Jack Hammer Vehicle Suspension Idea, Unsprung Mass, Tire patch load Calculation, Jack Hammer load?

[kaazx9r]

This is just something I was curious about, I'm sure a lot of users here has had some kind of idea related to their field or vehicles in general. I'm not a fully seasoned engineer just a car guy. So be kind.

But I wanted to know what would happen if I put a jack hammer system right on the unsprung part of the suspension.

Would it create any kind of load on the tire patch? Not only to increase grip but to also during a bump or droop, using the jack hammer system to accelerate the un-sprung mass faster. Excuse my crude illustration, typical jack hammer at Home Depot:

You can see the energy (excuse my wording) is rated at 46.5ft-lbs. This should be Kinetic energy.
1) How does one take the jack hammer rating and figure out the static load per blow?

Too many unknowns, so I did my own experiment using a leg press, I put a 25 lb weight on each side and let the press drop, a picture would help:

So the spring compressed 30mm. Then I put some weight on the press to see how much the spring compresses (see picture of spring) with different weight to find K, using F = Kx and found the spring K value is 20lb/mm. I used an old caliper mounted under the spring washer with a few clamps. Working backwards with my initial dropped weight, I found with a 25lbs on each side, the static load on the spring was about 600 lbs if the spring compressed 30mm. Given it has some rubber in front of the spring, I would say the load is probably more like 750lbs. But taking the 600 lbs.

Now taking that and finding the T = f*d, where F=600lbs d=30mm(.0984ft), and got 58ft-lbs. So this is kind of close to the Home Depot jack hammer. I then solved for velocity like 50lb x V X V/64.32 = 58ft-lbs, and got 8.65ft/second.

2nd question, let's assume the jack hammer has a rating of 58ft-lbs and a load per blow of 600 lbs. Would this work on a vehicle? Or would you have reaction loads on the car frame which would negate any benefit from the jack hammer. Here are the two cases, one directly and the other with a pivot since 600lbs is not that much.

What would be the reaction force on the sprung part of the frame? 600 lbs or something else?

Case 2, with a pivot:

Now you can see we've amplified the load to 7200 lbs using some mechanical leverage but the pivot has a reaction load of 7800 lbs, would this be the amount of reaction load on the frame?

3) What would happen with a jack hammer is put on the sprung part of the vehicle?

Kindly discuss.

 

[BrianE22]

Very interesting question. The car body on the vehicle springs would typically have a natural frequency around 1.5 hertz. I'm guessing (WAG) that the jack hammer is operating at a higher frequency above that. So attaching the hammer direct to the body wouldn't produce much load at the tire patch - you have an isolated system.

Other than that you'll need to wait for others. The unspring natural frequency is about 11 hertz (mostly tire and wheel and spindle mass on the tire spring).

 

[IRstuff]

Suspension travel is related to natural frequency; a high natural frequency implies a hard ride. Even with suspensions that are in the range of 1.5 Hz, there are already cars known for their hard rides; jacking up the natural frequency will make for a bone jarring ride.

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[3DDave]

There are already active adaptive suspensions that use sensors to detect road variation and react to wheel response accelerations and use actuators to apply force to the suspension in response. The jackhammer is the actuator part of the system.

https://en.wikipedia.org/wiki/Active_suspension

The main focus has been to simply modulate the damper/shock absorber to decrease the force imparted to the chassis as it merely requires management of hydraulic valves.

See the end of

https://www.carexpert.com.au/car-news/adaptive-suspension-explained

 

[kaazx9r]

Thank you for the fast replies, i thought about the current electro-hydraulic, to my knowledge these systems are to control how the suspensions compresses or rebounds and are faster than regular springs.

So I’m guessing what I posted is no different but want can the electro hydraulic systems continuously put loads on the tire, to me it this is different than a sway bar which lift the inside wheel to keep the car level(or electro hydraulically if using that kind of system), I want to Jack hammer the inside wheel to create more grip on the unloaded tire, how would you calculate that and would it negate due to the reaction forces it creates on the chassis.

 

[3DDave]

Whatever gets put on will be soon removed, like in dozens of milliseconds, decreasing the force on the road.

 

[IRstuff]

Suspension does not change the weight of the car, which is what the contact patch represents.

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[kaazx9r]

My thoughts are exactly what you guys are posted, that whatever you put on you will put on but what confuses me if I was using a hammer for tampering or breaking concrete there’s no way I would be able to put 600 pounds of static load from each blow from my body weight alone this is what I’m trying to calculate the static load that would act on unsprung on the suspension. And if it would also act in the opposite way on the on the chassis.

 

[3DDave]

A jackhammer provides an impulse; over the short contact time with rigid concrete the force goes up as the duration goes down. A small force over a long time is converted into a large force for a short amount of time.

 

[GregLocock]

You've sort of reinvented active suspension. I can't see much advantage in this approach, a very brief increase (or decrease) in vertical load is not really going to help much.

Typically wheel hop (unsprung mass bouncing on tire stiffness) is at 10-12 Hz, and the device shown is at 18 Hz. Depending on the type of vehicle this could excite structural resonances in the car's body.

Cheers

Greg Locock


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[MintJulep]

Seems to me that the same result could be accomplished by replacing the springs with solid blocks.

That would also allow removal of dampers and anti-roll bars.

You could go further and change to solid tires. That should bring the stimulus and response forces almost perfectly in phase.

Much simpler than trying to add all this other clap-trap.

Might have some undesirable consequence on ride quality though.

 

[kaazx9r]

Thank you, this is still different than an active suspension which to me just changes the dampening force but active suspensions, far as I know, put active loads back onto the suspension, all they do is increase the force so the wheel doesn't float off the ground.

What I'm asking, during a turn, the inside tire is unloaded, the active suspension will compress the unloaded tire so the vehicle weight will shift back on that unloaded tire, that is great, but I don't think active suspensions send impulses to the inside tire, that, to me would just raise the car up on that side. Given this weight transfer illustration:

An active dampening suspension would just control roll to keep the vehicle weight on all tires but I'm asking, if you take an external force, like a jack hammer, and let it act on the inside tire to further increase load on it. It doesn't have to be the slow jack hammer I suggested, it could be one with a very high frequency but I'm asking about the reactive forces on the chassis and how much additional static load it puts on the tire. And how to calculate it given the jack hammer specs above.

So consider a completely smooth road and a vehicle taking a turn on it, I want to use the additional static loads produced by something like a jack hammer so I can increase tire load and hopefully grip, and take the turn faster. (Obviously an active suspension (or passive anti-sway bar) will level the car to reduce roll, that's not my question or idea).

 

[3DDave]

The jack hammer goes bang then the recoil lifts the tire for a while so it no longer touches the ground. In addition the reaction to the recoil increases the body roll, making the weight shift worse.

 

[MintJulep]

Your error lies here:

if you take an external force

 

[kaazx9r]

The jack hammer goes bang then the recoil lifts the tire for a while so it no longer touches the ground. In addition the reaction to the recoil increases the body roll, making the weight shift worse.

I guess it depends on where the hammer is acting, on the sprung or un-sprung mass, I believe the dampers will handle the recoil if you’re letting the hammer act on the sprung part of the suspension. Which I don’t want.

But you’re right the reaction is what I’m concerned about. If the hammer acts with 600 lbs blows at a super high rate on the unsprung tire (neglecting tire compression) then the ground provides the reaction, which is what I want right?

 

[3DDave]

Just like a pogo stick - bang, bounce leave ground, wait to come back down. Add a damper to the recoil and a greater force is added to rolling the car over.

Just add 600 pounds of lead to the axle. Better than the jackhammer.

 

[MintJulep]

If the hammer acts with 600 lbs blows at a super high rate

How does your hammer get set for the next blow?

 

[kaazx9r]

Just like a pogo stick - bang, bounce leave ground, wait to come back down. Add a damper to the recoil and a greater force is added to rolling the car over.

Just add 600 pounds of lead to the axle. Better than the jackhammer.

Thank you, your insights are very helpful.

 

[TugboatEng]

Look at the "Bose Active Suspension". Most active suspension systems today use airbags and ferromagnetic fluids. Airbags can make height and rate adjustments but aren't responsive. Ferromagnetic fluids are responsive but can only affect compression and rebound.

In the 1990's, Bose (same as the speakers) developed a suspension system that used magnetic actuators to fully control every aspect of the suspension, instantly. At the end of their advertisement they used the active suspension to actually hop the car over a speed bump.

 

[hydtools]

"You can see the energy (excuse my wording) is rated at 46.5ft-lbs. This should be Kinetic energy.
1) How does one take the jack hammer rating and figure out the static load per blow?"

At the time of impact with the gad(breaker bit) the energy of impact is all transferred to the breaker bit so there is no force(load) applied to the breaker suspension. There is a body reaction as the hammer is accelerated down and when it returned to repeat the cycle. At 1100 bpm, the cycle time is 0.055 sec. The impact energy is determined be the product of hammer weight and its velocity at impact. Or 0.5×m×v^2 kinetic energy.

The hammer is not mechanically connected to the motor crank and driving piston so there is no pogo stick driven action.

Ted