All right, I tried to do some measuring of scale speeds, but I can't find the tables needed and never got past Algebra in math. Could someone help me figure out the scale speed formula. I know that 1 scale mile = 33 feet. That means 60 seconds to cover 33 feet is 60 mph. But what if I measured 10 feet? Because that's what I did. Clean wheels, clean track, several laps of warm up for each engine, and then ran them through the speed trap. Unfortunately, the N-scale police department was unwilling to allow the railroad to borrow their radar gun. Everyone in town suspects its off anyway. Seriously....I have searched a number of places, and can't find an appropriate table. I found one for a four feet speed trap, but not 10. Theres got to be the formula out there. Thanks for the help Jeff Augsburg & Concord R.R. (a fictional shortline in Central Illinois) http://www.pegnsean.net/~revnjeff
Well....I don't have a formula, but I might be able to help out some. Using basic algebra, if 33 feet is one scale mile, then 10 feet is .303 scale miles. Finding the scale speed would just be a matter of dividing at that point. If my calculations are correct, it should take 18.18 seconds to go 10 feet at 60 scale MPH. Knowing that....you can set up a ratio. Measure the time it takes to travel the 10 feet and then use algebra to solve for the speed. You ratio would be (18.18 seconds / 60 smph) = (time to travel 10 feet in seconds / X smph). Solve for "X" and "X" represents your scale speed in MPH. Now....someone beat me over the head if that's wrong. [ 02. January 2003, 13:45: Message edited by: Hoss ]
Jeff, mark a 10 foot long section. Time the nose of the locomotive, in seconds, as it passes the endpoints, then divide the seconds into 600. The result is MPH in N-Scale. For example ... 14.3 seconds into 600 gives 41.95 MPH. Helps if you have a calculator. It's quite surprising to see how slow 45 sMPH actually is. [ 02. January 2003, 14:25: Message edited by: Hank Coolidge ]
Dunno ... I'm trying to figure out how I came up with my formula five years ago .... ? Kinda hard to resurrect a formula by solving it backwards. I'll get back with you on this after I've sorted my brain out! [ 02. January 2003, 14:43: Message edited by: Hank Coolidge ]
Hmmm...I came up with my formula this morning, but that doesn't mean I didn't mess up. [ 02. January 2003, 14:58: Message edited by: Hoss ]
Trying again... (1 mile / 33 feet) = (A / 10 feet) Solve for "A" and get that 10 ft = .303 scale miles. (60 seconds / 1 scale mile) = (B / .303 scale miles) Solve for "B" to find that it takes 18.18 seconds to travel 10 ft. (.303 scale miles) at 60 MPH (based on the fact that it would take 60 seconds to travel one mile going at 60 MPH). (18.18 seconds / 60 MPH) = (# of seconds to travel 10 feet / X) Plug in the time it takes to travel the 10 feet in seconds and solve for "X" to find the scale MPH. Using the 14.3 seconds stated earlier, we'd use the ratio (18.18 seconds / 60 MPH) = (14.3 seconds / X). Solving for "X" would tell us that if it takes 14.3 seconds to travel the 10 ft. then the train is traveling at 47.19 scale MPH. Am I right?
I think you guys are making this too hard. The formula would be: 1080 / x = scale MPH. First you need to know how many scale miles 10 feet is. That was correctly worked out as 0.303 scale miles (rounded to 0.3) Then the units have to be converted to scale miles and seconds since that is what you are recording. This is the same formula with the units attached: (.3 scale miles x 3600 sec/1 hour) / (x sec) = scale miles per hour Where x seconds is the amount of time the locomotive takes to travel the 10 feet. The .3 x 3600 becomes a constant of 1080 scale mile seconds per hour. When you divide, the "seconds" unit cancel each other out. Therefore, if the locomotive takes 14.3 seconds to travel the 10 feet, the speed of the loco would be 75.5 scale MPH. Hopefully this make some sense . . Shoot me an email if you need further explaination.
I love math - especially challenges. I came up with a formula that I think is accurate for your scenario. I found that the solution your problem requires 2 steps: 1. calculate the feet per second for each measurement 2. convert feet per second into scale MPH I made a Microsoft Excel spreadsheet to crunch the numbers and return values, but I will assume that you don't have this program so I'll give you a straight forumula you can use to figure out with any old calculator. My formula assumes that a constant in your measurement is a length of 10 feet of running train. Assuming a constant value of 10 feet, the only input you must provide is a measurement in terms of seconds. This is the total length of time in seconds that it takes a locomotive to run on that 10 feet of track at a fixed speed. Here's the formula: 10/number_of_seconds/0.00916666575 That's 10 divided by number_of_seconds divided by 0.00916666575 If my calculations are correct, the number returned should give you SCALE MPH. If you have Microsoft Excel, then play around in my spreadsheet at the following URL: http://www.boche.net/trains/nscalemph.xls This particular spreadsheet is ONLY FOR N SCALE. Try it out and let me know if it is as accurate as I think it is. Jas
Maybe I'm wrong, but unless you're using a fast clock the time will still be the same. Only the distance is scale. Did I miss something??
33 feet in 60 seconds is 60 Nscale MPH 10 feet in 18.18182 seconds is 60 Nscale MPH ..therefore.. 0.549999945 feet per second is 60 Nscale MPH Divide 0.549999945 by 60 0.00916666575 feet per second is 1 Nscale MPH Jas [ 02. January 2003, 22:22: Message edited by: jasonboche ]
I hate to seem arrogant, because I'm not trying to be, but I think y'all are wrong and I'm right. If 18.18 seconds is 60 MPH at 10 ft., how can 14.3 seconds be 70+ MPH?? That dog don't hunt.... We've got four different responses on how to determine the speed over a 10 ft. length of track and four different answers. None of them come up with the same number. We need a math genius to come along and straighten us all out...
I can also further simplify this formula and state that: X = (# of seconds to travel 10 ft. / .303). Just plug in the number of seconds to go the 10 feet on the right side of the equation and "X" will represent the scale MPH. So....using the 14.3 seconds stated somewhere up the line, we could say: X=(14.3/.303), which again gives us 47.19 scale MPH. Keep in mind that this particular formula only works for the 10 ft. length and it does not take into account a "fast clock". This is based on real world time. [ 02. January 2003, 22:44: Message edited by: Hoss ]
I can also further simplify this formula and state that: X = (# of seconds to travel 10 ft. / .303). Just plug in the number of seconds to go the 10 feet on the right side of the equation and "X" will represent the scale MPH. So....using the 14.3 seconds stated somewhere up the line, we could say: X=(14.3/.303), which again gives us 47.19 scale MPH. Keep in mind that this particular formula only works for the 10 ft. length and it does not take into account a "fast clock". This is based on real world time. </font>[/QUOTE]I disagree with your formula. elapsed seconds is INVERSELY proportional to MPH. In other words, as elapsed seconds decreases, MPH increases as elapsed seconds increases, MPH decreases You have it so that elapsed seconds is directly proportional to MPH so that when elapsed seconds decrease, so does MPH. That's the opposite of what you want. If elapsed seconds decreases over a fixed length, that means the train is moving faster, thus, MPH must be greater. Jas [ 02. January 2003, 22:52: Message edited by: jasonboche ]
You're right. Y'all just ignore everyhthing I said. I've confused myself now. Time to go home and beat my head against the wall again....
That's quite alright. I'm sure you still have a better looking layout that what I have. Right now I have nothing but benchwork.
At the bottom of this page is a nice calculator: http://home.cogeco.ca/~trains/rrsoft.htm I use a 30" length myself. Cheers, Rod Schaffter Leominster, MA --May the power (DC or DCC) of "N" be with you--
If your lazy like I am, just put up some mile markers at a distance of two feet apart, then use the printed out chart found on this site: N Scale Speed chart ...Eddie [ 02. January 2003, 23:43: Message edited by: eddelozier ]
