I hadn't, but yes, the video shows what I needed. Thanks!

Thanks! Yes, removing the couplers did the trick.

But now I have another problem. How do I remove the trucks?  I need to check the attachment of the wire to the wiper contact on the truck. Is the truck just a pressure fit so all I need to do is pull it out? Or is there something that needs to be opened or released to get the truck off?

Oops. Posted in wrong forum. Please delete. I have reported in the HO forum.

The rear truck on my HO S2 switcher has no power pickup. I suspect a detached wire. It would be an easy fix, I think, if I could figure out how to get the shell off. I have the exploded view drawing that came with the loco. But that's not that much help. It looks like I need to detach the handrail between the main body and the step, both front and back, But it seems the handrails may have been glued on. When I try tugging on the handrail to remove it, it will not budge, even using a fair amount of force. In fact, it feels like if I pull any harder, the handrail will break.

How do I get the shell off?

The rear truck on my S2 switcher has no power pickup. I suspect a detached wire. It would be an easy fix, I think, if I could figure out how to get the shell off. I have the exploded view drawing that came with the loco. But that's not that much help. It looks like I need to detach the handrail between the main body and the step, both front and back, But it seems the handrails may have been glued on. When I try tugging on the handrail to remove it, it will not budge, even using a fair amount of force. In fact, it feels like if I pull any harder, the handrail will break.

How do I get the shell off?

Quote from: Joe Satnik on August 31, 2014, 06:10:21 PM

Isn't trig fun?

Overall Length = 2R x [1+Secant(CRDG/2)] + TBW, where Secant is the same as 1/Cosine

Yes Joe, trig is fun! Very glad to see you got the same result for the overall length as I did. It's good that we are on the same page with the math.

Hopefully this is not beating a dead horse, but it is possible to simplify the length calculation by applying some trig theorems to the 3-terms sum I gave previously. An exactly equivalent expression (that is, one that gives exactly the same result as the longer expression) involves the sum of just two terms:

1) The radius of the loop
PLUS
2) The radius of the loop times the secant of half the crossing angle.

Again the result of this calculation needs to be doubled, then the width of the roadbed added to it, to get the overall length.

Sorry if this is a bit too much, but I enjoy doing this kind of math.

Quote from: James in FL on August 30, 2014, 10:28:58 PM
P.S. I think we have lost many lurking, I hear something frying.

I think you're probably right. But please, let me confess. In my other life, in addition to doing trains, I am also a physicist. I use trig all the time. And I would like to thank both you and Joe for the insight into how the Figure 8 works. I don't know if I would have ever realized that the sum of the angle of the loop plus the crossing angle is 360 degrees. That is the key result that makes everything come out right. With that clue, working out the details of the Figure 8 was not hard. Without that clue, I did not know where to start.

Thanks again!

Quote from: James in FL on August 30, 2014, 06:22:20 PM
I'm not liking AnyRails numbers if these are the dimensions you are listing.

You are right James. The values I listed are all from anyrail. And I agree, I am not looking for exact precision. Just want to get values that work.

I realize you asked Joe about the length calculation. But the trig is sitting in my notes, so let me tell you how I am calculating it. The distance from the center of the cross to the top of the loop is the sum of 3 terms:

1)The radius of the loop
PLUS
2) The radius of the loop times the sine of half of the crossing angle times the tangent of half the crossing angle
PLUS
3) The radius of the loop times the cosine of half of the crossing angle.

The full length is double the result of that sum.

The above calculation gives the length from track center to track center. So the width of the roadbed must also be added to the final result.

OK. I took another look at the 90 deg crossing case. For that case I came up with a straight length needed between the end of the crossing and the start of each curve as being 10.6875 inches. And combining straights of lengths 4 15/16 + 4 1/2 + 1 1/4, I get exactly the needed length! The part numbers for those sections are 44811, 44829-1, and 44899-3, respectively.

So, assuming you agree with the results Joe, it indeed looks like both the 30-deg and 90-deg figure 8's can be made just fine using E-Z track.

I think I may have been able to answer my own question again Joe. I had not noticed that way down on the bottom of the track sections shown in anyrail (down below where the crossings and switches are shown) there are more straight sections. Part number 44899-2 is 7/8 inches in length. Two of these give a length of 1.75 inches, which is very close to the 1.7153 inch length that is needed, assuming my math is right (and I hope you will comment on that).

I'll bet even the 90 deg crossing case will work out using those special short lengths shown at the bottom of the E-Z track library!

Quote from: Joe Satnik on August 29, 2014, 10:04:35 PM

Keep playing with AnyRail.  You'll learn a lot, and it's a lot of fun.    

OK Joe. I set up the figure 8 in anyrail using the 11.25 inch curves and 30 deg crossing. Then I worked out the math to see what I would need in the way of straight sections. It seems things are even worse for this case than for the 90 deg crossing case.

I find that the distance from the center of the crossing to the end of each curved section is about 3.0144 inches. This value is the tangent of 15 degrees times the radius. Subtracting half of the 2 19/32 inch crossing length from this gives the needed straight section length as about 1.71753 inches. Unfortunately, none of the E-Z track straight sections shown in anyrail is very close to this. The closest straight lengths I see are 1 29/32 inches (part 44841-1) and 1 1/8 inches (part 44829-3). Considering both sides of the cross, putting in two of the first parts comes out too big by 0.38 inches, and putting in two of the second parts comes out too small by 1.185 inches. If there is another E-Z track straight section that I missed that would work better please let me know.

This is what motivated my original question. Even though it is possible to work out the math of what's needed, that doesn't mean the sections exist that will get it to fit properly.

Of course I may have not worked this out correctly. Please let me know if you come up with something different.

Quote from: Joe Satnik on August 29, 2014, 10:04:35 PM
TM,

Not sure what you mean by "blossom out", but an 11.25" radius circle is still the same width no matter which degree crossing is attached to it,

What  I had meant Joe was that I had thought it would be necessary to go to a larger radius curve to use the 30 deg crossing. But from what you say, I see I was wrong. So I'll give the 30 deg crossing and 11.25 radius a try in the software. I just hope things fit better than they did with the 90 deg crossing. In that case, it seems you wind up with quarter inch gaps on the straight sections.

Thanks for the heads up on AnyRail, Joe. The 30 deg crossing may minimize the length, but it causes the width to blossom out. No problem though, since the software will let me try out different arrangements.

Quote from: James in FL on August 27, 2014, 10:39:22 PM

I'm not totally confident in my formula.
Twice the square root of twice the radius, I get 9.486.

Hmm. I think we are using different formulas. From what you say it sounds like you are using a square root of the radius. The formula I worked out for the length of the rectangle, from track center to track center, looks like:

(2 +2 sqrt(2)) X radius

Or approximately 4.8284 X radius = 4.8284 X 11.25 = 54.3198 inches

And oops, I see I accidentally typed 53.3198 yesterday.

As for the roadbed, I don't have a piece of track with me. I am basing the 1.5 inch width on the fact that the description of the Thunder Valley train set says that the supplied track produces a circle of 2 foot diameter. Assuming a track radius of 11.25 inches, the diameter center-to-center would be 22.5 inches, so it appears the roadbed adds 1.5 more inches.