TRANSCRIPTIon

Jonah: I am Jonah Hobson, Marketing Manager at ITI, and welcome to the showcase webinar series. Today’s webinar, titled Winch, Drag or Roll? While that may sound a little bit like a 1980’s game show, what you’re in for is a few problem solving, rigging workshops. So you’re definitely in the right place. Keeping these aside, this webinar should be a lot of fun. I hope everybody received an email, containing a link to the 2 reference cards and a workbook that will be needed to follow along the presentation. If you registered late, you can still access this information by visiting ITI.COM/SHOWCASE-WEBINAR-MATERIALS. I apologize to those who have heard this introduction in the past, but for folks joining us for the first time, I’m going to give a very brief introduction of ITI before handing things over to Mike.

ITI is a training and consulting organization, specifically in crane rigging and lift management training courses at both client locations and at our three state of the art training centers in Woodland, Washington, Memphis Tennessee, and Edmonton Alberta. We currently have a presence in the United States, Canada, and Brazil. Trainers based in all three countries. We’ve done work in every continent except North Korea. ITI is lucky enough to work with a very broad industry base. As the nature of crane rigging activities go, it’s not limited to any industry, as you can see that on the slide here. We’re really proud to have worked with some of the best organizations in the world. The fact that the very successful companies you see on the slide there as well as others who aren’t listed, trust us with their rigging operations, it is something we’re really proud of and really reflects well on ITI. Many of your have been here before. This is the showcase webinar series- it’s a free monthly seminar that we put on to help educate and better the industry. You can see a number of our past showcase webinar topics on this slide. if anything there jumps out at you, all of those presentations are available to watch, the recordings as well as the files for download. So whenever you get a chance, something there interests you, like I said, you can access that by going to the showcase webinar website.

So our host today is ITI president, Technical Director, Mike Parnell. I’m sure many of you on the line are familiar with Mike’s work in the industry. He plays a very key role on two American Society of Engineers, or ASME, committees, and those committees set the standards for lifting, rigging, lift planning, and Mike is actually the chair of the ASME P30 committee, which sets the standards for lift planning and that should be released sometime in 2014. So one last thing before I turn it over to Mike, everyone has a questions pane on their dashboard, type into any question that come on during the webinar. Please enter your questions in there. I’ll field the questions at the end of the presentation, we’ll have a Q&A session and I can give those to Mike, and we’ll get answers for you.

So that’s enough out of me, I’ll hand it over to Mike. So whenever you’re ready, Mike, please take it away.

Mike: Okay, very good.  Thanks very much, Jonah, I appreciate it. I’m going to pop my screen up here, and let’s see if we get a good hand up. We got, let’s see - Jonah, you see a Winches and Blocks up there? I’ll start that, show my screen.

Jonah: Yep. You’re good.

Mike: We good there? Let’s see.

Jonah: Yes sir.

Mike: Okay, and let me make sure if we hit any bumps. Jonah, i’m going to put a new page up on the screen. Just, to let everybody know, you mentioned a P30, everybody seeing that?

Jonah: Yep.

Mike: Okay, excellent. The P30 is scheduled to release, it’s a new ASME document, there will actually be a short webinar on it. Mike Newell's is actually the Vice Chair, and then Kate Hyatt, is the secretary for P30 and B30. Kate will be joining us and let’s see, the release date, if you can please make a note of this, if all printing happens correctly, should be March 23rd, 2014. Then I believe we’re looking to make a webinar on that, a couple of nice piece webinar will be about, I believe it’s April 10th or 11th. We’re going to send out an announcement to everybody that’s on this program and we’re going to also then conduct a P30 lift planning workshop in Baltimore. That’s going to be in May. I believe it’s May 22nd. We’ll also send that out, it’s a 4 hour, integrating the P30 into the actual lift plan development project. That is also the week of the ASME B30 meeting in Baltimore. We’re inviting guest, we’ll send out announcement, registration and all the opportunities for everybody that’s online here and that are registered, we’ll make sure you have an opportunity, if you’d like to attend that. We’ll have most of the P30 committee will be there. It should be from 8 a.m. to 11 or so in the morning. We will get all the information up as where that is. It should be a terrific roll-out for it, and would love to see you there. It’d be a terrific opportunity to see how it engages the document into live planning project and then for, I believe we’re using the 77 sound compressor installation project will be married up in that.

I’d sure like to welcome everybody here. This is going to be a fast-paced workshop. People have been asking me, Mike - when are you going to get back on the air and do some interactive workshops? This is going to be one of those in 2014, we’ll have a couple more as we go along. We’ve had a number of guest presenters and we really appreciate your attendance and those really specialty subjects as Jonah has identified. On the line, as today, we have Advanced Crane Technologies, Randy Strane, Randy’s son is new to the industry but we look forward to having him work with us downstream on some activities. We have Al Perez, from the middle east, Jimmy Chang is here with us, I think, and J.D. Barnett, from BHP Billiton, and I think Jim Jacobe from BP is with us. We’ve got a bunch of folks from CNI, Chicago a core of engineers at Navy Crane Center have logged on with us. It’s nice to have you with us from JB Drivers Alberta, and Jim Warole, thank you for being with us on the lift presentation. Jim was a presenter, and we hope he’ll be a presenter with us in the future. He has a super background in crane and rigging  activities. Matt Work Cranes is with us, and Phelps is actually with us today too, a number of other folks from Saudia Electric, Sharpe Group in the U.S. here, I think we’re still batting about 40 different U.S. states, six provinces, and somewhere about 20 countries around the world, have logged onto this presentation. So this is going to be fun. We’re going to have a good time and get things rolling along.

Let’s open up, I hope you have downloaded the link that was provided by Jonah for the workshops that we’ve had. So we’ve had about a 6-7 page document, PDF document that came out to everybody when you registered and you got notification of it yesterday, and then also, you had some rigging cards. I’ll put that on the screen, we’re going to be working from a multi-panel rigging cart and I’ll put that up. Also, journeyman card, and we also have a master rigging card and we also have a master rigging card we sent, provided with the link. The PDF link document that we put up, this is the first one. It’s really about Winches, it’s keynoted as Winch, Drag or Roll? Some of the workshops that we’ll be going through today goes through those subjects. In this first example, we’re just going to look at the one from the left on your page, this should be on page 3 of your document. We’ll just focus on this one for right now and move on to the other one.

We’ll do a couple on static loads, friction resistant, because that means hold is perfect and nothing is hurting us from friction standpoint. We may toss in that, and like to go over a couple of things with Friction, built within the system. We may do that with it here, and sort of keep rolling in a static mode. So, here is the idea that the loads, in this case, the load is 6 tons, and we got a 5 part line and for those that are new to multi-part systems, what we want to be doing is, an easy way to do that, I’ll blow that up for a couple of minutes and go back. So what we do is send, put a nice blade going across the lines that enter or exit the load, how many lines we enter or exit as an actual flight to that load or to that direct point, that is the multi-part, lifting or pulling the load and that is, sorry about that, maybe I can get Jonathan in here a minute and help me out. The number of lines in this case has 5 parts of lines against the 6 tons of load. So the first question is, what is the load to the winch line? Let me back this out just a bit. so we have 5 parts reeved up, and we got the dead end right here, on top of that block, and if you chase those lines up, I’ll just display that for a second, so it’s up one, two, three, four, five down to our lead line, if you look up in a multi-line book, you’ll see a factor lead line stress attention, it’s always going to the winch. Or, in North Carolina, it’d be the winch, for my friends back in the east and south.

So we’re looking for the capacity, making sure the winch is going to be super, and going to work for us, and we’re going to be able to lift the load unimpeded. Let me get rid of this for a second, we got a number of parts on the line, so, let’s resolve the load to the winch line on mechanical 5, and we’ll give you a second to do that calculation and on the next page, we have a fill-in the blanks section. I’m working on a two PowerPoint slides per project, so one page will be illustration and the other is fill-in-the-blank for your questions. So let’s go back to the illustration, I’m going to depend on you all to keep the answers on that fill in the blank page there. I want to resolve the question, we’ve got six tons and divided by 5 parts of line, so we’re going to get 1.2 tons to the winch line, each line make a little baby notation over here, 1.2 tons on this line ,that one, this one going across, that one, and that one. So all those 1.2 tons gets us back to 6 tons. So, for static load, we have 1.2 ton as showing up at the winch, so that will be item 1B, is our 1.2 ton.

Then the question gets down to, what is the load at the main, or big beam right up here in the top? what is the weight imposed? and can we rig heavy to accommodate that? Then we get some kind of surprise question, then we’ll get this cleared off on my screen here and okay, Thank you very much. So we’re looking at the load, the downforce on the entire system, that is going to impose on that big beam up ahead. So that is the idea, so we want to evaluate, what it’s got? It’s sort of acting like a crane and this might be something very similar, if I was just able to  draw it in, that crane boom right over here, going up and then encompassing that block on a mobile crane, boom cylinders and outriggers, so on, so it’s sort of acting like a crane. What would be on the head of the crane for that load? So there, let’s resolve that because we set the wind system up and what we’ve got for downforce there. So we know we’ve got 6 tons, we know we’ve got 1.2 at least, over here on that winch line. We need to add that together and our answer should be 7.2 tons right up here at the big beam. So now I’ll get some other information here real quick on dynamic load. Just for grins, I’m going to ask an off the wall question here, what is the shackle, what size shackle do we need to put right in there to support that block? There’s one up, one down, both with load, so what size shackle would it be?

Let’s open up that journeyman card. It says Journeyman Rigger’s Reference Card right at panel one, let’s go down to panel 5. On panel 5, we got a whole host of rigging hardware option there. You’ll notice the size column on the left, then we got Eye Bolt, Turnbuckle, Master Link, Shackles, and so on.  Look down the shackle column, what we had 7.2 tons, was the load times 2,000 and that’s going to give us 14,400 pounds. So let’s find the first shackle that gives us that 14000 and we’re right there at 1 inch. So we’re going to need a couple of 1 inch shackles, one up and one down that will connect that upper block to that big beam up there. i think that probably it will sustain us even for dynamic activities. So I’m going to swipe that away. So we got the shackle there. We were in a bigger workshop, we’d be working our way through identifying all the rigging components, all the lock sizes and so on to secure that. So we got 1.2 tons to the winch, 7.2 tons up there at the top end, and for those, we are just going to do this one time, for those interested in understanding the computing dynamics going around here, I’ll just do a little off the side review here, so i’m going to put the load on the first part of line. Here’s the deal, if you’ve got nothing else to go on, you don’t have legal line stress table, which can do some of these computations faster, there is one in the writer user’s manual. There’s one in the boil maker’s handbook, there’s one in the IBT, redhand book, there’s a lot of resources for lead line stress, jump block’s has one, and so on. what we’re really trying to find out is, lead line stress, tension to the lead line can increase as the friction increases at every shiv, every time a rope drives around the shiv and turn it, it’s going to get some additional resistance.

In the simplest way, it’s the most conservative approach, is to start with our initial 1.2 tons, which is our dead end, that is our single line coming up here, dead end goes over that first shiv, and as it turns that first shiv, we’re not going to, there’s no additional friction on that single part, so that’s basically times 1. on the next one, on part 2, and we’re going to say part 2, 1.2 times 1.04, and what I’m using here is a 4 percent friction for each shiv as i turn, so 4 percent, take 1.2 times 1.04 and we’re going to get 1.248. That is our second part of line, new tension. It’s really compound interest. For each part of line, we’re going to account for that extra pound of friction, we’re going to add every new part of line all the way to the winch. So we take that 1.248 times 1.04 for that third part of line, and we’re going to get 1.298, so that’s P3. P4 is 1.350 as Part 4, I’m going to write all over paper. 1.404 is part 5, and down to the winch line, I’m going to do that one more time, 1.460 is WL, winch line - we’ll call it that. So when you add the values up, really, while the thing is turning and hoisting, I’m going to identify that that’s really my load to the big beam. So real, instead of 7.2, we might get 8 and a half ton. We could be up to, I think I added it all up and got a total load of 7.92 ton, instead of a 7.2 ton dynamic. So, we’re still good for a 1.8 ton shackle, but just realize, dynamically, we’re going to be creeping up in the overall anchorage rating. Certainly, overall winch load happens down here at the tugger, winch. So every time we’re going to do that, there’s dynamic static not moving, dynamic is moving, and as we continue to impose and hoist, we’re going to see that dynamic number go up. So 7.96 ton dynamic, and 7.2 static.

Alright, so, the rest of the workshop we’re going to deal with, will this be static? Let’s go over here to number 2. The right side, so, let’s take a look, we’ve got a 4 ton load, and we’re supporting it by two different support points up above. One is a low beam, one is a high beam, for obvious reasons. We want to know what we’re going to get in a variety of locations, but first one on your page, page 3 is, what loading at the winch? So let’s see, we’re going to use the same process to work through it. How many arms are lifting that 4 ton load, and if we put that across there and bisect all those hoisting lines, we can find that we have a mechanical advantage or in this case, so we can account then, that each of part of line, in static condition, has 1 ton. So sometimes when I’m doing a quick estimation back that one up, I’m just going to do 1, 1, 1, 1, and then guess what, 1 right there. That 1 right there, on the winch line, has got to pull the other right behind it. So yes, it’s going to be right because of the dynamics but we’re getting the whole general gist here. So we have 1 ton plus friction, I’ll just put a plus F right there for the winch line, then our next question, I’ll get some things out of there, getting kind of our way for a second.

The next one is, what is the loading at the high beam? right up here? So if you’ll just mark your page, those values a part of line, the key is to arrive at the approximate value up at that high beam. I had 1 ton on that line, and 1 ton on that line, so really, that is 2 tons on the high beam. What is the load value for load beam? We just follow it up, there’s that one, this one, and this one, so there’s three of those, three tons up on that load beam. So I go to my answer page, go to my other screen, most of you have your fill-in-the-blank page, so we just update this. This one was B, this one was A, 7.2, this one was C, Charlie, this one is on the high beam, 2 tons, and this one on the low beam is C, 3 ton. Excellent, okay. Great job. We’re just going to keep rolling here. We’ve got another workshop, this is sort of interactive. If you’re working ahead or working together, all good, whatever works for you works for me. So, on the next one on page 4, have 4 different example of loading moving: skidding, and in that regards, we’ll just pull on this puppy sideways, so this way winch, drag and roll approach for dragging. It does happen, we just got to skid it, slide it, move it, but we’ve got different values that are going to throw us high and low for the force required to move those loads to surfaces. So what I’m going to ask you to do is take the most advanced part you can get in front of you now, listed in the table as mark C, that’s Master Rigger’s Reference Card, and I’d like for you to go to Panel 6, and that’s the 6th page, 6th panel, let’s take a look at this guy. At the bottom end right there, you’ll notice we’ll be using this formula later on, but this is a series of coefficient or friction that is listed down below in the bottom part of the panel. Notice we got writing material now, when we put these on the card, I want you to know - for the engineers on the line with us here, these are really, let’s take one for example, and you’ll see that it’s got a value of 50 percent. So in a strength materials engineering book, a number of tests have been done, what we’ll see in some of those books is wood on wood, and we have a range, the range may be from 70 percent to 30 percent depending on whether it’s sanded or not, sanded, rough, we may have hard wood, soft wood, wet, all kinds of conditions that can enter into that. The average, the breakaway average for the two materials on that surface, wood surface and wood block on it, the average breakaway will occur, and it will be on a slower grade, between 30 percent grade and 70 percent grade. Really equates to the value we use for the coefficient friction. At what percent will this surface, this wooden block surface down here, breakaway, from the subsurface. So what we’ve done is we’ve taken the average that we’ve all known from books and testing done ourselves, and what we’ve found, we’ve taken the average number 50 percent, and that’s the number you end up with on your card. So I’m going to card back up and note the average number that you see on that card is 50 percent, and that is the midrange number that comes from a variety of tests, some will be higher and some will be lower, that gives us a starting point. We all see that concrete on concrete is pretty high at 65 percent, and it is back from 80 percent, somewhere down to 50 percent, or 55 percent. The value we have is the average of those numbers, just as the byline here, we don’t put it on the card, but I can tell you from aluminum to aluminum is somewhere in the 1.50 range, it is extremely high, yet two aluminum surface, put them back in the back of a truck, get an aluminum gear box in the back of a truck, it takes more to slide it than it actually weighs by 50 percent. It’s huge. The atomic bonding between the two is huge and very high friction. when I put it on, you don’t see a lot of that happening, and you slide it high enough, you’re going to melt and you’re going to weld it together. Didn’t think it was much further labeling that and listing that on the card. It gives you an idea, that it’s just a huge value. The atomic bonding is alert and working right there.

So we’re going to use those values, and noted, up in the right hand corner lifting and dragging the level surface, on the legend, what we’re instructed to do to take the C up, CoF - Coefficient of Friction, times the weight equals the force required to move that load. So we’re going to use that formula to solve that page 4, in our book, handout. Remember, I’ll just put the formula back here: CF x W = Force to move that load. Let’s find the value for coefficient of friction for each one of these A, B, C and D and let’s apply that formulate to the 4 cases that are on the screen. You got page 4 in front of you and work that out already, we’re going to work right behind you here. So we’ve got a value, our load is constant, but with different materials underneath it and different surface to bear against. In the first case, we’ve got the load’s 12,000 pounds, W is going to be constant throughout workshop portion, so the secret here is concrete to concrete and that is going to be 65 percent times 12,000 and we’re going to a value of approximately 7800 pounds for that first example, so that will A, 7800 pounds. The next one we’ve got wood on steel, that one will be .30 times 12000 and that’s going to give us 3600 approximate force move that load across that floor. Let’s do that two down below, a lot of you are working on that already, moving with me. The next one is still, very light coefficient of friction, 10 percent times 12000 and that’s going to be 1200 pounds, horizontal force more or less to get that one. The last one, we got wood on wood, remember that on our first introduction, so we’ve got a 50 percent coefficient of friction times 12000 and we’re about 6000 pounds to move that way. It might cost, really 55000, no problem breakaway might be higher, that’s ok too, kind of figure you might get 10 to 15 percent higher, or 20 percent extra for breakaway just sliding it goes easier. Reduce that friction over time by putting lighter lower atomic bond type materials in there, UHMW, grease, soap works well on surface, so there’s a whole lot of ways to lower the friction, do you want the friction to work for you? So increasing and decreasing is always a big help. Let’s take a look and go forward. we’re going to combine two workshops into another workshop, before that, let’s transfer our numbers into our page here. So we’ve got, I just want to make sure we’re dotting the i’s and crossing the lines, so we’ve got 7800 there, 3600 there, and 1200 here, and we’ve got 6000 here. So here are the answers for those that are following along. Next page, I’ve got page 5, the handout in front of you, well, we’ve got a complicated kind of system here.

This isn’t unusual in a type of Ford plant, where you’ve got a winch set up, one location, and that winch is dedicated to that location. From there, they pick up the variety of locks. We’ve got them just rigging around columns right now but 4 blocks, sealing blocks, each line, within 2 or 3 adjoining rooms there, from their system transfer to plight transfer out and so on, equipment change out. They’ll use one winch as a central location and work it in 3 or 4 rooms by re-routing the different diversion blocks. In all those points that we’ve pre-rigged, put a block in, say I want to put a block there, we need to check with the building engineer to see that column, that beam, whatever we’re going to anchor to, sustainability’s robust enough to take the loading that we’re going to see when we start moving that object. In this case, we’re looking down on this load and it’s a 100 tons press, it’s right down here, we’re moving this press forward, I’ll just make some initial drawings here. So we’re moving it forward on rollers to that little x box there, box squared. That’s the idea. We know the load weights, the load itself weighs 100 lbs, so let’s get back to that rigging card, the Master Rigging Card would be very helpful. Notice you got that load on wheels, so we’ll take a look at that, load on wheels here on that right side, and I’ll kind of highlight that, get rid of this other stuff. Highlight that, and I have something to share with you, so we got a load on here and it’s about 5 percent, we can set up some rollers, that and help move that a little lighter. Going to bring here, take 5 with me, we certainly do a lot of jacking and rolling instruction, I just pulled some shots up here really quickly, three little pictures. We have an incline at all of our training centers. Bringing the load right up the incline, and then up onto a flat surface. So we have an incline plain calculation, winchers, tuggers, bong. We want to show this is a 3 point system right here, and that is a steer man, that’s just a trades name, steering in and then we’ve got two ends. We want to share with you as well, there is a, we’ve got, for our .5 here, let me take a look at that with you, don’t worry about the transition. Take a look, when we’re moving loads sideways, we’ve got to be real careful. I’m not going to draw the buttons, we’ve got a lot going on the incline here, load weight center gravity what subsurface dictates system we’re going to use, is it incline or level, what’s the floor surface look like? Wheels, pipe, skid or air casters, we got a new hydra system up, 400 ton unit up there we’re using for Master Rigger courses. Down here in page, we got pipes we actually will use, get back to caveman rigging, just to help teach and in great shape, the principles, participants, so we use wood rollers, steel pipe, steel heel man rollers, multi-ton steel and synthetic wheels, air casters and so on. So we want to always take these things slow. Just watch out for where center of gravity is first thing, make sure that our support themes are in front and behind, and by the side, and we’re supporting there, just humming through just something to share with you.

When we’re jacking loads, by the way, two one end or one side at a time, never jack a load on four corners at the same time, really dangerous practice. We can helicopter that thing down and hurt or crash and flood somebody bad. We chase blocking up and down and don’t get more than 4 percent grade out of level. We only want to sideload the jack by getting too aggressive and trying to raising one end too high or lower too much, we want to chase that up and down the blocking. This guy is using jacks here. Synchronize and keep clear and sure that the jack and CG location is good to prevent overload to either the roller side system we’re about the place for the jacks. This is that three point system that we can use, and we have a number of sets of these, I think this one is the 5 ton unit. we got a 25 ton unit I think we’re trying to configure. Notice we got steering at one end pops, on the other end. On the right hand picture, we’ve got the parts cross to get them squared and true, they only go straight, but the steering can make it steer and literally drive this around from the steering end, pivoting one corner right here to make that turn. So, to really tune the system, you need to make sure you’re not butting or going into a workshop on this moving load on wheels, is make sure that center of gravity is contained within that triangle of support. So, craning and pop in to that south and north and far east side there you got the steering unit, and then, if the center of gravity was further to the right, that’d be pretty curious, the stability could be an issue, in that case, we’d want to flip the whole system around. So you want to have that center of gravity, midline or closer to your double pop-end down there. Keep it as far away from your tipping access as possible, it’s reasonable achievable. Watch out for the height of the center of gravity, the axes, x, y, and z. We always want to make sure vertically, that the center of gravity is contained and stable. So we have to put our footprint out. Put out underneath the load, you might say, put some beam extensions up that create the better stability. This really is working on the same principle as a pallet jack. Nothing new.

The three point rule is a great system versus four point. We can go into that later in a workshop webinar. Here, you got a single down there that’s, you got a pop there and a pop there, and we’re containing the center of gravity. It really is the same thing. Not new thinking, just applying it in different ways. Let’s go back to, by the way, we do have an air caster system, in this case, we’ve got 4 air casters and because we can actually adjust their elevation by adding pressure, then you can use force on a load like that. There’s two on the front, the back, these guys are bringing a load back into the building, and they’re actually having to go over obstructions. Pretty challenging move there for getting it done. We’ve got a variety of ways to move loads. I’ll close this out here, and say no to that.

So we’re going to load on wheels, and we said about 5 percent. On synthetic wheels, we are experiencing about 2 and 1/2, somewhere to 3 percent friction friction. Nice big wheels, standards wheels, nothing to hurt you on 3 to 5 percent. That’s what we’re going to use for the workshop. So we’re going to our page 5 there on the book, and the first thing we want to do is arrive at, the dead weight load is 100 tons, and what will be the force to move that 100 ton wheel? So using our previous workshop right across the screen here, if you remember, we’ll take a coefficient of friction times CW equals force, so our coefficient of friction is .05 times the load of 100 ton, is going to give us 5 tons to move that load to the north. We’ll get rid of that in a minute. We got 5 tons anticipated to get that pump moving. Then let’s see, we got 5 tons, and that’s my magic number there. My answer slide, which everybody is using to record our information here, so our first answer weighs 100 tons, sitting just on rollers, what is the anticipated force? That force is going to be 5, B. We got a variety of questions, various connection point, call and winch to see which is that for the anticipated load. Now let’s go back to our drawing. I’ll put that screen here. Hope everybody downloaded their link. Otherwise, we’re working off my screen which is a poor substitute, so hope you got your workshop and your cards with you.

What we want to do, remember what we did last time, stop trying to find these values, we’re going to be searching for loads that call up TSRQ, and what is the load of the winch, North Carolina Version. Want to see what we have for loads. Remember we just talked about cutting the hoist lines, which lines are entering or exiting the load, and in this case, there are four of them. So we’re going to take that 5 tons, divide by 4 parts of line, and we’re going to find what? Our line fold is going to be 1.25 tons per part of line. So I just do this. One, two, three, so each one of those has got 1.25 and I’m just going to write 1.25 tons back here at the winch. Now, for those of you that were chasing the dynamic value, I did a little pre work. Just add 4 percent to every part of line after the first 4, and that 1.25, just put an S there, put down a D number for dynamic, and I’m going to arrive at 1.65 tons. We won’t do any more of that, so just to let you know. So, dynamic, all those shivs, breakaway is not really counted into that, but add the 1.65 ton on to the winch. 2 Ton winch minimum is certainly advisable in this case and higher would be better. So 1 and a quarter ton will be our working number, we’ll use for the balance of this presentation. Let me get rid of a couple things here, alright, excellent.

So our first question. If we’re going to use 1.25 tons, the weight of - question set up on the book, we’re working backwards from the winch, back to, so that winch is 1.25 tons, and it kind of goes backwards. QRST, we’re going to work back to the load. So those are the two things we’re going to act as the discovery points we’re going to use. We’re going to go right back to Q, and I’ve got 135 degree included angle right there. What does that do to the load in that column, and that block, and sling and shackle that are  in play? So, it’s noted in our head, 135 times degrees. I’ll take you to a card, the Journeyman Rigging card, JRRC, let’s go there and let’s go panel 4. There’s an example there on panel 4, and we just stuck in here for educational purposes to put in line of 6000, load is 1000, and then gave a variety of block setups at 180 degrees, 150, 120, 90, 60, and 0. So we set this example up to the right, example certainly follows right over here, line pull and follows to the block loading. Special notice here, I want you to take a look with me, the formula. The formula says we want to take the block here, times the line pull, for the block loading. The block loading is what we use the backup slings to restrain or hold that block in place at column. Now you’ll notice, so let’s work this one at 90 degrees. I know I have one of these in the workshop, so at this 90 degrees setup here, what we’re told to do is take the block setup factor, so draw right across here to 90 degrees, 1.41, in this case, 6000 is going to result in a block of 8460 block factor times Line pull equal block line. We’re not adding dynamic winch line pulling right now, adding just to get us there. So now I want to get to, everybody please switch cards with me. I appreciate everybody being so easygoing with me. I don’t hear any screaming, so that’s good.

Let’s go to the MRRC,  Master Rigger’s Reference Card, panel 1, reduce this just a little bit. I’m right on panel 1/2, you’ll notice that you have a whole bunch of numbers in the table, a whole bunch of columns. I want you to do is, go over here to the full included column over here, you’ll see that whole included column goes from zero to 180, and it goes in 10 degree increments. The column right next to it is the block load factor column, and those two columns work together. We pack these, these are rigging cards used for almost all training programs. What we have is these two are the values from that block, and that block value, let me get it highlighted here, that block value of Q is 135 degrees, that’s the included angle, so we’re going to search that card out, and see where that puppy comes from. Betwee here, I got a 130 option and a 140 option. My diagram shows me 135. Just like operating cranes, we always go to the one, in that case, produce the lowest gross capacity for a crane, we would find that on a chart. In this case in rigging, same concept is, which would produce the highest block load? So at 130 degrees, I see a block load factor of .84, and for 140, a block factor of .68. So between those two, which one is going to produce the higher block load result for me? Just using those two as the option. If we pick .84, that’d be right. We'd use the higher one. We know it’s a little bit higher than the one at 135, we’re not going to delay into this thing, but we’ll just pick the .84 and we’re at solving our problem.

Right here, I’m going to get rid of a lot of scratch off now. We know our values, 125, let me get rid of that, I appreciate everybody’s patience - I’m just an old man having it here. So, alright, got a .84 times line pull, so that’s block factor line pull, lp, so we’re going to put in the value for line pull right there is 1.25, and that’s going to give us what? It’s going to be lowered, isn’t it? So that’s going to be 1.04 tons. So as we open that angle up, we’re going to get less and less load to the block. If that were running right through the block, it’d be 0. There’s no load to the block. It’s fairly just kind of doing nothing. So as we make that angle 120 degrees, that’s going to become, 120 would be 1, as it get closer and closer back, it gets back degrees, then our load factor block factor is going to be 2. We all know that, you guys saw that on the card. So got a value right there of 1.05 ton is the answer for Q. 1.05 should be Q. Let me give you a minute here, let’s work out what it’s going to be at here, S, and then T. So I’m going to quietly work my way over to the card again and see what’s my next angle.

45 degrees, the line comes in and out, 45 degrees pulling that angle there. I’m going to quietly go over to my card, take a look and see what do I have in the 45 degree department, and let’s see what we got. We’re at 45, we got 40 there and 50 there, so I can see we got 1.87 and 1.1, so I’m going to use what? the lower one or the higher one? Excellent choice, we’re going to use the higher one. Excellent. We’re going to use 1.87 S block factor times line pull of 1.25, and we’re going to put that value down here as 2.34 tons on R. Okay? So 2.34 tons on R. Okay, Excellent. This one you’ll know in your sleep, because we use that for a whole lot of rigging stuff. Next is S. We're going to go find the factor at 90 degrees, so go down here, 90 degrees, our block factor is 1.41, we’ll use that. Solve this puppy over here, so my block factor is 1.41 times line pull of 1.25 and my load is going to be 1.76 tons. When I find that block loading, I need to get the sling that will go over and protect the sling, I got to get approval from the building engineer. Can I put down here, put 2 and half tons on that column? Can I put almost 2 tons on that column pulling on it? Will my holds work? So, it’s a whole lot more than just calculating, we got to confirm the building still stands up after we’re done with it, since we’re going to put so much load. We’re going to find, we’re going to put an awesome amount of load up over column T over there because we got a whole bunch of lines and force going in, pulling out, from that column, not all of them are the same direction. So, my goodness, we’re going to induce a ton of pull on that thing.

Now we got Q, R, and S, now we’re going to work on T. I’m going to wipe out all of this stuff out and get to that page up. Let me up this page a little and get it into view. Sorry about that. Alright, we’re going to work on T right here. A simple way to look at this, and I say simple as the only way, I start out like this. I know I got three straightaway pulling lines, 1,2,3 from right to left that are going to show up at 1.25 each. Now, this fourth line, this guy is going to go up, go into that line, and go to our left. And he’s working at a 90 degree block, so that’s going to result in something unique, pulling that way on that column. Its motor is pulling south direction, but that’s still a load. It’s still a force pulling on that column, all that rigging going around we’re just going to make sure it’s accountable for it, and can we get it to the building engineer to get his approval? So we remember, we’re going to use the block factor times line pull equals block load, that’s our formula. So in this case, what was the block factor for 90, excellent. That was 1.41 times lin pull of 1.25 and that gave us a block load of 1.76, so I’m going to take this puppy and add them over here, to these three. At the end of the day, those combined are going to make 5.51 tons of force, pulling against that column. What we all appreciate is, it’s not all in the same direction. But, we have slightly over 5 tons, we can sit down and arrive at the exact final vector for that column, it may be down in that column, in that range, the final vector is essentially this guy. We want to take this number one, and can we rig up accordingly? Can we get the right sizes and shackle, can we get the right dead end connection right sling? Can we rig the slings up to it, protect the slings on end account for the block? the final load to that column? and get the sign-off from the engineer we got that accounted for? So that’s the idea. My goodness, we’re piling a bunch of it, so it wouldn’t be a big shock when we step back to the big picture here for a second. Let’s put this into view here.

Shouldn’t be a big shock. Got a whole bunch of our initial project was to pull this puppy up and forward, we know that already, 5 tons rolling force forced to move that. 5 tons to move it. So if we end up at 5 tons load, I’d say we got off pretty lucky. You know, about the worst thing we can do for this would be the bad idea of the day, would be to somehow arrange our winch system to cut a big hunk and hole to the right of the walls here, take this and try to do this with it. Now watch what we’re going to come off this block right here, I know the block is sitting in the perfect location, we’re going to run right through these walls and put a BA block and in Missouri, where I grew up, that’s called a big ass, so put a BA block right there, and we’re going to come down and put a big connection right there to the load. Now, my goodness sakes, what do we have here? We might call that angle there 20 degrees, I got 5 tons here, let’s go to the master rigging card and see what 20 degrees is for block factor, oh my goodness, look at that. The 20 included, my block factor is 1.97. I got to get my calculator out for this. Let’s take a look then, 1.97, wow. I’m going to take 1.97 times our 5 tons, get that, 1.97, we got 9.85 tons on that column right there. So, that’s part of the advantage of the mechanical advantage. Adding lines reduces the amount of force, and everything you want to put into that system, whether mechanical converting blocks, you can always help reduce those anchorage points, because if we just want to go straight into that column and right back to the load, get rid of the monster pull, we’re going to put massive load, this block will be reoriented. Kind of bigger. But we’re going to put that same pump and load, put 9.5 tons to that, 9.85 to that angle, they’re about the same angle, goodness. We’re going to put almost 10 tons to a column. There’s no way to do that, that’s what they’ll say. So we buy smaller blocks, get high capacity tugger, and multi-part system to reduce that line. Reduce the respective loading to that anchorage. Just doing a single line pull, oh my gosh, that’s huge.

So, any time we can multi-part, we’re saving our bacon and lowering that load to those anchorages. So let’s look and double check then, our answer section. I don’t want to leave anybody hanging. Everybody all working from the answer stubs. I’m going to replicate our answers from the other page. This one should be B, 1.05, R is C Charlies, that was 3.24, item 5, this A, and item 6 is B. Excellent, very good. So let’s take a quick breath, okay. Ready.

Let’s take a look then, at pulling the load uphill. Of course we don’t want to put wheels on it, that’s too hunk and easy. So we’re going to skid this guy up. We got a BA dozer up there, and you can feel free to mark your page there. You know what BA stands for. So we got BA dozer up there, with a winch, skid this guy right up the incline. It’s 12000 pounds, load, and significant information here, it’s on a wood skid base, and it’s on a concrete ramp. The concrete ramp is there, and here, and there, so we’re going to slide that puppy right up there. This is part of that winch, drag or roll. Right now we’re dragging this. We’re going to see what we get for resistance. What force is required to get that up the ramp? One piece of information here, need to get started with. As we look at our fill-in-the-blank information down below, I’ll get to the next page later. We know what our weight value is. What is the coefficient friction for, let’s see - we got wood on concrete, where can I find, let’s go back to our Master Rigger Card panel 6, that’s the Master Rigger Card panel 6, and our average value for that guy, can see that down here, average value showing up, wood on concrete looks like 45 percent. At 45 percent, so when we do our formula, it’s 45 CF times W equals force. Look what’s going to happen here, we’re going to get a new formula. Stay on that panel 6 with me for a minute. we got three total formulas in the upper right hand corner. One is comparable, one is going up hill, and one is moving down here. So we’re going to use this uphill formula and incorporate it into this workshop. You’re going to want to get that page, that page 6, panel 6, to be the workshop page 8. Get that formula right there, we’re missing - let’s understand what we have in the legend. In the formula, this says, coefficient friction times the weight, times R, which is horizontal run, divided length of ramp, that’s our incline, plus height, vertical height, again divided length of ramp times W, which is weight, equals the force. So let’s take that formula over to solve this problem. We know our CF, coefficient friction, is 1.45. Excellent.

Okay, let’s see where we’re at. Let’s go to that page. Our run, based on our drawing, is at 10 ft. Our height, based on our drawing, is 4 feet, we don’t have is ramp. I’ll put the drawing up. We just found out that the run is 10, height is 4, we don’t have the length of ramp based on that. What I need to do then, is figure out, take the pythagorean theorem and let’s get, so if we remember, A squared, plus B squared, equals, our unknown length of ramp, D squared. So let’s just do that while we’re here. So 4 squared times 10 squared, equals our unknown C squared, and we know that’s 16 - 4 times 4 plus 10 times 10 is 100, equals C squared. We take 16, 100, and 116 equals C squared. I get rid of those by squaring both sides, so this square root on the right takes away that guy, and this equals, get that down, and I got 10.77 equals C. Alright. So that’s 10.77 feet. So let’s get rid of that stuff. Are we having fun yet? That’s the important part. I’ve got 10.77 feet for that length of ramp there. I can put that in my next page, I’ve got 10.77 feet. So the big question on this, looking for the solution, what is the winch tension during that uphill movement, that’s what we want to solve. So let’s put that formula up on the page. I’m going to flap a new page, so if you were working from this let’s do that. Freeze this a little bit. Let’s try this. Give this a whirl. I’m going to run out of space.

Alright. Let’s put it up there. We’ve got our formula, CF, I’m going to replicate the formula , CF times W divided length of ramp plus height, divided length of ramp, times W again equals force. So let’s put in all the things we know. So we got a .45 times 4000 times 10 divided by 10.77, I’ve got to work all that out on the left side. Plus, let’s work this out, I got 4 divided by 10.77, times 12000, equals my force again. So work the left and right side, then add them together. Give you a chance to move right through with it. We’re doing pretty well with our time right here, at about 1 hour and a half. So we have one of the final ones right at the end, something to work with while you’re working that out. See what we’ve got here. Let’s see if we got to the same place. I’ve got 5400 times .93 plus .37, times 12000, let’s see if we agree on this. I got 5022 plus 4440 equals F. I’ve got 9462, more or less. 9462 is what I come up with for value for force to move it, yes there’s some breakaway load up and down dynamics and all that, moving and skidding and working. So let’s transfer that. Got it right over 9462 to put that puppy up the hill. If we look at my dozer, do we have a 10,000 pound or 12,000 pound capacity winch? How good is the dozer for staying up that hill up there? There may be all kinds of opportunity to see if that’s going to work as our anchorage point and so on. He may actually, we saw this on a grooving project forment that our guy would run his winch line down, come down, and then put another dozer back up the hill. As he is working in grooming, he work the third dozer up on right over here. Run a block here. Split the load and certainly gets tired, he gets tired, but to pull this guy up and down the hill, work it that way. Put a wheel and so on. We just got a straight line pull, should be at 94, most of you have been with me on these and I just wander off onto these bunny trails. I appreciate your patience with me. I just have a ball doing these, and I appreciate you coming back for these second, third, fourth and fifth times. So we’re just having a blast here.

So 9462 is our anticipated tension there. Let’s take a look at one, put them up on the screen, back it down a little bit, what got us into this, and we’ll just do this - we don’t want to run too far over time, what we did was - with a client, they were working on an underground vault, and you can envision here, they got a hatch, and this big vault, this is all underground, the big thing is it’s an energized vault and this wall here is working that way. They got conductor that run underground and this location, 85 percent of their power is all underground for this city. So they got a ton of this underground wire running all over creation. Sometimes it turns up, things happen, and they get stuck, they may have a 4,000, 6,000 run of it or more and big wire, and it gets burned up, it sticks up in the conduit. Ugly things can happen. They’ve got to pull this wire out and replace it. So, breaking it free is the generally biggest problem. We’re working with them with big bonders, big winches, trying to anticipate - know what your load is. Where it is, what the load is, and understand the anticipated resistance and trying to be good for it. So, it’s a lot of work. It’s really dangerous. I’ve been in vaults down under working with these guys and energized stuff around in some cases, oh my goodness, just really dicey. Don’t want to fall over and grab anything, it’ll all lit up. So we worked with them. We designed some Floor Plates. Their engineers went off on us, but these are 5,000 pound capacity floor plates, and what they are is plate, they’re 5 8 plates, they’ve got solid round stock welded right here. Then they’ve got a hole, that hole is drilled in tact to upload a hoister of the appropriate size for 5,000 pounds. So that what you’re seeing here is the bail. Then these are all anchored for this vault that is set in place. So to help give them - inside a concrete vault there really is no rigging point. You’ve got to build your own rigging anchorage points and then we worked with them on designing and developing strength short slings that can go up and using what roller blocks that will adjust back and forth as the angle changes, you’ve got to have the latitude to roll back and forth here, roll back and forth there, and as this block may change its position, or something’s working in and out of it, you got to be able to have the freedom. It’s just really like tilt wall construction. So share that load out, and pin it down low as possible and reasonable level down to the floor.

Thank you all for staying with me there for a minute, this is what it’s all about. This is really tough, dangerous, tight confines rigging activity and we need to come up with good solutions, and this is what we came up with. You can see on the page there that the value is 4500 for the dynamometer for showing that load being pulled out and we want to make sure that’s anticipated, that is what will be. Of course, it reduces as you continue to pull it. Once it gets through, we have a way to pull that load up and out of that patch and up and out of it really quick. Here’s the question, we need to know what is the loading at block A, so we hve questions on our page based on the 4500 pound, line pull, we can already note in here, 4500 for B, that’s line pull. We can definitely put it right there. We know there’s a little bit more dynamics and turning for that shiv and so on. So we’re going to put D for 4500 pound line pull going up to the winch. So, here’s the question, what is the loading up at this block, A? what is the loading at the baby block B, and what is loading at floor plate, C? So now start breaking that down and see what we got for panning out and redistributing that load down to those smaller blocks. I’m going to increase the size of drawing here, bring it up to view. Get rid of some of these things here. Alright, having a blast.

Okay, so we know we’ve got a line pull. What I need to do is go back to my block factor column on panel 4 from my card, I can see I got a 60 degree angle right there, a line comes straight in here, comes around that block, right up to the winch. So I’m going to go back to my Master Rigging Card panel 4, panel 1 I’m sorry, I need to find the block loading for 60 degrees. Let’s find that guy, block loading comes down here, and find out 60 degrees, my block factor is going to be 1.73. Let’s then go into our diagram. Remember, it’s block factor times line pull equals block loading. so we know our block factor is 1.73 times line pull of 4500 equals block load of about 7785. I know a lot of you are crunching numbers, putting that in, pushing and talking together. So block loading at A is about 7785. So that’s at this block here. Funny that it looks, but we got a shackle behind it, another shackle going into and another block there. That load is transmitted right to that block. So our 7785 close up at the lead block, right here. It is transmitted at a straight line right into the receiving block on the far side that’s trying to help equal down the floor plate. So we’re going to write that in here, 7785 pounds, block loading at A. Let me get rid of some of these now. What’s going to happen here is we’re going to have a block load of 7785 and we’re going to divide it by 2 arms, you might say, we’re going to transmit that down to 2 sub-blocks, or baby blocks. It’s sub zero, let’s just call it zero. So the simplest thing to do there is what? Let’s just cut this guy in half, divide it by 2, and that’s going to give us 3893 per block. So this one on this block is 3, and its twin block brother is right there, 3, 3. So I’m going to use that and put that in 3893 there, and take that down, 3893. Break that block and break it down one more step, it sure doesn’t look like a 0, but believe me, it’s a Wednesday. Okay, no problem.

We’re going to break that load down to, that’s the worst load anyway, in best condition - I’ll back that up, if this skid is wide, let’s get the value first - if you keep the number on your calculator, we take 3893 divided by 2, and we’re going to get 1946,47 depending on your rounding and that’s our number at the anchor plate, hoisting is 1947. I think you got 1946 on your sheet there, but it’s just a rounding issue on your calculator step down. Let’s solve this. Put these values on answer page. Doing great, hang right in there with me. We know that we got block load at A is number 4 right there, B is 3893, that’s B, number 2, sorry, get back out, block load anchor plate is 1946 or 47, that’s a one there, and a line pull at a winch that is number 4, 4500. Okay, excellent. Now here’s the question of the day, go back to that drawing and I’ll let you know how to work that last workshop in just a little bit. We’ll get rid of a bunch of these stuff. Alright, let’s see. Here’s the question: if we get rid of these anchor plates, and since I have the magic pen I can do this, get rid of this one, and get rid of this one. Get rid of these one right here. We’re just going to take it to 2 anchor plates, let’s just do these first. Alright, this is the bunny trail, hang in with me, we’re going to get right in there. If we were restricted for the location of anchor plate, something like this, we might have an oh my goodness moment that can get into a big world of hurt here. If we were really flat, like this, hoist strings there, center, hoist strings here, slings and etcetera. Now, know what happens here, we’ve gotten rid of everything else, 785 and just winging it here for a minute, you know what can happen. Instead of, there’s no mechanical advantage in these things, we’re just able to step that load value down by submit to walks and anchor plate, but the flatter that that blue sling gets in its, what we call angle to horizon gear. The flatter that gets, what happens to it? Does it increase in tension or decrease in tension? Excellent, it increases in tension. So you want to be really cautious, we would do a lot of rigging here, we would be anticipating. Ideally, we can go down to a place we can divide that by 2, bat it, and effect. Straight away from each one, we got 3893. At 30 degree angle from horizontal, what happens? Remember the load factor for 30 degrees? I can take you right to the Master Rigger Card, here’s the Master Rigger Card front panel 1, here’s 30 degrees. Our length to height ratio is 2 to 1. We’ll take length of 2 times height of one times share of load and we’re going to be 3800 times at 30 degrees it’s 2 divided by 1, we’re clear back up to  h my goodness. So there’s 7785 in that line that’s going to those anchor plates. we just blew up our 5000 pound anchor plate. See how easy it was? It was max rated at 5k. We’re not over almost 7800 pounds. We’re just grabbing two of them. Because of location and spread, obstructions and all kinds of stuff, we can only put two anchor plates so far apart. Really bad idea. There are a lot of solutions we can come up with on how to resolve or lower that, but I got to tell you, when you put up multiple blocks, new plates and just stretch one sling to go over from this anchor plate up and through that 30 degree angle right in there. That’s a huge mistake. We’d be up almost approaching 8000 pounds. We’d be pulling plates off the floor pretty quickly.

Alright, very good. There is an answer key right into this. Jonah will be back with us in just a second, there is a last workshop and it’s fairly involved. Nothing that we couldn’t do or have done in parcel. The answers for all of these are on the link that Jonah will have and always provided for you. We’ll be able to resolve or solve all of the elements for the very last workshop. Really getting out of time here and I don’t want to overextend our welcome. I appreciate everybody hanging in there with us. So, get the link, for the answer key, and I want to leave some time for question and answer. So as we go along here. Jonah, can you help get this back online with me? Jonah, any questions as we got along here? Can you help us with the answer key for the workshop?

Zack: Yes, this is Zack, by the way. I want to announce, prior, that Jonah’s got - there were not too many questions that came in online to ask. But, a few on P30, we’ll answer several question as they came up. I want to announce, before we got on to the question, that Jonah mentioned the answer key. I’ll take control of the answer key. This is really because all of you were interested in this subject. I want to make sure you’re aware of a new program that we’re rolling out here. This is our fundamentals of rigging engineering. We actually just press released today, this is a website we built for the program. But a lot of this you’re covering today, like horizontal rigging system are covered in depth in these courses, as well as our Master Rigger Course. But this is one of a kind, there’s nothing like this yet in this general industry. We have about 14 industry leading engineers that are involved, Keith Anderson recently released a book called Rigging Engineering Basics. Marc Van Daal on local transport, a lot of people involved here. I want to make sure you were aware of it, you can check out the website. The courses, materials, you went through today - we have an alternative load handling equipment course, that’s about 3 days. It involves a lot of different types of loading, like strain jacks and other entries. The horizontal load moving evolved pretty heavily as well. There are about 5 required courses, and then several electives to choose from. ASME is involved in this from a CEU’s point of view. Again, we want to make sure you’re aware of this because of the subject matter today. Last thing I mentioned, it’s all going to be online and on demand. We’re doing lecture recordings with all of these gentlemen you see here, From Fluor and Bechtel and Wireco, Barnhart, General Dynamics, Electric Boat, a lot of great people involved. I’m going to turn it back to Jonah, there’s one question you got and he’ll get you out. Again, I see questions coming up on this program now too. Jonah, why don’t you cover on the workshop. Looks like Bechtel workshop on Mike, there’s a question from John regarding the angle on the lead line, does it come back or in the back. Are you going to pull previous workshops and we’ll make you presenter again?

Mike: Jonah, why don’t you turn it over and work to do that. Let me skip my screen up and make sure that my screen is in view, guys? Okay. Let me go back to number 1. We’re back, is that the one in question? I think it’s the first workshop.

Jonah: Yes, that’s the one.

Mike: Question is, does the angle affect the line pull on the big beam? Is that the sort of question?

Jonah: Yes, the lead line - the angle on the lead line effect at all?

Mike: Yes, it does. What happens is if we can get this lead line, increase this angle, let’s just say it’s 15 degrees right now, if we can get it out to further, 50, or further 90, we’re constantly lowering the downforce or load to the high beam. So, as we’ve already discovered, straight down, it’ll interact with that block and be a direct downforce. But if we get it further and further, went over to a column and then down, a block and over, if we can do that, definitely be reducing that one part of line line pull. It’s definitely an effect on the big beam, any time you can open that up to a wider angle it will lower the force on the big beam. Great question.

Jonah: Okay, any questions on our work? or were there any questions on the Rigging Engineering Program?

Zack: No, I don’t think so. There were a couple of questions from, regarding ASME P30. What Jonah is going to do is reach out to individuals after the webinar.

Jonah: Okay.

Zack: I’ll just make an announcement. In the chat box, I put in there rigging engineering.com the url for that program. We’re actually running a special promotion. It won’t be available until January 2015, but we’re going to intake group of about 30 to 60 people that were heavily discounted. If anyone is interested, shoot Mike or myself an e-mail. Or, visit the website and you’ll be directed into what to do. The idea is we want to get some feedback from different users and how they enjoyed it. How we can improve it and whatnot. How to work that program. Mike, I’ll turn it back to you to close it off. Thank you all for joining us today.

Mike: Okay. Very good. I just had a marvelous time today. I appreciate your participation and everybody hanging in there with us, having a fun day? Little brain teasers, working on our applications and problem solving muscles. It’s been a fun time for me. Be looking for the next webinar published to be soon by ITI and Jonah. A team will be getting a notice out for that. I really appreciate you joining me on these things. It’s a lot of fun and I get to meet a lot of you on the road, and please, if you see me somewhere, I’m going to go to the AMP deal on Las Vegas part expo this week. I’m going to speak at the panel. If you bump into me and if I happen to not know your name, just introduce yourself and I’d sure love to put a name and face together. Just been a great relationship and look forward being in the business together. Got a long ways, and be talking to you soon.

Jonah: Okay, thanks Mike. Appreciate, Mike. Everyone, have a great day and we’ll talk soon.