Technifold Bindery Success Strategies for MyPrintResource

Technifold USA, a manufacturer of equipment to help with scoring/creasing equipment, and attachments for paper folding machines, has put together a book of tips to improved productivity in finishing/bindery operations mpr

Envelopes printed on a Digital Printer

This post will mostly have appeal to those in the printing industry, who are considering a device to print envelopes (in color) using a digital device. These are our experiences with two machines, as of December 2013.

We have the Xante Ilumina (since 2009). It is based on the Oki engine. We have the DPP version which has a heavier duty fuser and is designed to run stock up to 500 grams per square meter. We have yet to run anything that heavy, though. The thickest we've done is some 16 point stock. 

We have run 6 3/4 remittance envelopes on the Xante. I have not noticed any problems affecting the drums or fuser.

I looked back in history, we have run about 9000 remittance envelopes on it (4 over 1) plus another 1000 single sided on the inside(gum side) On the outside of those envelope was only the address, which I decided to run on our standard inkjet addresser instead.

In addition to that, we've run about 7000 A7 invitation envelopes with the flaps opened. We addressed them on the front and printed the return address on the flap at the same time.

Overall we've run about 280000 pieces on the machine. Total supplies cost during that time period is $8600. And we too have never needed a service call. I have called a couple of times, and one time it turned out that had to replace a drum when it stated that it still had 10% life remaining. They told me that it kind of estimates remaining life based on usage, but they can get spots, etc which cause it to have to replace sooner. They told

That works out slightly more than 3 cents per piece. For this higher supplies cost I attribute to some business cards which we used to print on that machine for a customer (14 point with heavy toner coverage). We have since moved that project to ordering outside.

We had the opportunity in the Summer 2013, to participate in Printware's Try to Buy program, in order to evaluate their iJet press based on the memjet.

We got to run almost 30,000 pieces on that machine, and the ink cost us about $900, so that's right at 3 cents. We did have at least 10,000 of those were 8.5 x 11 size (nothing near full bleed, though - like letterheads and some graduation programs: plain text in brown ink).

I would say the speed is not 3 times like they advertise. On 8.5x11 we'd get about 28 per minute throughput with no interruptions. That was 70# text for the grad programs.

So what about that speed?
The iJet has two speeds. They call it Best & Fast. The 'best' is what I call normal, and Fast is like draft mode. Depending on the project and the colors, Fast mode may work perfectly fine. This particular project in question, I didn't think the Fast mode was be suitable, so I ran it in Best.

So Fast mode is 12 inches per second. Best mode is 6 inches per second. Letter size is 11 inches and you have about 2 inches gap between sheets, so each sheet requires about 13 inches.

Using 13 inches and Best mode at 6 inches per second gives 27.7 per minute for letter size. That works pretty close to the 28 per minute (which I got from actually timing the machine). So the gap is probably slightly less than 2 inches.
As I got more experienced with the machine, the feeding got better, but doubles were still an area of concern.

When the trial was over, we decided not to buy: we already had the Xante, and we did not want to drop down an additional capital investment. The lease option was not appealing to us either. We looked at our volume and just couldn't see enough to support the machine.

An advantage is that you can run Normal window envelopes, since there's no heat. Also this means that the prints are safe to run in copiers and laser printers.

But we have not had any problems with our Xante outputs smearing in any of our (or our customer's) laser devices. When we have something that we want to run on the Xante, we inquire the customer of the use. If they indicate laser overprints, we run them a test batch before full production. So far all have passed. As far as Window envelopes, we can buy some from Envelopes.com for about $40 per thousand (including shipping). I've heard that Western States has some, as well. It'd be nice if a local supplier carried them. We don't do a lot of window, on that machine, though.

Also, coated paper doesn't do well on the memjet. It will run, but the ink doesn't dry fast enough to prevent smearing. Also, NCR will run, but lots of doubles.

If we were making the decision today (like if we didn't already own the Xante) I'd definitely need to weigh our applications to see which would be better.

One thing, after running both machines, the Xante seems a little easier to operate. You just print to it, or print to the iQueue then go there to make settings (that's usually the way we do it).

With Printware's iJet (at least the way they set it up for us), we had to import a PDF rather than printing to it as a printer. That worked OK, unless the PDF came from the customer and they did not embed all the fonts. When the fonts are not embedded it substitutes fonts. Not always the desired result. I've heard there's a way to set it up to act like a network printer. They said you lose some capabilities when printing to it that way, though. I didn't push the issue during the trial period, because it wasn't that critical for our testing. If we had moved forward in considering the purchase, that would be something that I'd definitely want.

The 9.5 inch maximum width was also a drawback. The image width is 8.7 inches, but if they could find a way to at least run a wider sheet (11x17 or even 10x13), I feel they would have an advantage.

A feature that I did like about the iJet was the ability to full bleed a #10 envelope. I think Xante and maybe some others can bleed off 2 or more sides, now. That capability isn't available on our model, though.

I cannot think of anything else to add, right now. I hope this helps anyone else, who's considering.

David Scott

Math Puzzle - Addition problem

You ever see one of those puzzles like this: 

AHAHA

+ TEHE

TEHAW  {OK so the font's don't allow them to line up nice}

Each of the letters represents a digit in a standard addition problem. How to find out what substitutions were made.

We can break this up into a system of equations. First look at it like you'd add it 'by hand' :

So if we do it like regular math, we have to carry over (or regroup or however you want to call it) the ‘tens digit’ number from the sum.

So for example in the rightmost column: A+E = 10x + W {Where x represents the ‘tens digit’ number.}

Now we add the two H’s along with the digit that we carried:

H+H + x = 10y + A {the y is the carried digit}

Now keep going:

A+E + y = 10z + H

H+T + z = 10q + E

A + q = T.

Well we have 8 unknowns and only 5 equations.

What else do we know?  We know that the ‘carry digits can be either 0 or 1, in this scenario (9 + 9 = 18 would be the maximum sum in the first column, then 9+9 + 1 in other columns is 19, so still a 1 is the maximum carry).

Also, we know that since A does not equal T in the left column, then there must have been a carry of 1, so q = 1.

Also, since A+E + y equals a different number than the A+E + [zero], in the first column, then the y carry digit must be a 1.

So now let’s look at our equations, and see where we are:

A+E = 10x + W

2H + x = 10(1) + A  {we made y=1, and combined H+H}

A+E + 1 = 10z + H

H+T + z = 10q + E

A + 1 = T  {we made q=1}

We still have more unknowns than equations, but let's see if we can do something with it, anyway.

 Rearranging the 3rd equation as A+1 + E = 10z + H, then seeing that A+1 = T from the last eq.

T + E = 10z + H, rearrange this for H = T + E - 10z.

Plug that value of H into eq#4: (T + E - 10z) + T + z = 10q + E.

2T + E - 9z = 10q + E, the E's cancel: 2T - 9z = 10q.

Since carry digits can only be 0 or 1, let's guess for a moment that q = 0, that means 2T - 9z = 0, and 2T = 9z. Since z can only equal 0 or 1, let's look at 1, then we have 2T = 9, but since these are all whole numbers, 2T must be even, so it cannot be 9, so that would mean 2T = 0, instead. If that's the case then we have a problem, since T is the most significant digit, and we'd already figured that A + 1 = T, but that would make A = -1.

So by contradiction, we found that our guess was wrong, and q must equal 1.

Now that gives us 2T - 9z = 10. Now let's play guessing game again. For the moment guess that z = 1, so that gives us 2T - 9 = 10, or 2T = 19, which is an odd number and we know that we guessed wrong. So z must be 0 and we have 2T = 10, and then T = 5, our first number.

Since A + 1 = T, then A = 4. Plug A = 4 into equation # 2: 2H + x = 10 + A : 2H + x = 14.

So if x = 1, then that means 2H = 13, but 2H must be even, so x = 0, and then H = 7.

Plug this into equation #3:  H+T + z = 10q + E, along with our other findings:

 7 + 5 + 0 = 10 + E, so E = 2. Note that this also satisfies equation#3 True.

Plug [E=2] into eq#2: and we have: 4+2 = 0 + W, so W=6.

I'm not sure about a 'Formal Proof', but using our knowledge of how the carry digits work along with constraints of possible values of the digits and systems of equations, I'm confident that this is the Only Possible Solution.

A=4, E=2, H=7, T=5, W=6.