Ok, I posted up a couple of weeks ago about this and I thought i'd give an update and see what you guys think so far.

https://www.ford-trucks.com/user_gal...id=134350&.jpg
https://www.ford-trucks.com/user_gal...id=134351&.jpg

I have a video of it in action that I will also post up a little later.

We're adding a foot to the bottom of it to help stablize it at full reach because right now it sways around quite a bit.

What we're not sure of is the equations to prove that this thing works. Obviously these should've been done before we started, but we were confident that it work.

We have a pump that puts out 2700 psi.
The left side arm is 35.5inches long from pivot to pivot.
The ram is 24.5 inches from pivot to pivot when fully compressed and 36.5inches when fully extended.

So if anyone could help me out with this it would be grealty appreciated.

 

Jeez, talk about tippy! I'd be scared half to death tryin' to take a tire off that thing...

You're going to need to know the bore of the hydraulic ram, to find the force it is capable of exerting with that pump. (Alternatively, you could work out the force required to lift the car, then find the minimum bore that would do that with the given pump....)

 

The amount of work required will change with the height of the car (harder at first, easier nearing max height), because of the leverage disadvantage.

Say you have a 2 square inch bore, that would give you 5400# of force. If both 'levers' were perfectly horizontal, there would be 2700# of force pushing at each mount in horizontal directions, essentially trying to push the car apart. if they were perfectly vertical, you would get the full force of 5400# of lift. I imagine there's a complex formula for the equation to give you the answer from the first angle of interaction with the ground to fully extended angle but I'm a HS dropout so I can't help you there. At 45º angle you should have 2700# of ourward force (1350# at each mount) and 2700# downward force.

You also have the vehicles springs that aid in the beginning of the lift which helps the disadvantaged levers.

BTW, when you talked about lifting the 'side' of the vehicle, I didn't know you meant front and rear sides. That could pose a problem when doing the rears, as vehicles typically don't have front parking brakes and although typically required by law to be functional, rear parking brakes may not always be up to snuff to hold the car in place. I'd say that is a pretty dangerous apparatus.

 

Yeah, it changed quite a bit since when we first talked about it. Originally it was going to be side-side. It is rwd, so it has the tranny and the parking brake holding it along with the weight of the car coming down onto the lift point. Obviously you wouldn't crawl under this thing without using some jack stands and something to block the wheels, but for changing a tire on the side of the road it's no more dangerous than the jack that comes with the car.

Here is the link to the video, btw: http://www.youtube.com/watch?v=le9MsBiVT7c

oh and what is the definition of the bore of the cylinder; is it the ram that comes out? or the outside diamter of the tube?
either way:
the ram that moves/comes out has a 1.5inch diamter
the outside diamter of the ram (the part that remains stationary) is 2 inches.

 

Neither. It's the diameter of the piston the fluid pushes against, which is also the INSIDE diameter of the tube. It's usually bigger than the rod that extends.

*Furball, you've oversimplified the relationship between horizontal and vertical force. (Hint: in order to do these calculations, you're gonna need sine and/or cosine functions...)

 

You're working on a project that was addressed and marketed decades ago. Go to an RV dealer that sells motor homes and look at a motorhome with automatic leveling jacks. Or go to your local tractor dealer and look at a tractor with a hydraulic blade and imagien replacing the blade with a bar. My old John Deere 318 L & G tractor can use the blade (which could easily be a bar) to lift the front end. The hydraulic cylinder is powered by the hydraulic pump that operates the transmission and power steering. Simple and safe.

 

Yes, I stated that in my reply.

Also, the values of the forces would be fractions of the cars actual weight, whereas I was using the maximum theoretical capacity of the aparatus.

 

Yeah, thats what I was thinking, since it IS a triangle. But I don't really remember anything about the cosine/sine functions and how they're related to triangles.

How do we determine the inside diameter? would it be safe to assume half way between the two.. 1.75? Now that I type that I'm pretty sure that one of the guys in the group said this morning that it was probably a 1.75 bore.

edit: osbornk, Have you ever seen one marketed for use on a normal passenger vehicle? We're not really concerned with originality and I'm pretty sure the professor isn't either. I think it is more of getting something together and getting it to work type thing. Either way, at this point there's no turning back.

 

And they let you into engineering school?

I can't explain it over the internet, but I STRONGLY recommend finding a textbook or teacher/professor who will refresh your memory.

It is a triangle, but you're getting the relationship between side lengths wrong. That's where the trig comes in.
That would be the number I'd use.

 

Sin<angle> = opp/hyp
Cos<angle> = adj/hyp
Tan<angle> = opp/adj

You need to calculate the forces in the members for a few select conditions of interest; the forces are calculated by using statics and FBD's (free body diagrams). Then you need to look at the connections and brackets, such as at the frame (or wherever) and the pin C/T the hyd cyl and the link on the RH side. Gotta wonder how the frame rails like the side load that they'd only see in a vehicle collision, but it might be ok, but who knows for sure. I truly do hope you include both vertical and horizontal forces in your calcs at the appropriate joints. If any joints are bolted and use an array of fasteners forming a pattern, you must look at that joint as an eccentrically loaded bolt pattern, if this applies. With all forces known, axial stress = P/A, bending stress is Mc/I, and shear stress is V/A. *Hint: in short members bending stress is often negligeable and can be neglected in your calcs. From Mohrs circle you can calculate applied stress and then compare that against material allowables to determine your margin of safety. On the bolts, you compare the force (shear, tension, combined) against allowables provided by the fastener supplier. On members that are long and have a substantial load, you should run a buckling and/or crippling calc. Finally, knowing the force at the hyd cyl required to provide lift, you can compare that againt the max force that can be applied with your hyd cyl and pump setup. If you know someone with a solid understanding of structures, you can simplify your analysis by making sound decisions on load paths based on experience. I won't be offering that advice here. Don't forget that you want a good margin of safety to account for loading that is above and beyond the car itself, sitting as shown, and perhaps without the hand on the fender.

*The pistion diameter of that cylinder is supplied by the vendor, get the model number and try to look it up on their website.

**If you used welded connections anywhere, I hope you took into consideration load paths when deciding where to weld rather than just slapping beads down in random and hopefully good places.

 

Let me enlighten you on the next little *surprise* before it happens..............if you extend the piston of that hyd cyl much more than say, oh 50% of the stroke of that cyl, you have a member that has a rapidly diminishing ability to take a bending load. Ditto on it's ability to resist buckling. Beware of potential collapse and personal injury, both without warning.

 

who said anything about engineering school? Its an engineering program at a community college, I'm in my first semester and obviously don't have the knowledge to be building these things, but the professor wants us to just 'know' how to do something. Obviously we're trying to learn and get something out of this, I'm not sure why you would respond at all if you're going to have a condescending attitude.

There is a piece of angle bracing the frame and that is what the two pieces are attached to. I have a drawing from auto CAD if you have the program and are interested.

and you're saying that that cylinder is going to(possibly) collapse when fully extended?
now that you mention it, it does appear to be bowing in that second picture up top.

 

That was not my intent but instead I attempted and hoped to help you. Sorry I wasted my time trying to do so, but good luck on your endeavor.

 

Oh no, that wasn't directed at you cowboy. I appreciate the help you're providing and what everyone has contributed, I just don't understand the need for something like 'and they let you into engineering school?' Don't get me wrong.

 

No, it was directed at me, and I deserved it. Based on what you said the professor was asking for, I assumed (wrongly) that you were taking a somewhat advanced course, by which point you would certainly need to know the trig. If it's an intro course, I'm not sure how the prof. expects you to do the calculations, but that's beside the point. I jumped to conclusions, and I do apologize.