How to power a dump trailer?

[oilwell1415 563]

Judging from the pic it looks like there's about 6" from the hinge line to the deck of the bed.  That's also about the vertical distance between the cylinder mounts.  If the cylinder is 30" long from mount to mount the vertical component of the force will be 20% of the total, so 440 lbs.  There is also a horizontal component of the force that's equal to about 98% of the total, so 2156 lbs.  It is acting about 6" above the hinge line and generates a tilting moment of 1077 ft-lbs.  If the bed is 6' long the center is 3' from the hinge and the moment results in 360 lbs of force at that point.  360+440=800.

[Goofey 89]

What angle do you calculate on?

[oilwell1415 563]

No need for angles if you have the dimensions.

[Skipper 1,788]

Oilwell1415 is absolutely right, there is a rotational force, however the arm that the horisontal force is pushing on, is a function too, between the two lines between the 3 hinge points. To find the arm you go Tangent to the angle times the distance times the force horisontally, then divide by the distance to the center point. If we look at the pics, I guesstimate like Oilwell1415 (and we may very well be wrong) it to be a 6 foot trailer with the cylinder lift point at 2 feet from rear hinge point, and center mass at 3 feet. Now until we have measurements, we can only use OP's initial bet at an angle, and since that is 2.5 deg. the resulting upwards force from the rotational force will be very small. Again, when the angle gets bigger, so does this number, but until we have actual measurements, we can't really do much more. 

 

Just for the illustration, if we assume the 2.5 deg are right (which I doubt very much), this is the horisontal force's contribution.: Tan 2.5 deg = 0.04366094 x 2 feet x (sine 87.5 deg = .999 x 2200) divided by 3 feet. equals 63 lbs.

 

This again illustrate that it is the angle that determines how it should be constructed. The way it is right now may be sufficient, if the angle is indeed bigger than OP's initial guesstimate, just as Oilwell1415 too suggests.

 

Now back to contemplating how massive these forces are, and how they affect the frame work. We are talking 2200 lbs (1 ton) of push on hinges and crossbars etc, and that's only at 700 psi. Imagine going up to near full pressure on that little 2" cylinder. Say 2800 psi. That would give you 8800lbs or 4 metric tons of pressure. Please try to understand why I suggest that it might be prudent, strengthening that old frame up just a tad, if OP is right at the 2.5 deg, and he wants to lift, say just 1000lbs. That would take about 4.5 tons of cylinder pressure directly to the frame. 

Edited June 8, 2020 by Skipper

[Alex175 784]

14 hours ago, Skipper said:

Now until we have measurements, we can only use OP's initial bet at an angle

 

Yeah......I was wrong.....Like really wrong on my early guesses.

 

Here are the actual dimensions drawn out.  I did not include actual bed length in the photo, but it is 72".

 

16 hours ago, oilwell1415 said:

Judging from the pic it looks like there's about 6" from the hinge line to the deck of the bed.

 

You were perfect on that one.

 

Edited June 9, 2020 by Alex175

[Goofey 89]

What did I tell you. Them people are smart. 

[Goofey 89]

Strangely enough I find this math thing fun. I searched the net and used a online formula and calculator. Am I right if I say 10 degrees? 

[oilwell1415 563]

42 minutes ago, Goofey said:

Strangely enough I find this math thing fun. I searched the net and used a online formula and calculator. Am I right if I say 10 degrees? 

Yes, but you're just adding unnecessary steps.  In order to find the forces using angles you have to find the sine and cosine of the angle.  Finding those requires using the measurements, so why not just use the measurements in the first place and leave the trig on the shelf to use later?  If the total force in the cylinder is 2200lbs, the vertical force will be 2200*6/35.5 and the horizontal force will be 2200*35/35.5.

[Goofey 89]

Huh. Strange. I dont get it. It gives me sort of the right numbers I guess but I dont understand why. Tried calculating the total using the arm force thing Skipper talked about. Does about 750 lbs total tip power sound about right?

[oilwell1415 563]

It's in the ballpark.  I'll have to sit down when I've got more time and crunch the numbers.

[Skipper 1,788]

There are still some guestimates, as we don't have 100% solid numbers on the rear overhang etc, but we are close enough to shoot from the hip and not miss entirely. Based on what I see in the pics and the measurements given, I estimate that there is a 6" overhang on the rear. Now based on the type of load, that is likely not going to have much significance, so I do a trade off to compensate, and move center of mass 1" backward, and just simply skips calculating that overhang with its impossible to guess right load. Now based on those guess measures of the overhang, center mass will be at 29" forward from rear hingepoint. Cylinder lift point will be at 24" forward from rear hingepoint.

 

Now to clarify. Oilwells method of getting the forces to appear are quite legit. We do it in different ways because I used to teach this stuff, so I tend to pile on the methods and the "why things work the way the do" talk, just to spark a heureka moment in those interested. But in reality, it boils down to the same. The relations between angles and lengths in that triangle. So no matter the way, the result is the same.

 

Now a quick crunch with a few cut corners and rounded numbers, gives a total tipping force of well over 700 lbs at 700 psi. Lets call it 1 lbs per psi, and then there is a little margin in our favour to work with. So now we have to talk relieve valve. That valve determines the max pressure in the system, and the max is specified by Eaton to 700 psi. Some Eaton 1100 powered systems are set to 700, some to 650, and some have drifted and/or landed a bit outside those numbers too. So in reality, you can expect 600-700 psi, equal to 600-700 lbs. Then take the beds own weight out of it. Lets call that an even 100 lbs, just for the easiness of it. That means that on the current setup with a 2" cylinder the OP can expect to tip 500-600 lbs of load.  

 

Now it is up to the OP to give it some thought wether that is enough or not, and if not, then what route to go? More pressure meaning non onboard easy hydraulics, or a smaller (ish) depending on desired tip ability, redesign of the mechanism

Edited June 9, 2020 by Skipper

[Alex175 784]

So I'm figuring more out, and I feel like I am getting close, but I am also still lost on what I believe is the final step.  I think at 500-600 pounds, that would be on the small end/under what I would like.  What I would like to do before looking into other power systems, or altering the trailer, is to calculate the power I could get by increasing the bore of the cylinder instead.  I've calculated vertical and horizontal force using @oilwell1415's formulas, and I've attempted to calculate the actual force using @Skipper's formula of 

On 6/8/2020 at 6:08 PM, Skipper said:

Tan 2.5 deg = 0.04366094 x 2 feet x (sine 87.5 deg = .999 x 2200) divided by 3 feet. equals 63 lbs.

 

 

So using the information of a 10 degree angle for starting, and using the 24 inches (2ft) and 29 inches (2.42ft) instead of 2 and 3 feet respectively I am coming up with the formula 0.176 x 2 x (.999 x 2200) / 2.42 which comes out to ~315 lbs.  which skipper describes as the horizontal force contribution.  When you guys did the math do you take the Horizontal force contribution and add that to the vertical force to get the lifting power, or is there another step that I am missing(like a vertical force contribution)?  If that is the case my calculations are coming out to around 686 lbs, with you guys both coming in at over 700.  Below is a breakdown of the numbers that I have compiled, to maybe see if I messed up somewhere along the way.  Now I know how Alice felt when she jumped down the rabbit hole.

 

Component	Formula Sign	Value		Bore	Force	V Force	H Force	HF Contribution	Lift Ability?
Side A	A	6		2	2198.00	371.49	2167.04	314.91	686.41
Side B	B	35.5		2.5	3434.38	580.46	3386.00	492.05	1072.51
Side C	C	35		3	4945.50	835.86	4875.85	708.56	1544.41
PSI	PSI	700			Formula Used
Tan	T	0.176			(3.14*((BORE/2)^2)*PSI)	F x A / B	F x C / B	T x HP x (.999 x F) / CM	VF + HFC?
Sin	S	0.174
Cosine	C	0.985
Lenth(Hinge Point)	HP	2
Length(Cental Mass)	CM	2.42
Force	F	Column F (ironically)
Bore	BORE	Column E

[oilwell1415 563]

I'll give this another go now that we've got actual measurements.  The dotted lines show what you would have to do if you wanted to just use a single force and lever instead of two.  In this case I think it's easier to just do two moment calculations.  The two triangles are proportional, so it really just comes down to preference.  I have neglected any overhand at the rear of the trailer.  If you make the overhang 6" and still measure from a point 36" from the very back of the trailer the force (W) goes up to 607 lbs.

 

What we've got going on here is a summation of the moments about the hinge, which I've designated as point A.  The players in the game are the force of the cylinder, Ftotal, the vertical force, Fv, the horizontal force, Fh, the vertical lever length, Lv, the horizontal lever length, Lh, the load of the bed, W, and the load lever, Ll.  The system is assumed to be at rest and steady state, so the sum of all the forces is zero.  Fv and Fh are calculated using the formulas to the right and are proportional to the lengths of the triangle formed by the cylinder and the and its mountings.  Fv=372 and Fh=2169.  We multiply the horizontal force and vertical lever together and get 13,014 in-lbs and multiply the vertical force and horizontal lever together and get 5,208 in-lbs.  The total moment generate by the cylinder is the sum of those two numbers, 18,022 in-lbs.  Since we assume that the system is stationary, the sum of all moments must be zero.  This means the distance from the back of the bed to the center of the load, shown as Ll and equal to 36", multiplied by the actual load, W, must also equal 18,022 in-lbs.  So we have Wx36=18,022.  Divide both sides by 36" and you are left with 506 lbs.  Assuming I did the math right.

 

[Alex175 784]

1 hour ago, oilwell1415 said:

Assuming I did the math right.

The math is now making more sense to me(I really am starting to wish I had paid more attention in school), I do have two clarifying questions however as I think two of the pieces of information you used in the calculation may be incorrect.  Some of the info I included in the text posts and not in the picture so it could have easily been missed.

 

Starting with LH, the overall length from the rear hinge to the front pin of the cylinder is 59", pin to pin is 35", so therefor wouldn't LH be 24" instead of 14"?

 

Secondly, the trailer bed itself is 84 inches long, for LL which is the load level, is this looking at the center mass of the trailer bed?  If so I believe that number should then be at 42 instead of 36.

 

I know the drawing isn't to scale, so maybe I misinterpreted something so I just wanted to check.

 

And I really am grateful for all of the help, you guys have gone way out of your way to not only help me find the info, but teach me how to get there and I cannot express how appreciative I am.

Edited June 10, 2020 by Alex175

[oilwell1415 563]

You are correct that Lh should be 24"  That explains why the numbers didn't make sense.  I put the numbers in a spreadsheet initially and probably fat fingered that one.  The 1 and 2 are too close together. 

 

The bed length was just a guess.  I don't remember seeing where it was ever stated, just that we guessed it was about 6 feet.  I used half of that.  But  LL is really just an arbitrary number pulled from the sky to define the point where the center of mass of the load would be.  Since you can put the load anywhere on the bed you want to put it, this really doesn't matter.  If you have a heavier load, put it towards the back and it can still be lifted.

[Alex175 784]

On 6/9/2020 at 8:30 AM, Alex175 said:

Here are the actual dimensions drawn out.  I did not include actual bed length in the photo, but it is 72"

 

Just went back to look for it to make sure I wasn't crazy, but it wouldn't have mattered because I put the wrong info anyway.  I meant to put 84" but put 72", I always mess that up for some reason.

[Skipper 1,788]

6 hours ago, Alex175 said:

Now I know how Alice felt when she jumped down the rabbit hole.

 

Don't sweat it. Happy to help, and you are getting the hang of it I see. The devil is in the details, and there are quite a few to keep straight, but you are doing fine. I also had a few "aw heck" moments when i did the crunching. Had to put it on a little sketch too, to straighten it out.

 

I would like to suggest a few other alternatives for you to think about.

 

1. Relocating the bracket that holds the cylinder in the front end of the trailer, to give it more angle. Would require a bit of welding. More upwards force, but less rotational. You would have to try your way to find a good placement. Remember the cylinder should then swing downwards in its closed position, to its new location, so your 35" measure will be altered.

 

2. Using two 2" cylinders, equally offset from the middle. That would give the same force as a single 3", but you would have no problem fitting it due to the placement of the square tubing in the middle.

 

3. Combine the 2 cylinders option with lowered front brackets, to go even higher if needed.

 

Also take into consideration that using a 3" (or 2x2") will double your cycle times.

 

[oilwell1415 563]

I was turning numbers over in my head last night because something just didn't make sense and decided to confirm them on paper this morning.  You just need to put a 3" cylinder on it and be done with it.  That increases the available lifting force by 125%.  The down side of a larger cylinder is that it takes longer to extend.  A 3" cylinder has a volume of 169 cubic inches, which is just under 3/4 gallon.  If the pump on the tractor pumps 4 gpm the extension time on a 3" cylinder is only about 10 seconds.  If you did a 2" cylinder it would be under 5 seconds, which in my opinion is way too fast to maintain good control of the dumping process, especially early in the movement when the geometry is going to cause things to happen in a hurry.  If you really wanted to speed things up you could do some creative valving to include a regeneration circuit that's is manually chosen by you or you can include a pressure limiting valve on the retraction side that automatically kills the regeneration at a fixed line pressure to give you full capacity.  Either way, a 3" cylinder is the answer.  How complex you want to make the system after that is up to you. 

[Alex175 784]

Well, after some heavy thinking and multiple options, I decided that I want to try to go the simple route at first with a single 3" bore mounted to the existing mounts.  Numbers wise it should do more than I need it to, and so long as fitment isn't an issue I think right now it's the best option for me.  I ordered a cylinder yesterday from Northern Tool, which will be delivered early next week.  From there I'll run to my local NAPA, have some hoses made up, and we'll see what we can do.  I'm also working on the wheels this weekend, new bearings should be delivered tomorrow, right now I am trying to clean the wheels so I can paint them up before I put it all back together.  Only problem was the wheels were powder coated and they are an absolute nightmare to sandblast.

[Skipper 1,788]

That is obviously the easiest solution, if it fits where the 2" would go originally. It is just on top of the draw bar, which is why I suggested alternative placement, but as long as you did measure first, you should be fine. 

 

Keep us updated on how it turns out. Remember pictures or a video of it working. We really like that around here

Edited June 13, 2020 by Skipper

[wallfish 15,919]

18 hours ago, Alex175 said:

I decided that I want to try to go the simple route at first with a single 3" bore

Rod size and stroke length?

Something to consider, the fluid volume in and out of a double acting cylinder is not 1:1. The mass of the rod in only one side of the cylinder when it's retracted needs to be accounted for. The rod mass will create a "swing" in the fluid volume of the reservoir. Calculate the fluid volume of ALL the rods on the tractor. Will the trans reservoir handle that fluid volume change when all cylinders are extended and retracted? Will that swing create an under or over volume of fluid in the reservoir? Or, will it still be within the parameters of the recommended amount of fluid? Could too much or too little fluid in the trans reservoir damage it?

It's not that much and typically not an issue with a reservoir tank but since you're intending to use the tractor trans, it's probably worth looking into.

[Skipper 1,788]

Correct. That was the first thing i checked before recommending this solution. A 3" cylinder typically has a 1.5" rod ( or a 1.25" ), and OP said it was a stroke of 24". That's worst case 695cc of fluid displacement. Out of the 4700cc the tranny takes, that is completely acceptable. IMO. But you are absolutely correct, it need to be taken into consideration.

Edited June 13, 2020 by Skipper

[Alex175 784]

On 6/13/2020 at 6:32 AM, Skipper said:

Keep us updated on how it turns out

 

Looks like it's going to be a few more days, the cylinder was supposed to be delivered yesterday, and when I checked tracking this morning it said that it was.  Only problem was, FedEx didn't deliver it to me.  Gave them a call and they said a person with the initials MB signed for it, well I'm AJ, and my wife is SJ, so that's a no go.  So now I am stuck waiting while they try to locate the package...