Deadlift holds and strongman auto holds

[Guest DavidHW]

I've got access to a Hammer Strength deadlift machine, the kind where you can hold the handles at the sides just like the Strongman auto hold events. My question is this: how much weight would you have to load on such a machine (approximately) to simulate the 2500 lb cars they use in Strongman? For some reason, I remember hearing the ratio was roughly 3:1, that is, 833 lbs loaded on the Hammer Strength machine should be roughly equivalent. But I doubt this is because I can hold 833 lbs on the HS machine for about 20 seconds, and I am *not* anywhere near being Strongman strong, if you know what I mean. So anyone have a guess as to my question?

TIA,

David

[OldGuy]

That ratio is almost certainly the other way around, and that holding up the back of a 2400 pound car is = to supporting 800 pounds.

[Guest el4n]

Because the Hammer strength machine wont be a true 800lbs or whaterver you load on (like lifting the car wont be a true 2500 because you are only lifting one end) since one end is anchored, you'd most certanly have to load alot more weight onto the hammer machine, much closer to the 2500lb end of things, then an 800lb one - right?

[Guest DavidHW]

Well, here's my thinking: in photos I've seen of strongman comps, the center of gravity of the weight (the car) is about 6 feet from the lifter. On the Hammer machine, the weight's center of gravity is about 1 foot from the lifter. Now thinking of physics here, fulcrums, levers, and all that, isn't it easier to lift equivalent poundages if you're further away? So that a heavier weight X at distance Y from the lifter will need the same muscle power to lift as, say, weight X/4 at distance Y/6 (or something like that)?

Help! Is there an engineer in the house?

David

[Guest DavidHW]

OldGuy:

We don't disagree. Sorry if I wasn't clear. Yes, I'm asking if 800 lbs on the machine takes the same effort to lift as 2400 lbs of car (with my additional comments on weight and distance taken into consideration).

David

[mobsterone]

I asked some competitiors about this when I went to the Blair Atholl highland games 2 years back. They said it was like holding 600 lbs on a normal bar for time. Try that in a power rack. I'd have to try a hammer instead of doing partials to know for sure. Of course you could just load it up as much as possible and if thats more than 800 then do that and then start extending the time.

[BarBender]

Statics 101

The sum of the forces are equal to zero.

The sum of the moments are equal to zero.

[Guest DavidHW]

mobsterone:

Interesting. For me, a 600 lb deadlift partial hold feels roughly equivalent to around 725-750 on the Hammer machine. I'd just do the regular deadlift holds with a bar, except that the Hammer machine has the grips pointed the Strongman direction -- wish I had a trap bar. :-(

BarBender:

What? Me no speaka non sequitur.

:-) :-) :-)

[BarBender]

This might help.

http://enrich.sdsc.edu/SE/physicsforces.html

In the final problem, F2 = 100 pounds.

[Guest DavidHW]

After reading that lesson, I'm more confused than ever, and I've actually taken high school physics (and obviously forgotten most of it).

I keep getting the following:

2500 lbs at 6 feet = x lbs at 1 ft

x = 15,000 lbs? I need 15,000 lbs on my Hammer machine to equal a 2500 lb auto hold at 6 feet?

That's not right.

???

Seems more logical that 2500 lbs at 6 feet would be equivalent to 2500/6 at 1 ft or @420 lbs?

[Guest DavidHW]

OK, I've found another page to confuse me:

http://enrich.sdsc.edu/SE/physicslever.html

So the Hammer Machine is a Third Class Lever because the weight is on either side of you, and the automobile hold is a Second Class Lever because the weight is closer to the fulcrum? I'm really having trouble getting my head around the comparison.

:-(

[Jett]

I remember reading this site a while ago. It shows step by step instructions to roughly figure how much force it is to lift a car up. Sooo.... what you do is this... 2500lbs * 3feet = x * 6feet, and you get 1250 pounds, enter the 1250 into the equation for the measurements for the machine.. so x * 1/2 = 1250 * 1, and you get 2500, which makes more sense than 7.5 tons to put on the machine, but, it still makes no sense since that is the weight of the freaking car! This is driving me nuts, why does physics have to be so hard? Took me half an hour to figure that out, arrg!! Im not sure if we are taking the right approach to this though, probobly gets into some pretty hard math, since the way I figure it, and what that site about the weight being closer or farther away from the fulcrum, the weight of the car is more spread out than the single place where the bar comes out to put the weight on, I think that plays a big role in figuring this thing out, any math professors out there?  :hehe . Ill ask my math teacher tommorow, maybe he can figure it out. You think something like this would be easy, but there are so many variables it just makes your head spin :0 . This would be a good question to put on a physics test!

[Guest Harlan Jacobs]

We have a deadlift machine ( Not hammer ,but sounds close ) at the gym where I train. We also have all the strongman toys. On ours, I would say about 1000 lbs is equal to our S-10 .

[Guest DavidHW]

OK, I found the following at the Ohio Valley Strongman site:

[http://www.ohiovalleystrongman.com/2002autohold.htm]

"The Dodge Auto Hold requires a contestant to dead lift a Dodge Neon weighing approximately 2,500 lbs. into position and then hold the auto for as long as possible. Realistically, the weight will feel like approximately 1/3 of the vehicle's weight due to the dead lift bar utilized for the event. The dead lift bar will extend approximately five feet out from the back of the vehicle."

Since the Hammer machine I use has a similar lever action with the above, and since the weights are just about on either side of the lifter (maybe 12 inches in front), the original 3:1 ratio I was remembering seems about right; i.e., @830 lbs on a HS machine should "feel" like the 2,500 auto hold, at least in this case.

I'm still trying to figure things out using the math and physics involved, but this little snippet seems to suggest I'm in the ballpark.

David

[The Mac]

David,

I don't think that 830lbs on the hammer machine is going to be that far off the mark - it is definitely in the region where you should be thinking that you are probably capable of this kind of feat. I think the most interesting discussion of how much force it takes to lift a car is on Juan Lancaster's SumoPages site (as linked above). I don't think you should get the results mentioned below - read his example and you will get some heavy, but very achievable Deadlift numbers to lift the car. I would agree with Mobsterone that if you can pull a long-range partial deadlift (to simulate lfting the handles) of 650lbs+ and hold it for a while you are probably not far away (if at all) from being able to hold the car, especially with the extra leverage of the handles out the back.

I think the most difficult thing about trying to work out if you can lift a car is that the car obviously doesn't conform exactly to any simple maths you can apply - the weight distribution is not equal (try lifting the back of two cars of equal kerb weight but with one having the engine mounted in the rear - I've seen it happen to the unwary! ) and so on. Sometimes trying to use maths to solve real-world problems can cloud the issue, especially when there are many variables involved.

At the risk of alienating some people, I don't think you have to be INCREDIBLY strong to do some Strongman events. Clearly, to dominate the car hold and hold it for a long time (60 secs plus) you need incredible grip-strength, which you of course should have a head-start on already if you hang around here! I have seen it so often that very good performances (in terms of time) have come from people who, for whatever reason (usual because they are extremely tall) have found it very difficult to lift the car from the ground (Forbes Cowan is an excellent example of a strongman to whom this applies).

[BarBender]

In all seriousness, the mathematics involved is not beyond high school algebra. With formulas given, the mathematics is reduced to arithmetic.

At the risk of giving the impression that I am being condescending, I provide resources as a starting point to allow one the satisfaction of learning. There is no favor done by giving out a solution. Spewing forth answers does not help one to learn how to solve similar problems.

[The Mac]

You do not come off as being condescending, BarBender. I was merely trying to point out that, even with the formula being given, the real world may NOT be as simple as the model would suggest. Unfortunately.

[Guest DavidHW]

The Mac:

Thanks for the extended discussion. You should write an article for a strength mag on the variables encountered in SM comps -- seriously. Two or three anecdotes from stunned firstime SM competitors would be most entertaining. That's one of the things about the SM sport that makes it so compelling, the "you don't really know what you got until you get there and lift" angle.

David

[Guest DavidHW]

BarBender:

Hey, as much as I simply wanted a straight answer, I appreciate the attempt to re-educate me. Thing is, I aced calculus and high school physics not two decades ago; yet being the liberal arts major I became, it appears it all went out the ears.

David

[BarBender]

There can be no argument that real world problems are far more complex than the ideally constructed problems posed by academia. However, one must understand the simplified problems where ideal conditions are assumed before attempting to solve other problems where additional conditions are included.

If someone wanted assistance in understanding the physics of a torsion spring gripper, an appropriate response would be to direct that person to resources that explain Hooke's Law, mechanical work, and torque. This would only be a starting point. The study of hand grippers has proven to be very complex.

For auto holds, it is important to understand the concepts taught in Statics. Leverage problems make up much of the course work. A thorough analysis of the auto hold would include a study of the dynamic element of getting into the lockout position after which we consider beam deflection, oscillation, and friction. I prefer to take the easy route by constructing a simple lever problem.

I was never upset with anyone in this thread and I hope that my previous responses did not appear unhelpful or cryptic.

[Guest DavidHW]

BarBender wrote:

<<<I hope that my previous responses did not appear unhelpful or cryptic.>>>

Unhelpful? Precisely the opposite. Cryptic? Perhaps, if only because I let myself forget everything I learned my junior year in high school. :-)

As someone above said, these sorts of problems would be great for a high school class in physics, one maybe with a few athletes (or anyone else) who needed convincing of the importance of math and physics. A class project where students convert Strongman events into quantifiable lifts on their school gym's equipment would be a fantastic learning experience.

David

[The Mac]

Not a problem, Barbender. I for one would welcome such a scientific evaluation of the auto hold (or any other event) - it obviously wouldn't make the hold any easier, but to me at least it would make it more interesting!!

Thanks for your ideas - I had never really thought of the practicality of trying to come up with accurate model of such things. Perhaps now I shall give it a little more thought.

[Guest el4n]

I think this question has gone further than it ever was proposed to. Someone simply wants to know how much weight should be put on the hammer machine to equal the same weight that would be used during the 2500lb car lift in a strongman event. Very straightforward question requiring only one answer, which is the weight needed to load on the Hammer machine.

BarBender you seem to infer that you understand the problem, physics, calculus etc. etc. etc. but don’t want to offer an answer. To me this makes no sense, if you knew the answer why would you not provide it. However, since this would be so logical I think you do not know the answer.

Maybe Wannagrip should start a math discussion section on this board if you guys want to keep throwing back and forth "thoughts" on solving this apparently extremely perplexing problem.

Someone dazzle us and simply give a numeric answer to this problem, not a philosophical one.

[crushjunior]

Find the centre of gravity of the car. (The distance from the cg of the car to the fulcrum (the front axle) divided by the distance from the handles (where you lift the car) to the fulcrum) times the weight of the car is the maximum amount of force you need to lift the back end up. Naturally having handles extending a distance behind the car will make the lift easier as well as having handles that are higher up. Shocks that have more give will make lifting the car easier allowing most of the weight of the car to bear on the lifter when the frame comes high off the ground will make the lift easier. Shocks that stretch only very little will make lifting the car harder particularly if the lifter has not cleared knee height before the full force has to be applied.

For a 2500 lb car I estimate the cg to be 2% closer to the front then the back end (engine in front) with a total length of 12 ft (I'm not sure how long the car would be but it sounds close to what it would be). The distance between the front axle and the rear bumper (no handles!) is 9.5 ft. The distance of the cg of the car to the front axle is 3.36 ft. 3.36/9.5x2500=884 lbs if you lift right off the rear bumper, if 650 lbs clears the bumper over your knees then it would be like a partial deadlift with 884 lbs with a 650 lb deadlift required to get it up to the knees. That's estimating that the tension in the springs will support 73.5% of the weight needed to lift the back end by the time you're at knee height. Just as you move over the knees that will become 100% at which point the rear wheels will clear the ground.  With a handle extending 2 feet behind the car, the distance from the front axle to to the point where you lift becomes 11.5 ft and so 3.36/11.5 x 2500=730 lbs to clear the back end and maybe 535 DL to get it up to the knees if the handles are at regular DL height and if they are at knee height to begin with then it would be equal to a lower (440'ish) regular bar DL.  The handle deadlift should be easier because you can place the handles in line with the cg of your body while a regular deadlift has the bar out in front of the body. Maybe 2-3% easier. Then again a reverse grip in a regular makes it easier but you can't reverse on the car DL so that would make it harder overall I think.

Oh yeah, as the frame of the car moves up more weight will shift to the front axle,as the entire car moves up again additional weight will shift to the front axle making the lift somewhat easier then what I described above.

I'm looking to lift a car so that's why I have figured this stuff out. My regular DL is too low right (500) now so I'm going to wait till I get stronger. Then I need to find someone with a light car who'll let me try my stuff on it.

[BarBender]

el4n, there was not sufficient data to provide a solution. The dimensions of the Hammer Strength machine can be found but the biggest unknown is the vehicle used in the auto hold. Even if two vehicles have identical curb weight and length, the center of mass may not be in similar locations. Since we do not have enough information about the original problem, all we can do is explain how to go about solving it.