# Turnout Frogs Numbers and Angles

MORE THAN YOU EVER WANTED TO KNOW ABOUT SWITCH FROG NUMBERS AND SWITCH FROG ANGLES

1. How does one calculate the frog number using measurements at the track?

2. Given the angle of divergence at the frog: how does one calculate the frog number?

3. Given the frog number: how does one calculate the angle of divergence?

4. Summary

5. Calculate the frog number (“N”)
Visualize a line of symmetry through the center of the frog with the rails diverging on either side of that line. Pick a point x units down the line from the point of the frog and measure, perpendicular to the center line, left or right to the rail; call that measurement y units. The SPREAD is 2y; and N = x/2y. Those of you who don’t have the time to wade through the following detail sections can skip to the Summary.

6. Given the angle of divergence at the frog, calculate the frog number
Call the frog angle “alpha”. The cotangent of alpha/2 is x/y (the adjacent side of the right triangle divided by the opposite side), and x/y = 2N:
cotan (alpha/2) = x/y = 2N
N = cotan (alpha/2) / 2.
Or using tangent = 1/cotangent:
tan (alpha/2) = y/x = 1/cotan (alpha/2)
N = cotan (alpha/2) / 2
N = (1 / tan (alpha/2)) / 2
N = 1/(2 tan (alpha/2))
(In case you don’t have cotangent on your hand calculator, it is the inverse of the tangent–1/tan; tan is opposite over adjacent; cotan is adjacent over opposite. The cotan of alpha is also tan (90-alpha). If you have an ancient and honorable analog computer called a “slide rule”, you probably have a cotangent “table” on it.)

Wait! There’s more! Actually, it turns out that the inverse of the sine of alpha
1/sin (alpha)
is very close to the frog number!
N = 1/sin(alpha)

1. Given the frog number, calculate the angle of divergence
From 1) above, N = x/2y so y/x = 1/2N.
The tangent of alpha/2 is also y/x. So

tan (alpha/2) = y/x = 1/2N,
alpha/2 = arctan (1/2N),
alpha = 2 arctan (1/2N).
As a check, solve 2) above for alpha:
N = 1/(2 tan (alpha/2)),
tan (alpha/2) = 1/2N,
alpha/2 = arctan (1/2N),
alpha = 2 arctan (1/2N).
Alternatively, for a close approximation, solve from 2) above:
N = 1/sin (alpha)
for alpha, getting:
alpha = arcsin (1/N)

1. Summary
Exact:
N = x/2y
N = 1/(2 tan (alpha/2))
alpha = 2 arctan (1/2N).
Close Enough:
N = 1/sin (alpha)
alpha = arcsin (1/N)

Randy Lehrian Jr. said:

Maybe this will hep:

It all just about the angle that the frog makes. A ratio between rise and run. 8 units long and one high is a #8 . The large the number the more gentle/gradual the switch. Which equates to running longer equipment.

This seams simpler.

I agree … it is simpler. Problem is … it’s not quite correct. It is, however, a close approximation; probably as close as any of us need in order to build a working turnout. I have nothing against shortcuts and approximations; I use them all the time. But I think it is important to understand the differences between shortcuts that work in the real-world and the true science behind them that makes them possible.

Thanks for the post Bob! I just learned something. I do see what the difference is. I’ll post back a revised visual in a minute.

Where is the asprin???

Well if we want to get really technical, real railroads use “theoretical” 1/2 point frogs. In other words they take the theoretical point of the frog, and and move a specific amount (I think it’s 1/2" or it could be 1/2 the difference between the theoretical point and the actual point? I’m not exactly sure, but that’s how the MOW people have described them.) to make a blunt ended frog. The frog is classified as a #22 (or what ever) but it’s actually a #21.XXXXX number frog. I’m sure someone smart could figure out the formula for figuring out the difference between the theoretical point of the frog and the actual point of the frog.

Oh but on the real railroads, that point is continually worn down, and rewelded… Craig Townsend said:

Oh but on the real railroads, that point is continually worn down, and rewelded… Craig,

Do you think that should be modeled? Perhaps on a siding that gets a lot of use, position the MoW guys who are a bit behind schedule.  With a sound module (Conversation between RTC and MoW foreman)

Hans-Joerg Mueller said:

Craig,

Do you think that should be modeled? Perhaps on a siding that gets a lot of use, position the MoW guys who are a bit behind schedule.  With a sound module (Conversation between RTC and MoW foreman)

Oh boy, now it’s starting to sound like too much work, and not enough fun. With enough slow orders for frogs being rebuilt, you might ‘die’ before getting to the end of the garden… Those 10mph slow orders for frogs add up quick.

I would think you would want to try and build good switches. The ones that don’t quite come out right would be placed on seldom used sidings, where slow orders would not be such a big deal.

The first one I built in large scale went on the mainline, and worked quite well, until the frog started chipping away. I had used Bondo for the frog on the first iteration of the switch. I rebuilt it with JB Quick, and new ceder ties after it had been in service for several years. The JB quick didn’t chip like the Bondo did, but I needed a slightly larger switch so that my Pacific and streamline passenger cars would be happy. So last year I removed my switch, still serviceable, from service.

My point is, your first switch can be useable, as long as you pay attention and keep the gauge accurate at all points along the switch.