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Obama's Flow Is Adequately Hydrated

IZwack

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1998 Ford Explorer
Not sure if this even deserves its own thread but I'm bored so I'm going to start this thread anyway for my Explorer's full hydro steer.

The all important hydraulic cylinder. 2" bore 8" stroke, double ended.
hydraulic_cyl.jpg



The CharLynn/Eaton steering valve.
steering_valve.jpg



Some hoses. 4,500 PSI rating, fire proof <-- ahah :confused:
hydraulic_hoses.jpg



I also have the "steering column" that bolts to the top of the steering valve.

I have the fittings that adapt the hoses to the steering cylinder. The other fittings needed I plan to get from the local hydraulics shop around here.

The 3/4" heims that are going to thread into the ends of the hydraulic steering are in the mail so hopefully in a few days I'll have em.

For cooling, I'm going to use what I'm already using for cooling the steering fluid -- which is a transmission cooler pulled from a 1st gen Explorer. I picked this cooler 'cauz its about half as thin as the 2nd gen Explorer's transmission cooler so it was easy to weld in front of the radiator support.


And since every project has to have a timeline, I plan to take this easy since I'm still mostly focusing on the Suzuki. So hopefully I'll have this installed and running in a month or so.

And since I'm a Senator, I know Brooklyn is going to ask where I got these parts from (read below)
 



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Which place did you get these parts from?
The cylinder, hoses and fittings to adapt the hose to the cylinder were from surpluscenter.com .

The steering valve was an E-bay auction.

The steering column was from a Pirate4x4 user.

Heims are from ExpertOffRoad.
 






Ah I almost forgot, I plan on using the stock 2nd gen Explorer engine-driven steering pump. If this doesnt work, I have two options:

1) I have a 12v electric 3000 PSI hydraulic pump I could use which would allow steering even if the engine died (which doesnt make too much sense since my Explorer has an automatic transmission)

2) I can adapt a larger volume hydraulic pump .. maybe from the larger F-series trucks or something.
 






I have that electric pump set up on my 88. It pulls a lot of power. You will need a second battery for this. Those pumps should only be used intermittently. They are very noisy. The brushes have to be replaced eventually. The brushes are similar to those which are used in a 12 volt winch motor. The solenoids go bad very quickly when the pump does a lot of cycling. I have a Dayton/Fenner pump from Grainger.
 






Ah gotcha, well that will be my last resort then.
I'll try to post up pics of the electric pump later just for reference.
 






Ah I almost forgot, I plan on using the stock 2nd gen Explorer engine-driven steering pump. If this doesnt work, I have two options:

1) I have a 12v electric 3000 PSI hydraulic pump I could use which would allow steering even if the engine died (which doesnt make too much sense since my Explorer has an automatic transmission)

2) I can adapt a larger volume hydraulic pump .. maybe from the larger F-series trucks or something.

3) Benchworks Steering, or something close to that. Power steering shop in Phoenix, AZ. They can hot rod the pump for you to match the specs of your ram. I talked to them on my last explorer when I wanted to do the pump to keep up better with the full hydro. I sold the truck before I could do it though.
 






3) Benchworks Steering, or something close to that. Power steering shop in Phoenix, AZ. They can hot rod the pump for you to match the specs of your ram. I talked to them on my last explorer when I wanted to do the pump to keep up better with the full hydro. I sold the truck before I could do it though.
Thanks for the info Michael :thumbsup: .


Unfortunately I dropped my digital camera tonight so AAAAHHH! Now I gotta borrow my brother's camera until I can replace mine.
 






Anyone know what fitting (SAE, JIC, etc..?) and size is on the 2nd gen power steering pump? SOHC engine if it makes any difference.

The reason is I'd like to just order a hose assembly online with the right fittings instead of going to the local hydraulics shop and paying twice as much to get a custom hose made.
 












At some point in the past I measured how much travel the steering link between the two knuckles does lock to lock and I think it was between 5 and 6 inches. So whatever that number is, I'm definitely going to have to limit the 8" stroke hydraulic cylinder or else the hi-steer arms are going to sheer studs.

So.. today I broke down the cylinder and did some measurements of its internals. What I plan to do is turn down some aluminum sleeves on a lathe to limit the travel.

hydraulic_cyl_open.jpg

From left to right: cylinder rod, cylinder body, home made spanner wrench, and finally one of the cylinder caps.
 






Tonight we're going to go over, very briefly, how much force we can get out of a hydraulic cylinder.

For now, we will focus on the type of cylinder I'm using which is a closed center double-ended cylinder - in other words, as one end of the rod is going into the cylinder's body, the other is pushing out at the same rate. Later on, we will breifly discuss a single-ended hydraulic cylinder and how it differs from a double-ended.

So on with the show!

Tonight's discussion will of course need a diagram so here you go
hydraulics.gif


D(b) = bore diameter
D(r) = rod diameter

Going back to ye ol physics book, we are reminded of the equation
Pressure = Force / Area
or simply put: P = F/A

Rewriting this equation to solve for Force (since thats what we're interested in), we get F = P*A .. or Force equals Pressure multiplied by the Area.

Pressure is a defined number given by the hydraulic pump -- which unfortunately we don't know the exact figure but we can guestimate :D. The only term we need then is Area. So of course Area = Pi * Radius^2 .. or.. in terms of diameters, Area = Pi * (Diameter/2)^2 . Plugging this Area equation back into the Pressure equation, we get

F = P * Pi * (Diameter/2)^2

Well from here we know everything so its time to plug in the numbers and finally see how much force we get out of this cylinder.

But wait, we forgot to take into consideration that the fluid does not push in the same direction on the cylinder's rod. The Force vector of the fluid pushing on the rod is 90 degrees from the Force vector of the fluid pushing on the bore. What this means is that the Force of the fluid pushing on the rod does NOTHING for us. So what we must do is subtract the cross-sectional area of the rod from the area of the area of the bore .. or A(b) - A(r) equals the new area, where A(b) equals the Area of the bore and A(r) equals the cross-sectional area of the rod.

So going back to the fundamental equation for Force:
F = P * A

We make a subsitution:
F = P * ( A(b) - A(r) )

And fill in the A's with the area equation in terms of Pi and the Diameters D(b) and D(r)

F = P * ( Pi * (D(b)/2)^2 - Pi * (D(r)/2)^2 )

Pulling out the common Pi to simplify the equation:

F = P * Pi * ( (D(b)/2)^2 - (D(r)/2)^2 )

Well then there we have it.

Now comes the part -- guestimating what pressure the hydraulic pump is rated at. I did a few minutes of searching online and I think most stock automobile pumps are around ~1,500 PSI.

With the cylinder I'm using, the bore is 2 inches and the rod is 1.25 inches (rather thick). So lets see

F = P * ( (D(b)/2)^2 - (D(r)/2)^2 )
F = 1500 PSI * 3.14 * ( (2/2)^2 inches - (1.25/2)^2 inches )
F = 1500 PSI * 3.14 * ( 1^2 inches - 0.625^2 inches)
F = 1500 PSI * 3.14 * ( 1 squared inches - 0.39 squared inches)
F = 1500 PSI * 3.14 * 0.61 squared inches

Subsituting pounds per square inch for "PSI":

F = 1500 pounds/squared inches * 3.14 * 0.61 squared inches

The "squared inches" of course cancels out leaving us with

F = 1500 pounds * 3.14 * 0.61
F = 2,873 pounds

So, assuming the hydraulic pump can flow enough gallons per minute at 1,500 PSI, then the hydraulic cylinder should be pushing just a little over 1 ton (not the "short ton", but 2,204 pounds).

So how does a single-ended cylinder differ? Well logically, since such a cylinder only has one rod, then the other side has a full bore area and therefore we wouldnt need to subtract the rod's cross-sectional area from the bore area -- which would give us more Force output.

Lets see what type of Force we would get if I was running a single-ended cylinder.

F = P * Pi * (Diameter/2)^2
F = 1500 PSI * 3.14 * (2/2)^2
F = 1500 PSI * 3.14 * 1^2
F = 1500 PSI * 3.14 * 1
F = 4,710 pounds

A difference of 1,837 pounds -- pretty significant!

Of course this extra 1,837 pounds of force is available only on extension when the hydraulic cylinder is extending. On compression, when the hydraulic cylinder is getting shorter, the amount of force is still 2,873 pounds because of the rod.
 






Wow!
 






yeah--he's a senator--he gots to be smart.:D
That sounds like almost enough to break a wheel off the hub if it didn't want to go there--
 






Yea enough to turn the wheel around the truck
 






yeah--he's a senator--he gots to be smart.:D
That sounds like almost enough to break a wheel off the hub if it didn't want to go there--
I'm not smart, I just play one on TV :D

But typically, if you dont limit the cylinder, either the hi-steer studs or the knuckle's ball joints will sheer off.
 






Since you are a senator and all, just don't drive drunk into a river and drown your girlfriend when you are finished like..........Teddy K! (No relation).
 






Can't be a real senator...he actually does what he says he's going to do.
 






What about trying something completely new. Use linear actuators with limit switches. Design a controller which uses a joystick, and make a drive by wire with a joystick instead of using a steering column. I once saw something like this in a magazine about 20 years ago. It was some concept prototype. I wonder what happened to the idea since then?
 



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That sounds interesting BrooklynBay -- but with a joystick, I don't think I'd be as precise in controlling my vehicle through very tight trails or when straddling a rock/log.
 






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