I wish to do this over a distance say about a metre. The method i use is ok but pretty crude and not very accurate. I use a stepping block and a 1 metre long piece of wood and a spirit level. This is ok but it is just point to point and doesn’t allow for the extra distance of the curve.
Rgds ian
See my thread at http://www.trains.com/TRC/CS/forums/965564/ShowPost.aspx for the easiest way to determine grade using a digital level from Sears.
JimC.
Hi Ian,
I’m sure there are lots of ways for this. On my gradient, which does a full circle in 8’ diameter and climbs 6" I use a small level that has an adjustable dial. I put the level onto a flat car at the top of the gradient, on the slope, set the adjustable dial so it reads horizontal and then go down the gradient slowly adjusting the track as I go to keep the level at the horizontal. I do this 2 or 3 times until I’m happy with it.
Cheers,
Kim
Just come back to this Ian, forgot to add that the adjustable dial is marked off in degrees for gradient measurement.
OK Kim, what’s the formula to convert degrees of grade into percentage?
If a 1% grade is the ideal we should be aiming to achieve, how many degrees of gradient does it take to get that percentage? I had to take algebra 2 twice just to squeak through it, so am not very up on that math!
Hi Bob
The easiest way to work a 1% grade is for every 100 mm the train travels it rises 1mm.this works for every measurement that the user finds easy to use whether its cubits or inches or mm.
Since the dia of the circle is known it should be a simple matter of marking the diameter and for every 300mm rasing the line 3mm or 1/8" roughly for every 12"
But do check the manufacturer has measured the track diameter the correct way using the center line of the track.
regards John
Thank you fo the advice and if you knew me you would know the answer would not be that easy.
I don’t believe a laser will do it as they don’t go round corners. And a small level will not do it as although the reading is accurate it is too localised. DannyS even has a device that you can buy for a very modest sum that will show it to you on a scale.
The problem is the curve and using stepping blocks and a 1 metre long straightedge with a spirit levl is pretty simple and roughly correct. But again the length will not be quite right unless the straight edge’s length is the same as the length of the gradient being measured. So in my opinion a curved straightedge of the correct length may do the job.
Using a jigsaw (saber saw) you measure out 1 metre length of plywood of roughly the ciorrect curve, cut a 4 cm wide length and use this as a straight edge.
The idea of using stepping blocks and a spirit level you will get to the right conclusion. a secondary stepping block of half the size of the main one to be placed under the centre, will greatly assist stability and accuracy. I am also looking at using flextrack or similar, bent to the right curve as wel,l for the straightedge.
If you use a 1 metre length straight edge, then the stepping blocks size in centimetres will be the same as the %. a 2 cm rise in a metre will be 2 % and a 1 cm will be 1 %.
Now getting on to converting from % gradient to degrees, for the likes of me one is pretty close to the other but if you must; multiply the % by 1.1 and it will come to very close to the degrees. I know this isn’t 100 % but for practical purposes it should work.
Rgds Ian
PS, I have not actualy done this and i was wondering what others thought?
Important note:
When using a calculator to do the conversions, make sure it’s in degree mode when working with degrees! tan(45) = 1 and atan(1)=45, if you get any other answers it’s not in degree mode.
Ready for some trig?
tan {theta} = rise / run
where {theta} = the angle
rise = the distance elevated
run = the distance between the start of the grade and measured elevation
For a 1% grade, tan {theta} = 1/100 solving for {theta} we get {theta} = 0.5729 degrees
(Plug that in to your calculator using either the atan or inverse tan (tan**-1) function: atan(1/100)
Going the other way, tan (0.5729 degrees) = grade proportion. The grade proportion = 0.0099 in this case, which is very close to 1/100 = 0.01
Multiply the grade proportion by 100, and you have your grade percentage.
If you want to use the length of the track instead of the distance traveled from the start of the grade to the point measured, use sine rather than tangent. However, you won’t get the grade proportion back, you’ll get the ratio between the rise and track run.
You can do more math to get all the numbers you could want, but by now you’ve probably got the information you want. 
(If your head hurts, just forget this stuff and go run trains for a while.)
It’s very easy really. A gradient is between 0 and 90 degrees, therefore 90degrees is 100% gradient, a 1 in 1 gradient is 45 degrees or 50%, a 1 in 2 is 22.5 degrees or 25%. 1% is therefore 0.9 degrees. I think the majority of garden railroaders want a gradient between 2 points rather than set out at a fixed degree of climb to get the right percentage from a fixed point to somewhere that becomes the right height. The real life surveyors did this but they had miles of countryside to play with and they could go where the terrain allowed their planned gradient. (a very rough description but you get the gist)
I plan a gradient so that the climb between the 2 fixed points is equal throughout, once the loco enters the gradient it pulls at a constant rate without meeting unexpected dips or bumps. Once the rough groundwork is done I set my spirit level to the requiered degrees, a full curve of 8’ diameter is roughly 20’, a climb of 6" makes it 1 in 40 or 2.5 degrees on the spirit level. I then run the spirit down the grade - no track as yet - and make adjustments as I go, when that’s done the track is added and the spirit is mounted on the flat car and the bubble shows level. The car is run down the track slowly and if the bubble moves out of level adjustments are made. This works equally on straights and curves and the percentage or degree can be worked out once length and height of climb is known. Please remember that this is a hobby and not an exact science where steepness of grades means loss of revenue etc etc.
Cheers,
Kim
actually a vertical rise (90º to the plane of the ground) has an unsolvable gradient (that, or it’s infinite…don’t remember which at the moment) - so a grade is measured between 0º and 45º.
(example below)
Angles and their corresponding gradients: (rise:run)
90.00º angle has an unsolvable gradient (1:0)
45.00º angle has a gradient of 100% (1:1)
22.50º angle has a gradiant of 50% (1:2)
12.25º angle has a gradient of 25% (1:4)
6.125º angle has a gradient of 12.5% (1:8)
3.063º angle has a gradient of 6.25% (1:16)
1.531º angle has a gradient of 3.125% (1:32)
0.766º angle has a gradient of 1.563% (1:64)
so by that, if you had a grade of 1.5%, you’d be rising 1 foot for every 64 feet of run.
If you know the verticle rise from the begining of the curve (point a) to the end of the curve (point b) and the distance (outside, centerline, inside, PICK ONE) from a to b (I use a walking wheel to measure) then it is the same calculation used for a straight line. e.g. 3" rise in 100" of curve equals 3%.
A Walking wheel to measure distance cost $40 to $120 depending on features… mine cost $99 about 2 years ago…puchased for some row work I was involved with now a good tool for track length and grade analysis. A laser level and yardstick…meter stick for Ian… to develop the vertical differences point a to b…well $30 to $400 (it’s a level versus a transit and 30 feet versus 300 to 600)…
arcs and tangents and cosines … just to complicated for me
mr
imrnjr, Yes, I agree with your method; however, the question was if you had a spirit level that you could adjust or rotate the vials in the frame, and the amount of rotation is calibrated in degrees, what was the math required to determing how many degrees will result in what percentage of grade.
So far, the math majors who have expounded on the subject are giving answers far beyond my mathematical abilities. Heck, I have to unzip my fly to count over 20!
Puckdropper has it dead on, with all the math for converting from any angle to any gradient.
I took that and applied it to a range of angles to give a rough “line” of where things come into play.
While Puckdropper’s method will give you nearly an exact angle to grade conversion (allowing, of course, for any calculator round-off error), it can get tedious.
If you don’t like the math so much, you could probably use my table to get a rough approximation of grade. For example:
You take your level, and determine that you have a 35º angle.
Take a quick look at my table, it’s between 22.5 and 45º
that corresponds to a gradient of something between 50 and 100% (it’s actually a 70% grade)
So, as you see, you can “cheat” in a way, and approximate how steep of a gradient something is just by using that table. I don’t know much about Garden RRing, but I would like to believe that similar rules apply as with smaller scales regarding grades. as such, you’ll want to keep your angles roughly 1.5º (which is a grade of 3.125%) or less.
Hope this helps some.
If at all possible you would want to keep your grades in the one to 1.5% range. Anything more becomes very limiting as to train length, and 3.5% is just about the max you can get awway with.
I would be careful of taking 45 degrees as being = to 100%, what about a climb of 75%? or 99%. A slope, or gradient, is between 0 and 90 degrees as any hiker will tell you, the top end being a climb rather than a hike. I don’t know about the states but here in the UK road steepness is now given in % rather than 1 in 4, 1 in 3 etc and cars can tackle very steep slopes, hence 0-90 degrees, and 1 degree = 0.9 %. In the end it doesn’t really matter, if you are building a gradient, make it gentle!
Kim
Your math is based solely on the 0 to 90 degree markers, and saying that each degree is a certain percentage of 90. HOWEVER, it has nothing to do with the actual mathematical formula for determining slope(gradient).
Puckdropper has alredy shown the math. If you would apply those formulas to what you’re trying to say - you will see where you are wrong. You will also see how I came up with my figures.
The 1:4, 1:8, etc. figures I put were the actual rise:run for a given angle/grade. Probably the easiest way to demonstrate this would be to take a sheet of graph paper and a protractor. You will see that an angle with 1 unit of rise for every unit of run (1:1) will have an angle of 45º. a 90º angle is unsolvable - you simply cannot divide by 0.
if you still don’t believe me, here is the math.
tan(angle)=gradient; atan(gradient)=angle
tan(45)=1; atan(1)=45
tan(26.5)=.5; atan(.5)=26.5
As you can see, the tangent of a 45º angle is a 100% grade; and the tangent of a 26.5º angle makes a 50% grade (so my math was off before, but not by much - 9% roughly).
a climb of 75% (grade) is an angle of 36º
a climb of 99% (grade) is an angle of 44.7º
Gentlemen;
I must say i am proud to assciate with all of you, this is one of those rare times on this forum where we have a considerable amount of dialogue about a decent subject and you can see who is not participating. The people i am referring too will know who they are.
The maths is too much for me and i really don’t care that much about degrees anyway, % is good enough. My method of using centimetres for height and metres for distance is very simple and accurate. you use a 1 metre length straightedge and have your stepping blocks in centimetres and each centimetre will be a %.
But on a curve a straight straightedge is no good because you have to go round a bend. My logic, have a straightedge that is straight vertically but bent to suit the curve horizontally. If you can; even simpler, use a soft tape and measure your 1 metre around the curve and it will not matter if your straight edge is straight horizontally as the the two ends of the measued distance have already been established and all you are establishing is the difference in height between the two and as you know the distance already the same formula will apply. length in M/ height in cM = %
I know this is not 100 % right but it will give a rough indication and in this hobby surely this will suffice. But if you have 100 points % and 90 points in a right angle, surely a % must be about 9/10 of a degree?
Rgds Ian.
PS Trigg if you read this, you will notice that the people you reagrd as experts, are conspicuous by their absence in this discussion.
I have to say that I disagree with this, you can not divide by 0. 90 degrees divided by 0 is 0, 45 degrees divided by 0 is 0. 0 does exist in degrees in that it is the horizontal. I agree a 1 in 1 rise is 45%, I’ve already said this, so what is a rise of 1 in .75? or a rise of 1 in 0.5. These are gradients and will have a percentage value as well as a degree value, the percentage value being greater than 45%. I have no idea about a tan, I do not know what it is or isn’t and as already stated it doesn’t really matter when all you want to do is lift your line a few inches over a given run, straight or curve. My main point in all of this, not exactly the answer to the topic, was to get a gradient with an equal rise (or fall) that does not punish the loco to much.
Festive cheers,
Kim
now that’s a nice incline
Interesting topic. All the math is very fine, but only John Busby came close to answering the original question, how to measure on a curve. All you math majors should realise that this exact problem is why calculus was invented. The question is how small a slice of a circle can you cut before it is not curved anymore? Using prof. Busby’s logic, the smaller straight segment of the curve you take (his expample was 100mm which is fine) the more accurate a measurment you will get. Take the segments and apply the mathematical formulas for gradient and continue on until you have completed the curve.
Practical application is very easy, use a shorter a piece of wood and take more measurments, OR use a longer piece of wood to measure from the origin of the rise to the terminus in a straight line to get the total gradient, then adjust the track down the length so it is smooth.
Without proper survey tools we have to suffice with string and water levels. But that’s OK, because that is all they had to build the pyramids with and it worked just fine.
Hello Jack,
Best regards from our techie:
Yours is the practical approach that will get quick and “close enough” results. Mounting one of the Sears modules on a flatcar will make it even easier.
Best regards
ER
actually anything divided by 0 is unsolvable (without the use of some really involved math). What I have been trying to point out to you is that your method gives what percent of 90 DEGREES an angle is (part/whole). it in no way, shape, or form, gives an accurate measure of the GRADIENT (rise/run).
If you were to do the math that has been previously shown, you will see that you are completely wrong.
If I was going to make a grade of 1%, the angle of the grade (as compared to the ground) would be atan(.01) which is .57 degrees. If I was to apply YOUR method, (1% of a 90 degree angle), you would be going up(or down) a grade of 1.5%, which is tan(.9). Now, for small gradients, this is pretty close, I’ll give you that - BUT if you were to make anything steeper than that, you will quickly realize that the part of a 90 degree angle method will give you a grade of mugh higher than you would have thought. for example: 3% grade → atan(.03) = 1.71 degrees; 3% of 90 is 2.7 degrees → tan(2.7) = 4.7%. thats over 50% steeper than the grade you were originally trying to create.