I Thought This Was Obvious!

(1) depending on how far you wish to move the load, why use a dolly at all? Why not use an air lift (like a hover craft), that would make turns really easily?

(2) fill in the bigger holes on the rough terrain, that should be the real no brainer.

(3) if you insist on the dolly design, then yeah, go ahead and put 10,000 rated casters for the first try, and then when you break that one, increase the rating for the next one.

Minimum working load capacity of each caster 120000 pounds because up to 4G can be exerted on any one caster because of the potential upward thrusting effects (worst case) of the other three wheels from movement over uneven pavement.

Fast answer, no CR4 click considerations.

Rate the casters for half the working load. Lean on the manufacturers MOE.

Buy 6 casters and keep 2 as spares, (one will disappear from stores stock).

Test and monitor the platform, review and inspect after one week of operation, (depending on duty cycle).

Post pictures to CR4 and tell them to stuff it.

-A-

<...tell them to stuff it...>

Stuff what, where?

There is nothing worse than being in a room full of people who think they are the smartest person in the room.

_{(How many of them, I wonder, are actually Engineers?)}

-A-

Actually there is one thing worse: being in a room full of people, one of whom claims to know what all the others are thinking whilst seeming not to know which ones are Engineers.

I hope this thread is not yet closed. First, 30,000 pounds load should be big, I imagine a 14-ft x 40 feet footprint for this load, like a steel module being carried. Next, I am not sure if casters can support at least 20,000 pounds each, because that's my hunch. To answer this question, that's the most obvious thing I would check first, as an engineer. But as a student, just to have a numerical answer, I would go for 30,000*0.7 (to account for uneven load distribution on at least 2 wheels)*1.5 (impact factor), which gives me 31,500 pounds. But I myself can't be satisfied with this math. To truly answer the question, we should know also how high is the center of gravity from the base. That will affect the weight distribution when it moves and when it stops. Based on the velocity, we probably can solve the force due to it, and how it gets distributed to the wheels. At some point we need to know how the load gets loaded, and how it gets unloaded. Practically, in 5 seconds, without math, just imagining the possible scenario, I would say the casters should have a rating of at least 30,000. It could be less only when I'm really sure that the operation, including loading and unloading are pretty much controlled, and we know what dimensions we are talking about. On the 6th second, I'd say, is that available in the market? For me, this question has incomplete data, to get accurate answer. For example, where are the dimensions? We're not probably talking about a refrigerator being lifted here, are we?

As you can see this is not a closed thread. As you also astutely observed, people will often provide the forum insufficient information to properly answer a question. Often this happens because the Original Poster (OP) does not know what is critical to properly solve their dilemma. Occasionally the ego of the OP gets in the way of us getting the proper information.

You've added some very useful insight and questions here that I suspect will never be answered.

Welcome to the madhouse!

I'll second Redfred's welcome, but want to add one thing.

What I've often observed- and been guilty of!- on this forum is replies that try to delve deeper into the whole scenario rather than just focusing on what a defined problem is. Guess that's what you get for asking engineers questions! While that can help when the OP is unsure of things, in this case we had a clearly defined question that also delineated factors that weren't a concern such as dynamic loads- although some of that was clarified after the OP.

Fair enough, we need to think of this particular one as a theoretical question (like a test! Which is why your comment of a student made me post this) instead of a practical exercise. Continually looking for more data can sometimes divert us from the original point, if we don't make sure we focus on the specific question instead of everything we can see and do from the information provided.

So please don't take this as being pointed directly at you ! It's meant for a few more, and a bit of a reminder that things sometimes get lost or confused well down in a thread. And welcome to the zoo!!

P.S.- did you bring bananas?? or cream cakes???

"Ask an Engineer the time, and s/he'll tell you how to build a watch."

Got one for an Englishman to appreciate! Not engineer related, but definitely a bit of practicality in it.

A shepherd was driving an extremely large flock of sheep across a Welsh road one day, and happened to block off a flashy new motorcar that was puttering along the road. After waiting a few minutes for the sheep to cross, the driver evidently realized it was going to be a long time.

He put the car in park and hopped out, approached the shepherd and struck up a conversation (please insert appropriate accents yourself!).

"Nice day isn't it?"

The chat continued for a couple of minutes, mostly the shepherd grunting replies to the other chap's comments and queries. Then the car driver said, "You know, this is a HUGE group of animals but I could tell you exactly how many there are."

"Nah!!"

"Sure I can!"

"Nah!! Na way!!"

"Okay, well I like a little wager sometimes. Would you place a bet on that?"

"Na way you could tell, with them runnin' all around!"

"Well, tell you what. If you're so sure, how about if I can get the right number in less than 10 minutes you give me one?"

The shepherd thought for a minute, but this guy just didn't seem to be going away. Finally he shrugged, "Aright!"

The other guy whipped back to his car and pulled out a surveyor's transit, laptop computer and some other doodads. While the shepherd stood scratching his head this guy ran around doing various things, entering notes in his computer and muttering to himself. Finally he came over to the shepherd, just as the last few sheep were about to cross the road.

"Done! 8 minutes and 16 seconds! And you have 1,442- you had 1,441 before I started but one just gave birth to twins and another died of old age while I was counting."

The shepherd stood staring at him with his jaw hanging halfway to his ankles. Finally he shook his head.

"Wallll,... I giess you got yourself a sheep. Just pick one out...."

The car driver walked around for a couple more minutes, but kept coming back to one particular animal. "I'll take this one!"

He hauled it over to the car and threw all his equipment back in. As he did so the shepherd walked over.

"Naw then, fair's fair. How 'bout if I can tell what you do for a living you give that back?"

The fellow looked up in surprise. "Sure!"

"Wal, you're a consultant!"

"What! How did you know that? I don't have any signs or anything on the car, you didn't see my cards..."

"Naw then. Ye showed up where you're nawt wanted and nobuddy asked you. Ye told me all 'bout things I already knaw an' things I already knew was gonna happen. An' before ye got here ye did'n knaw nuthin' bout this an' ye still don't knaw more than what ye just found out."

As the fellow looked on in shock he continued,

"Naw then. How 'bout you jus' give my dog back!"

A couple in a hot air balloon got lost when a thick, heavy fog suddenly blanketed the area, making it impossible to see well enough to navigate. They'd been drifting through the featureless gray for several hours when suddenly they spotted the vague outline of a tall building looming ahead in the dimness. As the balloon cleared the building the couple saw a man standing on the rooftop, looking out from the railing.

"WHERE AM I?!" the balloonist shouted down to the man on the roof.

"YOU'RE IN A HOT AIR BALLOON!!" the man on the roof shouted back.

"THANK YOU!!!" shouted the balloonist in reply, then promptly manuevered the balloon to the staging area about a mile to the southwest, where he landed it.

"How did you know where you were just from that?!" the man's wife asked, incredulous. "Easy, Darl! When that man answered my question literally but without actually telling me anything that wasn't already painfully obvious in the extreme, I knew right away he worked for Microsoft and that was Microsoft's building and so I knew where I was, and given that I know which direction they are from here, here we are, safe and sound!" :-))

Tha ballonist would have been shot down as a competing cloud.

"...You've added some very useful insight and questions here that I suspect will never be answered"

You'll get your answers. Stay tuned. I'm a little busy now but will post something later this evening.

I'm still really pissed at Karmarat. His comment was uncalled for. I've have no issue with sarcasim, trash talk, and all the usuall banter that goes on but calling someone an asshole for no reason what so ever is wrong. I deserve an apollogy. If he thinks that's too much to ask than at least an explanation for the outburst.

I look forward to your answer, this is really an interesting and important topic. I have been tinkering over an idea of how to know exactly how much load each support carries, by using instrument or using spring attached to those supports and measuring the change in length due to loads. I actually have been doing lifting calculations for loads around your value, and to do these calculations I needed a lot of input, the change in any of the parameters would affect the final results of my calculations. The most important item affecting it is the eccentricity of center of gravity. I've done some physical experiments, and they showed that the worst case scenario, when the CG was too eccentric, all the loads would need to be supported in only one support. But designing for that worst scenario is often cost-prohibitive. I would like to know how different it is, when the loads are being carried from the ground, instead of getting lifted.

Since you made computations can you show how and in which situation the load is on ONLY one point ? Is it a new static you dveloped ? It could be for all of us of great interest!

Transient effect on a rough, pitted surface.

For instance: You are zipping merrily along at 2 MPH with your loaded-to-the-gills-dolly and then, for just a split second, three of your four wheels find themselves mid-air over potholes whilst the remaining wheel hits a pebble. SNAP!!! goes your underrated caster and Michelangelo's Pieta - the Real One, that is* - the one in that unmarked crate on your so-so dolly, the priceless artifact you didn't know you were towing, topples off your now-worthless, three-legged contraption and smashes to bits. Your next thought will not be about casters at all, but about killing the nitwit who put crap casters on that dolly you're towing. Trust me.

And finally, if you thought one caster was inconceivable, then what outta really bake your noodle is that, on a rough surface, there is a slight probability your load will be airborne on occasion - yes, airborne! - that is, is no wheels at all! Not a problem in itself; it's the landing part that hurts. It's those transient loads, the ones those hard, un-sprung casters will feel on impact, that have the bad habit of making insurance carriers, among several other folks, very, very unhappy.

And, so, as the old story goes: The Kingdom was lost for want of four, suitably-rated, suitably-designed casters.

-----

* An identical copy having been discreetly placed on display so that no more lunatics will try to smash the real one to bits with hammers - a job best left professional dolly designers.

I think Alberto was correctly stating that there can be similarities between supporting things from below and above. In either case, it is easy to imagine a dynamic load all being on one of four supports.

For example, the OP left many things unspecified in his rush to impress us with his ability to perceive the obvious. One of the more critical is the ratio between the CG circle size and frame size. Another critical element is the vertical location of the CG. So it is easy to imagine a condition under which, (with a high CG, and with a relatively small frame size/CG circle size) all the weight shows up on one wheel during a transient, such as a swerve and braking maneuver.

This sort of weight transfer happens too frequently with fork trucks, which have four wheels, but can capsize over a single wheel during a transient. Such has been the cause of many deaths and injuries.

The OP did not specify the material carried in the container. If it can shift (water being a good example of such a material) then the static CG circle looses much of its its significance for design purposes.

I would be glad to share it in the future when I have time, still so busy right now. It's not a new static. Perhaps it is just a simple application of the fact that ropes does not practically carry compressive load in lifting, it can only work in tension. Simple fact, but sometimes forgotten.

Actually I am using steel ropes, cables and or chains in my lifting calculations. To visualize the scenario, make an equilateral triangle, the bottom being a stiff object or bar, the other made up of strings or rope. The bottom stiff member represents my lifting bar, the strings represent the cables. The apex of the triangle formed is the lifting point. Now attach an object to the center of the lifting beam. You'll see that the loads are evenly distributed, and the lifting beam is horizontal. Now, move the load toward any edge of the beam, you'll see that the lifting beam is leaning towards the end closer to the applied load. The load distribution on the strings are uneven now. The one closer to the load gets more tension now than the other. The lifting "system" will automatically move and configure itself such that the lifting point always coincide with the center of gravity of lifted load.

Continue moving the load, you'll see that at some point, only one string is carrying the load. This string gets all the tension load now, the other string does not take any more tension load. When this happens, suddenly you will find, you don't have a triangle anymore, it's now just a straight vertical line! This is where accidents happen. If any of the ropes is not designed to carry all the loads, the whole thing could fail.

But we don't normally design for that worse scenario, instead we always try to keep the center of gravity of our loads within reasonable limits, in such a way that we can be assured that all ropes still act in tension, even if they are evenly loaded. That's why I said the center of gravity calculation is very important. I am still working on the question of how much eccentricity may be allowed. But people working on this same thing should pay close attention to it. The load doesn't need to be at one end of the lifting bar for this straight vertical line configuration to happen. This configuration happens when one rope loses any tension on it already, and stop its function in the system. This is not easy to visualize using paper and pen only... As you may realize also, this is not just a case of a frame or truss analysis that can easily be done by some software. Your geometry is very much dependent of the nature of your load. But the solution is simple, once you find your correct configuration.

I hope this fills your curiosity for the time being. Cheers!

-P.S. My take is that if you only have less that 5 seconds and money is not an issue... just get the rope that can take the full load to be lifted, or spend a day of calculations to get a more economical size. (Although I can probably do that in less than an hour, with help from a little homegrown program...)

I know the problem for loads up to 4000 Tones (metric) I was involved in the weighing of Off shore modules especially with the goal to determine as well weight as COG position for the design of the hanging "ropes". Their length was so computed that the hook was always almost on the vertical of the COG when the module was lifted and maintained an horizontal position.

As soon as the loads are important and the distribution of partial loads cannot be controlled the cables are specific and not all of same length.

I dare say that between lifting and carrying there is a little difference.

Yes, I agree, they are different, although we can see some similar principles involved. In offshore, we have seen (in the news) some very costly disasters caused perhaps by so much eccentricity of the center of gravity.

Worst case scenario would put the 32000# load 6" closer to two of the casters. Assuming 8000# per caster, the two casters would have to carry a load greater than (8000# x 2). I would guess a caster with a rating of at least 10000# each would be a safe choice.

Hopefully this thread has finally set you on the Path of Salvation, having pointed-out the Real Problem here, and one having nothing at all to do with casters, CoGs and ladder dollies: your thinking things are obvious. Ha!

Things are seldom as obvious as you might think!

This thread reminds me of poorly-thought-out math questions, in which the teacher has a particular "right" answer in mind, and cannot fathom that there may be other valid answers.

What is the next number in this sequence: 31415

Little Johnny practically pees in his pants as his had shoots up.

"Nine!!! It's the first several digits of Pi!!!"

Thoughtful Sally, however, says "Six. The sequence I have in mind is 31415161718, in which a 1 is stuck in between each digit in a counting sequence."

Suppose you were born on 6/28/73 (making you a young whipper snapper). What digit follows 62873? Obviously, 1. That is the digit that comes after 62873 when it shows up the first time in Pi.

I am anxiously awaiting TerraMan's return, when he will teach Tornado why the location of the center of gravity relative to frame size makes no difference. It is probably obvious to Tornado that, (we'll take this to the extreme to demonstrate the point) if the frame is small enough to put the CG directly over one wheel, then that wheel will be have to accept essentially all the load.

Two of my favorite parts of pi is when you reach the place where all the digits of pi repeat themselves - all of them - and then, at some point after that, you find an infinite string of 9s.

But my very favorite part is that all those irrational numbers - including pi - take up no space at all on the Real-Number line. None whatsoever. If you were to throw an infinite number of infinitely sharp darts at the Real Number line, you'd never hit an irrational number. Ever.

Insofar as that one-caster-bearing-the-whole-load situation arising, I did mention it early on (but we know these things never happen in Real Life, right? Maybe if sometime later the insurance adjuster tells 'im, 'ee might actually consider it a Real Possibility. 'Inevitable Possibility', more like. 'Future Reality', and with a really expensive, one-of-a-kind load. Hard Knocks 101. )

The rational numbers are only countably infinite; the irrationals are uncountably infinite (greater cardinality). Thus any dart is more likely to hit an irrational number (indeed, infinitely more likely).

I just had a great time poking into nooks and crannies of number theory (Hilbert's paradox, etc.) and stumbled into the terms veridical and falsidical, both of which were new to me. So now, so that I don't forget them, I'll have to try to work them into conversation... perhaps when I order my next Subway sandwich.

The combination of CR4, Wikipedia, and a little curiosity can be a fascinating time sink. Math can seem extraordinarily simple and straightforward: huge spreadsheets are just an extension of counting on fingers and being able to tell that you the same number on your left and right hands. But the stuff that seems as if it should be really simple and concrete, like the ratio between a circle's diameter and it's circumference, can get fairly arcane.

Thanks for the diversion.

For Pete's sake! Stop distracting me with the facts already! I don't know what I'm doing and I'm not letting you stop me with well-researched, verifiable facts. Nice try, anyway.

Well, take solace in the fact that an infinitely sharp dart will never hit the line let alone a number on the line.

With an infinitely thin dart, I want to know how you get it to stop when it hits the number line. And how does one hold one to throw it.

You can't check that box ex post facto, but you can later rate your own comment as OT.

I see your 10:1 safety ratio and raise you a full Caterpillar D11Ls as the primary platform.

Guaranteed to never tip over with a 30,000# payload regardless of terrain or load center of gravity offest and should the terrain become that difficult it can even plow its own path to follow!

It's all because there is no limit on how much of someone else's money can be spent on safety overkill you know.

You're welcome.

45,000 lbs should be enough.

So this could be anything and it seems I should obviously know that this the base frame of a very large shopping buggy. With that bit of knowledge I know that there exists nowhere a caster capable of handling the load that will be put in it, or at least any that will allow it to travel in a straight line. I will try, however, to ride it through a parking lot.

I haven't googled it yet, but I tend to believe what you said. Maybe that's the obvious thing to the OP? I hope he confirms it. :) Or there's some magic math or a rule of thumb about casters that he originally wanted to share. I've never designed a caster yet, nor anything that involved casters, though I see it everywhere on stuffs smaller than our current subject. Anybody here can give a good link for casters? Let's check capacities and availability...

Good synopsis and explanation.

(My 17,000 lb guesstimate was not calculated, but based on the COG circle being relatively small compared to the dolly frame, even though we didn't how small until now. Another factor was the coincidence of having used 17,000 lb wheels for a project. I forget whether the brand was Hercules or Colossus. My app was a 2-wheel scheme that had to be centered pretty closely to use a forklift on one end for lifting and traction of a ~30,000 lb brine chiller.)

Sorry I can't show you the entire dolly.

Oh come on, there must be a bit more you can tell ? If you don't tell, it'll just look like a fabrication of your mind. Not that I'm trying to goad you into telling stuff you shouldn't.

Best I can offer. Can't show you what's its used for!

Dang !

+ + = ?...

Pay close attention and you may see what your looking for.

lol - just a short while back I was reading an article on that very thing !

Nicely done explanation!

So are all of you just assuming that a caster rated for XX,XXX pounds breaks clean off at XX,XXX + 1 pounds?

Apparently it is also assumed that whoever is towing this rig is going to tow it at "2 mph" and stay "always" in the controlled environment. Well that ain't gonna happen, some knuckle head will run it into a curb or hole or whatever at a higher rate of speed. There will be an impact load. Heck, I saw a forklift carrying a GBU-2000 pound bomb run off a loading dock and I think they are supposed to be handled like real carefully. So, as to how much safety factor? It has to do with value of the cargo and the ramifications of a failure. Hauling chicken manure ain't the same as hauling a space shuttle.

I agree with you and all of your points should be considered in a design but...this thread has almost 160 posts debating only 2 factors to consider..a rigid frame and uneven ground. Can you imagine what would happen if I asked people to consider, frame twist, shock loads, abusive handling, CG height above ground, sudden deceleration/acceleration, caster type, etc

Keep in mind this was just a challenge for you to spot the isostatic system and the 2 caster sizing methodology. I could have just as easily shown a picture of a four legged table with a mug of coffee on it. I'm glad I didn't because I'm sure someone would have suggested what type of finish I should put on it in case the coffee spills when the table teeters on 2 legs!

I'd like to thank you, Sir. I think I've learned something today. I agree on your "theory" side. But of course, "deciding" really on the rating to actually use is a completely different matter. It's like after you analyze the forces on a steel beam, you actually have so many choices on which beam size to use, depending on so many compounding factors. So we may be looking at the same number, but still so many ways to treat it. Cheers!

"...But of course, "deciding" really on the rating to actually use is a completely different matter"

Agreed 100%. That's why the original question asked for "...minimum manufacturer's stated working load...". This is your starting point. The rest is up to you taking into account the other variables.

TerraMan, I think you did good, stirred the pot and freshen things up. I know you know the "what if's" or "if it can go wrong, it will go wrong", the stuff that makes designers lose sleep. Most of us are here just for the fun you've offered. So Thanks!

LOL! Yup, describes this place to a T!

Basic agreement, minimum assumption is 1/2 load per axle. But, you would choose as close to full load as possible per axle. (5 sec decision).

Real world #1 kicks in: what is available and at what cost (4 hour research and decision)?

Real world #2 kicks in: how long does this cart have to last and how often (distance) will it be used?

Real word #3 kicks in: what is the budget?

Real world #4 kicks in: how big is the safety risk personnel and contents if a wheel completely fails.

Real world #5 kicks in: how fast will it be towed and can you put a limit on it?

You need all of these assumptions 'decided' and so there while there is a technical calculations discussed by others, the more important aspect is the real world.

Where would a good design likely end up ~ 1.5x full load per axle....but that is just a guess!

Good luck!

Excellent work! GA

Europium also stated that it's possible for 1 wheel to see all the load when 3 wheels come of the ground. That's also incorrect! The overall COG would have to be directly over that 1 one wheel for this to be true. The green ball is the COG location. The dolly would have to do this:

Not very likely at 2 mph.

With 30,000 lbs and running at only 2 mph and without sudden change in speed, meaning acceleration effect is negligible (acceleration almost zero), I don't expect a disaster like what you illustrated.

But it it were lighter like a car and running 120 miles, and hitting a stiff wall, say about 3 feet, that would be disaster. Do you agree? The vertical load tends to stabilize, while acceleration tends to destabilize, and this is where the height of center of gravity plays is important.

I think I can just make a summation of moments around the front wheels. My overturning moment will be (mass * acceleration * height of the CG). My righting moment will be (weight * horizontal distance to the CG). If my overturning moment becomes higher than my righting moment, that means the the rear wheels will go up, the CG will move forward, and then you have your flying wheels.

Exactly! I'm surprised nobody suggested installing bars like this. Then we have nothing to worry about!

Thanks for the comment!

Every body looked at the presence of a hole and since nobody looked at the effect of a positive obstacle I thought to have a simplified look at it.

Again for simplification a considered that both wheels at the front meet the same obstacle profile at same time.

The mass will be rotated against above indicated axis.

If we consider F as the resulting force on one wheel (simple or double) its value will be :

F = 2/3*π^2*h*(V/lx)^2 with the Jy= M*L^2/3*(1+(H/L)^2)

Computations have been done for a 20' container in order to have a near to reality geometry. Results are in % of maximal force in the discussed configuration on 3 wheels and with the maximal eccentricity of 12".

Results show that a dynamic factor of about 10% at low speed should be considered.

Obstacle height has been set at ≈1" since such a dimension could be considered as small with respect to the others and disregarded as possible effect.

The form was approximated with a cos function since it was easier to compute and since all functions can be approximated with Fourier series The first harmonic was considered. One sees also the effect of lx which is very important.

Although a quite rough approximation this gives at least a realistic order of magnitude. The presence of elastic layers was neglected as well for the model, considering them will reduce the value.

Uhmm...you looking for a job or maybe freelance work on the side? I could use a guy like you. Haha! _{...seriously!}

Cheers.

I am a bit far but I never say no!

Thank you, nick name, for looking at (somewhat) what I threw onto the table in #63.

Since you can put numbers to this (I can't), I have a question...

You said <<Obstacle height has been set at ≈1" since such a dimension could be considered as small with respect to the others and disregarded as possible effect.>>.

My intuition says that the obstacle height applied to a 12" diameter wheel (with a small lx, which, I believe approaches a step-function) makes a significant contribution to overload. Did I misunderstand something here? What would be the overload if a step function (it looks to me like the lx of 0.05m might approximate a step function for wheels of this diamter, and more paractically the wheel hitting an uneven concrete pavement seam) were applied when approached at 0.1m/s, wheel diameter of 12", and height 1, 2, 3 inches, respectively?

Just asking....

I think that in a plant people are a bit more than a bit less attentive to such a situation. A seam of 2 to 3" is visible especially between two concrete plates. Even with a 12" wheel i doubt that nothing will be done. Do not forget that speed is low and people are around when such a load (15.000 kg) is moved from a place to an other and people look around.

In this approximation "h" goes at power 1 but the ratio (v/lx) goes at power 2 so that you can on yourself easy compute the amplification factors. For instance if h=0.05 m all values are for same speed multiplied by 2. For same "h" if speed is double and lx half values are multiplied 16!

In case of a step (true step) the wheel will come to the edge and "roll" over. The equation is a bit more complex and the accelerations depend on the ration h/R.

I shall have a look and give some values but I very much doubt that those are situations we should consider.

You're absolutely right: it never happens. Ever.

Nothing like one good experiment (and experience) to put to rest a thousand theories.

I gather you've never actually used a dolly in real life? Never actually used one on a rough or pitted floor? Never actually used a shopping cart on a rough, pitted pavement and had to struggle to keep it from toppling? Never sat in an office chair and tried to scoot across an uneven floor, only to find yourself face down on said floor as a result? All the equations and analyses in your Ideal world are no substitute for Experience and Common Sense in the Real one.

Go ahead, post all the equations you like, then argue your case to those casters in that pic above. Go on, explain those. You are the expert here, are you not?

Show wheel trajectories that terminate in the opposite side of a pothole and the acceleration curves that result from a head-on impact with that, all whilst your dolly is bearing a 30,000-pound load, and then show what happens to that. You're the graph expert and so it should be very convincing and a lot of fun. Go on! Do that analysis instead of ideal cases which merely support your argument. C'mon, be democratic in your analyses. Maybe even think about admitting a tiny bit of Reality in your scenarios. Better yet, put your money where your mouth is and insure those loads with your own funds. Make sure they're very costly loads, too, just to cut to the chase.

And finally, please give me a head's-up if you ever design any dollies, won't you? So that I will know to shop elsewhere? There's a dear.

Oh yes, don't forget that if your dolly is rated for 30,000-pound loads, some fool somewhere will drop a 40,000-pound load from his forklift onto your 30,000-pound-rated dolly and then tow it 11 MPH across his company's pitted, crap gravel car park to the warehouse, 100 yards away. He's got a hangover from yesterday's Superbowl party and he's in a hurry to get back home. Not only, that dolly is five years old now and this has happened before, following last year's Superbowl party, for instance. Not only, he does this all the time. Why? Because he's a fool and fools tend to be careless. Even better, lots of fools do stuff like this every single day, but you have to get out once in awhile to notice these things. Which reminds me, I forgot:

These things only happen in the Real World and therefore do not apply to your Ideal-World analyses.

Never mind.

"Oh yes, don't forget that if your dolly is rated for 30,000-pound loads, some fool somewhere will drop a 40,000-pound load from his forklift"

1. We design at 4:1 on yield.

2. Industrial casters already have a substantial safety factor built in.

3. "...he does this all the time". That's a management problem. Fire his ass after the first episode of drunken disorderly conduct handling heavy equipment and putting people's safety at risk! It costs less to get rid of fools and replace them with competent people than it does to over design everything!!

You're right. Never mind.

I sense a little

But wait a minute. From #66

"....but for me personally? I come here to get my fair share of abuse but nearly always leave disappointed..."

That was you, right? I take it you're now satisfied?

Who's the pansy now!

This has been a very good discussion for me, I am learning some new things here. And I agree with Mr. Nick Name's calculations, noting that they have all been for his particular problem only. So many people got involved in this discussion, and I read most of them, and I can see that no one was totally wrong. It's like that if we would err, we would err on the safe side, and that is a good sign. The main intention you had in your original post was just to show that the full load should not be divided by four, and I'd rather say that you've achieved that and much more beyond that.

As with much discussion on CR4, people love to debate the theotetical and practical answers. It would be a sad day if we ever all agreed. My post was clearly the best, and some day people will GA me for it. All part of the fun. Mrs K will defer to my expert knowlege and trust me to push the shopping trolly once a week - I'd humbly suggest that this makes me an expert .