Question 468828: On a 3% railroad grade, at what angle are the rails inclined to the horizontal, and how far does one rise in traveling upward 5000 feet measured along the rails? (The track rise 3 feet for each 100 feet of horizontal distance gained.)
Found 2 solutions by stanbon, Theo:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! On a 3% railroad grade, at what angle are the rails inclined to the horizontal,
tan^-1(3/100) = 1.7184 degrees
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and how far does one rise in traveling upward 5000 feet measured along the rails?
r/5000 = 3/100
r = 5000(3/100)
rise = 150 ft
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Cheers,
Stan H.
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(The track rise 3 feet for each 100 feet of horizontal distance gained.)
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Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! it looks like the 3% grade is referring to the slope of line made by the tracks.
the railroad people describe this as the rise to run ratio.
the grade is expressed as a percent which is the ratio times 100%.
you are told that the track line rises 3 feet for every 100 feet of horizontal travel.
this makes the grade equal to 3/100 = .03 * 100% = 3%.
you can use this information to find the angle of elevation.
draw a right triangle and label it ABC. A is on the left, B is on the top right, C is on the right at the same level as A.
The angle of elevation is angle A.
The vertical distance traveled is BC.
the horizontal distance traveled is AC.
the distance along the track is AB.
AB is the hypotenuse of this triangle.
AC is the side adjacent to angle A.
BC is the side opposite to angle A.
you use the tangent formula to find angle A.
tangent (A) = opposite / adjacent = 3/100 = .03
A is the angle whose tangent is .03
A is equal to 1.718358002 degrees.
That's your angle of elevation.
you can now use that angle to find the vertical distance after the train has traveled 5000 feet along the track bed.
the track bed is the hypotenuse of the triangle.
the horizontal distance traveled is the adjacent side to angle A.
the vertical distance traveled is the opposite side to angle A.
you will use the sine formula to find the vertical distance traveled.
sine (A) = opposite / hypotenuse = x / 5000
multiply both sides of this equation by 5000 to get:
5000 * sine (1.718358002) = x which gets you:
x = 149.9325455 feet
you will use the cosine formula to find the horizontal distance traveled.
cosine (A) = adjacent / hypotenuse = y / 5000
multiply both sides of this equation by 5000 to get:
5000 * cosine (1.718358002) = y which gets you:
y = 4997.751518 feet.
your figures are now:
angle A = 1.718358002 degrees
distance along track bed traveled = hypotenuse = 5000 feet.
vertical distance traveled = elevation = opposite side = 149.
horizontal distance traveled = adjacent side = 4997.751518 feet.
as a test, take the vertical distance traveled and divide it by 100 to get 49.97751518 feet.
we then multiply that by 3 to get 149.9325455 feet.
this is exactly the vertical distance we calculated earlier using the sine formula, only this time we used the 3% grade formula of 3 / 100 to get it.
answer check out so we're good.
answers to your question are:
angle = 1.718358002 degrees
vertical distance for 5000 feet of travel along the track bed = 149.9325455 feet.
round to whatever number you need.
a picture of what we just did is shown below:
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