but then that's why the ski-tip is curved up, no?; so that in contact even with a bluff face, the tip of the ski tries to move vertically.

i'd design the ski for all of the ground contact loads, then see how much fwd load the structure (the loadpath from the mass into th eski) can take.

or do a research project, like car impacts.

RE: ski impact

(OP)
hi rb, thanks for the advice. the problem i'm having is that the impact will cause a deflection, not a deformation. all the other threads i've come across involve a deformation.

thanks

RE: ski impact

"I'm having a bit of trouble calculating impact forces" ... force due to impact has been discussed often. it comes down to how long does the impact take to transfer the momentum, how quickly does the vehicle come to a rest ?

if your ski hits a step, then the tip of the ski contacting the step will see a force, balancing the change in momentum. the worst case (i'd expect) would ben when the ski tip does not deflect upwards, and has an off-set compression load applied.

RE: ski impact

(OP)
okay. i understand all of that.

what i'd like to know is if i went about calculating the impact force correctly. my final answers do not make sense to me.

RE: ski impact

i think your energy method is more like the vehicle going over a bump, rather than impacting a step.
since the change in PE is independent of speed, then so too the change in KE, then so too the force calculated from the change in momentum.

so, no, i don't think your method is calculating impact forces.

RE: ski impact

(OP)
thank you! that makes a lot of sense.

any idea how i could make the change in potential energy dependent on the initial velocity?

RE: ski impact

"any idea how i could make the change in potential energy dependent on the initial velocity?" ... no.

if you are considering going over a bump, wouldn't the suspension absorb most of the energy ?

if you're running into a step there will be an impact with the step (even if the vehicle doesn't stop), a change in fwd velocity resulting from an arresting ground force.

RE: ski impact

If you read ALL the threads on impact forces, you'll find that the prevailing approach involves assuming some level of spring-like behavior, which would then allow you to determine a timescale, which then allows you to determine an acceleration.

TTFN
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RE: ski impact

"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."

If the object you are hitting is a step>.4 you have a horizontal impact problem whose solution depends on the model of the ski to attached mass using energy and momentum conservation equations,
including material energy losses which is very difficult.

If step<.4 you have a different impact problem where the force has a vertical component and again you use the energy/momentum equations with an energy loss model.

But the problem is you must deal with different obstruction shapes which lead to multiple force configurations.

So, in summary, these problems do NOT lend themselves to analytical solutions. Your best bet is to do the empirical stuff using accelerometers to get a realistic answer.

RE: ski impact

[quote myself]you can quote people like this[/quote]

Quote (myself)

you can quote people like this

just a small fyi

NX 7.5.5.4 with Teamcenter 8 on win7 64
Intel Xeon @3.2GHz
8GB RAM
Nvidia Quadro 2000

RE: ski impact

On the other hand, if you are looking for a worst case problem, then simply model the ski and attached mass connected by a "spring" and maybe break up the ski into a series of springs and masses. and allow the system to hit a wall.
Then, the solution to momentum/ energy equations are could yield some worst case data.

RE: ski impact

Thank you Walter, I never knew.

RE: ski impact

At v0=20 km/h=5.55 m/s the velocity of your mass, assuming it will be able to pass over a step h=0.2 m high, will become
v1=√(v0-2gh)=5.18 m/s
So, if this is a method to reduce the speed of the AGV, it isn't very effective (not considering shock acceleration effects).
Note that when the tip of the ski is 0.4 m past the step, this doesn't mean the mass has lifted 0.2 m, more likely half of that, depends on the length of the ski and/or the mass. The mass will be at 0.2 m height when the CoG of the mass is at the step start.
Apart from this, your results are not necessarily wrong. In formula you did this:
(L is 0.4 m for you, but possibly wrong as noted above)
impulse I=m(v0-v1)
time τ=2L/(v0+v1)
force F=I/τ
Now rearranging a bit we find:
I=2mgh/(v0+v1) (decreases with increasing speed)
F=mgh/L (independent of velocity)

prex
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes and launchers for fun rides
http://www.levitans.com : Air bearing pads

RE: ski impact

(OP)
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.

it would be ideal if i could make the change in potential energy dependent on the initial velocity. there must be another term that i am missing out somewhere.

RE: ski impact

To make the change in potential energy dependent on the initial velocity you must send the ski over a climb, not a step. The vehicle will stop at changing heights with varying initial velocities.

prex
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes and launchers for fun rides
http://www.levitans.com : Air bearing pads

RE: ski impact

(OP)
okay. so i have modified my calculations.

i stopped assuming that the displaced height would remain constant at 0.2m. this assumption meant that there were no solutions for speeds less than 5km/h as there wasn't enough energy.

instead, i calculated the theoretical maximum height that would be reached at each speed. kind of like projectile motion. with this, the potential energy is dependent on the speed.

both the impulse and force increase with increasing speed.

any thoughts?

RE: ski impact

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?

RE: ski impact

(OP)
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.

RE: ski impact

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

RE: ski impact

(OP)
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.

RE: ski impact

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.

RE: ski impact

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

RE: ski impact

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: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

RE: ski impact

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: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

RE: ski impact

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: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

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