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.

 

[kaazx9r]

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

Thank you, to determine reactive forces, what I need to do is setup a pendulum hammer where the end the hammer (handle side) rotates but is also suspended on a spring, then let the hammer swing and drop against the ground, the ground also has a spring on it and measure how much the hammer end compresses the suspended spring and how much the hammer heavy side compresses the spring on the ground, EDIT: kind of like this:

This is my guess on how I would go about checking the reactive forces. Can I hypothesize the spring on the ground and the suspended spring on the hammer handle end will compress the same amount?

Also is this considered a closed system?

 

[Compositepro]

This concept and discussion is very similar to perpetual motion. When you understand how a jackhammer works and how a suspension works you will see that they effectively work to cancel each out.

 

[kaazx9r]

This concept and discussion is very similar to perpetual motion. When you understand how a jackhammer works and how a suspension works you will see that they effectively work to cancel each out.

That maybe true but what makes me curious is this quote "The conservation of linear momentum is not valid in such cases for the impulse applied to the particle by the fixed surface is an external impulse; hence the momentum of the particle is changed but the momentum of the fixed object is not changed by an equal and opposite amount." from "Mechanics For Advanced Level" by L.Bostock & S.Chandler , pub:- Nelson Thornes.


This is why I asked if what I proposed, where you have a jack hammer on the un-sprung part of the vehicle suspension with it's own source of power, is a closed system.

 

[hydtools]

The ground isn't very springy, it's more absorptive and will remove energy from your breaker system requiring more energy be put in to the system to keep it going.

Ted

 

[lucky-guesser]

Equal and opposite. You can use the hammer to fire the tire back down to the ground, but that same amount of force pushes the rest of the car up. This increases body roll, and you end up with even more weight coming off that inside tire. The only way I could see anything like this coming in handy would be in very rough off-road terrain. Even then, this would take tens of thousands of dollars to get something that sort of works, and then you would be selling to a very niche market.

 

[kaazx9r]

Equal and opposite. You can use the hammer to fire the tire back down to the ground, but that same amount of force pushes the rest of the car up. This increases body roll, and you end up with even more weight coming off that inside tire. The only way I could see anything like this coming in handy would be in very rough off-road terrain. Even then, this would take tens of thousands of dollars to get something that sort of works, and then you would be selling to a very niche market.

I appreciate the replies, but the quote I posted earlier:
"The conservation of linear momentum is not valid in such cases for the impulse applied to the particle by the fixed surface is an external impulse; hence the momentum of the particle is changed but the momentum of the fixed object is not changed by an equal and opposite amount." from "Mechanics For Advanced Level" by L.Bostock & S.Chandler , pub:- Nelson Thornes.

It explicitly says "the momentum of the fixed object is not changed by an equal and opposite amount."

This is why I ask, with what I proposed, would it be a closed system? if yes, then equal and opposite applies, if no, then I should have an additional 600 lbs of static load on the un-sprung part of the vehicle, which is basically just the tire.

 

[lucky-guesser]

You are hoping that this will work as it will function as an open system. An open system requires there to be another chunk of mass hanging around somewhere that this momentum can be transferred to/from.

This is a closed system, the unsprung suspension goes down and the rest of the car goes up. Its not open because there is no spring or stack of bricks hanging out above the car that you can con this equal and opposite force off on. Early on in the thread you show this system being attached to an arm that's attached to the chassis. That's it, your force is going to the chassis and thus the rest of the car. There is nowhere else for this impulse to go.

 

[3DDave]

It is external if there is a person standing on the track with the jackhammer ready to hit the suspension as the car goes by. It is internal if the jackhammer remains attached to the car.

L.Bostock & S.Chandler is saying the fixed surface is of infinite mass so changing the momentum of the fixed surface is impossible. That is not the case for any part of a car. When you jump up, the planet moves away, just a tiny and non-zero amount.

 

[hydtools]

The "fixed surface" is fixed and cannot independently apply any force, it can only react to some input force. If a particle hits a fixed surface, the force applied by the particle impulse is f=m×v/t .

Ted

 

[kaazx9r]

thank you all for clearing this up, I may have some followup questions but for now this is great input.

 

[kaazx9r]

I had a followup, so I setup a rig kind of like my leg press, and varied the location where I want to mount my so called jack hammer, basically right near the unsprung part of the suspension seemed the best.

What I noticed the rebound of only 25%, so basically the chassis moved up.

So from here, if I had a 600 lb static load, and the rebound was 200 lbs, could I assume we have net load on the unsprung tire by 400 lbs or am I missing here?

Secondly, I'm assuming jack hammers have some built in spring to reduce the rebound.

Also, rebound or lifting of the chassis isn't that bad, if I take that 200 lbs of rebound and send it to the rear of the vehicle, to uncompress the rear of the car, so the load will return to the front, basically interlinking it with say a 1:1 hydraulic circuit.

Can someone clear this up? I believed I would get 600lbs of rebound into the chassis but it doesn't appear that way, what do I need to fix?

 

[Compositepro]

You still have not bothered to understand how a jackhammer works. You have read in some specs that it has a "rating" of 600 lbs force and are making simplistic assumptions of what that means. The average force that the hammer provides is limited to its weight. It can provide moments of higher peak force during the impact. The stiffer the object being hit, the higher the impact force, and the shorter the impact duration. A soft pneumatic tire will average all the impacts to be equal to the weight of the hammer.

 

[kaazx9r]

Right now to make things simple, I’m assuming the tire is infinitely stiff. And even it the tire was soft, the blow should provide some extra load on the tire which is what I want.

If the only force the hammer can provide is the weight of the Jack hammer, say 35 lbs, then whats the point of the having a Jack hammer.

 

[MintJulep]

whats the point of the having a Jack hammer.

To hit things with a pointy stick repeatedly, until they break.

 

[CWB1]

...active suspensions, far as I know, all they do is increase the force so the wheel doesn't float off the ground.

There's quite a bit of functionality possible with active suspensions but like anything, it depends on the amount of time and money spent. Many systems not only stiffen the suspension during high-speed cornering but will also lower the ride height and/or lean the vehicle slightly to shift the CG. With a front-mount camera, some systems will also identify cracks, potholes, or other unavoidable issues and soften each corner momentarily to minimize wheel/tire damage. My personal favorite is maintenance mode, which creates more room beneath the vehicle by raising the suspension to max height. On low vehicles it allows jacks or bodies to easily fit underneath and prevents scraping the body on curbs/ramps. On longer travel systems like off-road it can save needing a jack - raise the suspension, place jack stands, lower suspension so vehicle drops onto stands and tires raise off ground.

 

[lucky-guesser]

Also, rebound or lifting of the chassis isn't that bad, if I take that 200 lbs of rebound and send it to the rear of the vehicle, to uncompress the rear of the car, so the load will return to the front, basically interlinking it with say a 1:1 hydraulic circuit.

Its equal and opposite, not equal and 6ft to the rear. You can't just send that force wherever you want. And any sort of hydraulic system is going to have a buffer time to send fluid (and force) back and forth. Also, you are lifting the front, so you try to counter that with lifting the rear? You have now successfully lifted the entire side of the car and sent a load transferring jolt through the entire car to the opposite side.

Quote (kaazax9r)
whats the point of the having a Jack hammer.

To hit things with a pointy stick repeatedly, until they break.

Idk if it was intentional but good point here. Many suspension components are cast which is brittle. Hitting them with that hard of an impulse will likely destroy whatever it is mounted to.

You also say you are doing all of this theory under the assumption that the tire in infinitely stiff, which its not, so all of the theory changes. Even if you did get some sort of meaningful downward force, the tire is going to rebound right back up with some force component of whatever you pushed it down with.

I will give you that its an interesting thought experiment, but if you are looking at this with any serious capacity you are going to keep hitting the inevitable barriers of physics, and even if it worked, the cost and complexity. Paying $5k and adding 100lb is hardly beneficial to a little bit of added tire force for a split second.

 

[Compositepro]

"If the only force the hammer can provide is the weight of the Jack hammer, say 35 lbs, then whats the point of the having a Jack hammer."

That is a reasonable question to ask yourself silently in your head to guide your thinking process. What do you think the answer might be? Simplify the question further and ask what is the point of any hammer and how is a jackhammer different from a plain hammer. These questions are easy to answer if you think about it. Until you do that all further speculations are a complete waste of your, and our, time.

"Right now to make things simple, I’m assuming the tire is infinitely stiff. And even it the tire was soft, the blow should provide some extra load on the tire which is what I want."

That is a really poor assumption, considering that the sole purpose of a pneumatic tire is to act as a cushion, which is the exact opposite of being infinitely stiff. Think what happens when a tire runs over a stone.

 

[kaazx9r]

All interesting posts, but lets stay on topic, it's just an idea, no one is spending money or trying to market this.

We're talking about a vehicle on a smooth surface, taking a turn, no bumps, the turn lasts less than a second or that the amount I need additional load on the tire is less than a second and I want to know the static force if you take a jack hammer type of machine to the unsprung part of the vehicle. Of course the tire will compress, but that's what I want. I only need it for a split second while I'm either losing grip or making a turn.

Even if you did get some sort of meaningful downward force, the tire is going to rebound right back up with some force component of whatever you pushed it down with.

So taking this, and from my observation, the rebound was only 25% into the chassis, my wheel was solid, and the chassis only showed a 25% upward force, but the downward force from the weight was a lot more, so I'm thinking either my experiment is wrong or if anyone can clear up why the chassis only rebounded 25% of the load I put on the wheel which was impacted by the weight I dropped on it.

I'm picturing a trampoline type of scenario, where the initial height of the person will be far greater than the rebound height the jumper achieves. The trampoline takes or like you all stated absorbs the force, but what is that force in my scenario like the trampoline. And what is the rebound force.

In addition, assuming the concrete was very hard, if I go by the fact a jack hammer load per blow is 600 lbs that means 600 lbs is getting returned to the operator, correct? Or do manufacturers build something in jack hammers to reduce the rebound? maybe some springs?

 

[kaazx9r]

Some topics that were similar that I looked at:

https://physics.stackexchange.com/questions/175956/good-way-to-compute-the-force-of-a-hammer-blow

"For example, if a 5 pound hammer is swung at 50 feet per second, then the kinetic energy of the hammer is 5 * 50 * 50 / 64.32 = 194 foot pounds. If this then compresses a forging by 1/8 of an inch, then neglecting rebound, then it is about 194 * 96 or about 18,600 pounds of force on average. This is because 1/8" = 1/96th of a foot, and the average force is foot pounds / feet over which the force acts."

https://www.physicsforums.com/threads/linear-impulse-jackhamer-rebounds.306283/

"During Operation the jack hammer develops on the concrete surface develops a force @ t=0 f=0, @ t=.2, F=90, @ t=.4, F=0. Graph looks like a triangle To achieve this the 2lb spike S, is fired from rest into the surface @ 200ft/s. Determine the speed of the spike just after rebounding"

Initial velocity: 60.96 m/s
Final rebound velocity:
Return Momentum = RP RP = mv = -24.799 = .907v therefore v = -27.341 m/s

Not sure how correct the above is, but I wanted to post what I looked at, it appears the rebound has far less velocity and given what I'm asking, it should transfer less force to the chassis but wanted to run it by everyone.

 

[kaazx9r]

Its equal and opposite, not equal and 6ft to the rear. You can't just send that force wherever you want. And any sort of hydraulic system is going to have a buffer time to send fluid (and force) back and forth. Also, you are lifting the front, so you try to counter that with lifting the rear? You have now successfully lifted the entire side of the car and sent a load transferring jolt through the entire car to the opposite side.

Interlinking is an old concept, if the front lifts, the weight shifts to the rear, but instead of shifting it to the rear, you have a separate passive hydraulic circuit so whenever the rear compresses, that circuit sends force back to the front to keep the front springs compressed and the car level as possible, this is my understanding, and the concept is so good it is banned in most major motorsports.

But that's a whole different topic, getting back on topic, I'm trying to figure out the reaction forces and what is wrong with my observations.