Wondering if there is a way to figure out mph on our layout! If it takes a train 25 seconds to go 44feet around our layout, can i convert that info to mph? Its for my boys science project. Thanks, Scale mph?
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44 feet in 25 seconds is 1.2 mph
(44 / 25) * 60 * 60 / 5280 = 1.2 mph
57.6 sMPH.
Big difference between MPH and scale-MPH. It would be a 1:48 difference.
It's easy. A mile per hour is 5280 feet/3600 second or 1.467 feet/second.
Your train went 44 feet/25 second or 1.76 feet/second. If you divide this number by 1.467 you get 1.19 real miles per hour. Since O scale is 1/48 of full scale, multiply 1.19 by 48 gives you 57.2 scale miles per hour (smph).
Another way to look at is to divide a mile (5280 feet) by 48 which gives you 110 feet. That is the length of a scale mile. so you went
((44 feet)/(110 ft/scale mile))
------------------------------------------- = 57.6 smph
((25 seconds)/3600 seconds/hour))
Of course this leads into a discussion of a "fast clock". We'll leave this for later.
Jan
Thanks to all who answered! Jan was spot on to what we are looking for the project! 5th grade science project with the lads trains front and center. He gets to bring a engine,transformer and some track to school!
Another way to do it that can be executed on any section of the layout with a definite start point is the following formula:
Inches traveled in 2.5 seconds equal scale MPH in O
(a number of other figures were published for other scales but I don't recall them off the bat)
---PCJ
Here's another comment. The greater the distance traveled the more accurate the calculations. So let the engine do multiple laps. Then take the total distance and divide it by the time.
Jan
Measure 11 feet on your track, time how many seconds it takes your train to travel the 11 feet, then use this chart:
SCALE SPEED TABLE | ||
Time to Travel | O Gauge | |
11 Feet | SCALE | |
(in seconds) | MPH | |
4 | 90.0 | |
4.5 | 80.0 | |
5 | 72.0 | |
5.5 | 65.5 | |
6 | 60.0 | |
6.5 | 55.4 | |
7 | 51.4 | |
7.5 | 48.0 | |
8 | 45.0 | |
8.5 | 42.4 | |
9 | 40.0 | |
9.5 | 37.9 | |
10 | 36.0 | |
11 | 32.7 | |
12 | 30.0 | |
13 | 27.7 | |
14 | 25.7 | |
15 | 24.0 | |
16 | 22.5 | |
17 | 21.2 | |
18 | 20.0 | |
19 | 18.9 | |
20 | 18.0 | |
21 | 17.1 | |
22 | 16.4 | |
23 | 15.7 | |
24 | 15.0 | |
25 | 14.4 | |
26 | 13.8 | |
27 | 13.3 | |
28 | 12.9 | |
29 | 12.4 | |
30 | 12.0 | |
35 | 10.3 | |
40 | 9.0 | |
45 | 8.0 | |
50 | 7.2 | |
55 | 6.5 | |
60 | 6.0 |
Sorry, I don't know how to make an Excel file reproduce better in this forum.
.....
Dennis
Thanks again to all , i do like the last one by Dennis, will save the info. The lads project is done. will try to get a pic from the science fair he in. Thanks again to this site, usually get great info from members.
Outstanding, good for him!
Paul
Thanks Railrunnin, the kids all happy about it! Hopefully they will give him a good grade for his work!
Most steam engines in Canada didn't have speedometers and the engineers would count the telegraph poles in say "10 seconds". They had a formula worked out to give the actual speed . There are 40 telegraph poles to the mile with mileage signs which start at the initial terminal for that subdivision... In CTC the signal numbers on the mask are also the mileage numbers.
I can't remember the formula. 40 poles in 60 seconds would be equal to sixty miles per hour anyone?