Sin x degrees = vertical distance / total distance

so Vertical Distance = Total Distance * Sin (x degrees)

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-Brian
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"The fastest hour run I've ever seen was done in 58:27."

I allways thought that was 10% grade

jaretj

I think so. If that is the case, than it would be .10 of 90 correct, or 9 degrees?

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Yes, I shave my legs. Yes, I am comfortable with this. I am enlightened.

I was thinking of a 1 foot rise every 10 feet or .3 miles for 3 miles which would be 1584 feet.

That would be for a X degree angle not X% grade.

Hugh

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Genetics load the gun, lifestyle pulls the trigger.

That's how I computed it in my OP, but I wasn't sure if that was indeed the case.

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Yes, I shave my legs. Yes, I am comfortable with this. I am enlightened.

rise/run == % grade = .1
hypotenuse = 3mi.
tan(angle) = rise/run
arctan(rise/run) = angle
sin(angle) = rise/distance traveled
rise = sin(angle) * distance traveled
rise = sin(arctan(rise/run)) * distance traveled
rise = sin(arctan(.1)) * 3mi
rise = sin(5.71) * 3mi
rise = .09949 * 3mi
rise = 1576'

(for moderate grades, the rise is very close to the grade * distance traveled)

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"Go yell at an M&M"

If I recall correctly percent grade = rise divided by run (horizontal distance). What the treadmill tells you is the hypotenuse of that triangle so...

X = horizontal distance
.1X = vertical distance


(3*5280)^2 = x^2 + .1X^2
(3*5280)^2= 1.01X^2
15761.3=X

Vertical Distance = 1576 feet

Did you graduate high school? College? If college, what major? Really, I'm curious.

I don't know about the OP, but I'll hazard to guess that you didn't take any courses in tact.

bingo --

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Have Fun ** Tri Hard ** Be Kind
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You're right I didn't but my question stands. How can a person have any education at all and not be able to calculate 10% of 3 miles? I know my daughter could do that in the 4th grade.
Bobby

Dumbass says what? Did you even read the OP's post?

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"What do I have to do, when I wake up tomorrow, to become excellent. Do that, and then do it again the next day, and the next, etc. One day you'll wake up and the time and the fitness will have taken care of themselves."

A 10% grade is equal to 10 feet of rise for every 100 feet traveled horizontally. This is calculated as follows:

(10 feet) / (100 feet) = 0.10 = 10%

In order to calculate the total vertical distance you traveled, all you have to do is multiply the grade by the distance run. Here's the calculation for your 3 mile run:

(3 miles) * ((5280 feet)/(1 mile)) * 0.10 = 1584 feet

Yup. He first shows what he thinks might be right (and was) but then goes on to say "10% incline on a treadmill for 3 miles is how many vertical feet?" So here is a person who does not have the math skills to be confident in a grade school math problem. All I am asking is: How much education does this person have and in what field? Is that upsetting to you?

Bobby, whether or not your daughter could figure it out, 10% of 3 miles is not the correct answer.

Sometimes the answer is not what it seems. Couldn't it be zero? Incline changes the feel of a treadmill, but there's no net vertical involved.

Well, you can take that route, but then there's no net horizontal involved either so...

Not exactly, but its within one half of one percent (1576 vs. 1584) and doesn't require the sin of the arcrtangent of the grade. I reckon < half a percent should be good enough for most work-out related issues.

LOL - most pointless workout ever!

Trig functions not required, just multiplication, division, squares and square roots - pythagorean theorem. Granted, it's not fourth grade material, but high school for sure.

Where is rappstar when we need him? He is an engineer, he would know this stuff.

Don't know about the op, but I have a doctoral degree and didn't know the answer. And no, I never took many math classes. Cryptonomicon is a great book, but not every " educated" individual has a background in math like most of those characters. I'll hazard that there are a great many areas you have a limited background in that would make you look like a dipshit to those that do.

While I'm sure we all tremble in the presence of your 4th grader's mathematical prowess, it's not exactly like the op asked for the answer to 4+4.

Fail

Precent grade in the united states is expressed as:

100 x RISE/RUN

Ok, my take...

Edited: When I was writing this post the diagram was not screwed up, but when posted my leading spaces were dropped.
.
. l /\
. l the rise, or the vertical gain, we'll call this "X"
. l \/
.____________________x

<horizontal distance > we know this is ten times the rise, so we can call it 10X. FYI, the grade in this example is easy to deal with, but if it was something other than a 10% grade we'd just divide 1 by the grade to come up with our coefficient for X. For example, if it was a 6% grade, 1/.06 is 16.66, so the sides would be X and 16.66X.

actual path that the runner traveled was along the dotted line, it is 3 miles.

now to solve for X:

Pythagorean theorem states: a**2 + b**2 = C**2

c must be the side of the triangle that is opposite from the right angle of the triangle. This is the same thing as saying C is the hypotenuse. In my crappy diagram above, where the "x" is located is where the right angle (90 degrees) is.

A is one side, doesn't matter which, I'll say it's X.

B must be the other side, 10x.

so x**2 +(10x)**2 = 3**2

solving the algebra...

x**2 + 100X**2 = 9 (note the 10 has been squared and shows as 100, the exponent now only applies to x)
101X**2 = 9 (since in both cases x is raised to the second power, algebra allows us to add the coefficients, when X just sits there with no number in front of it the coefficient is 1).
x**2 = 9/101 (I divided each side by 101)
x = square root of 9/101 (I've taken the square root of each side).
x = approximately .2985
x is stated in miles.

I don't think I'm full of crap, but I'm not 100% sure.

OK, and here is a real connection to all of this bs and triathlons. We can see that .2985 is very close to .3 ( the three tenths of a mile the original poster estimated the vertical gain to be). The reason it's just barely short of .3 is because the horizontal was not 3 miles, but rather the hypotenuse was 3 miles. If the horizontal would have been 3 miles than the vertical gain would have been exactly .3 miles. If these were the distances the hypotenuse would have to be approximately 3.015 miles. In other words if the runner ran 3.015 miles at a 10% grade, the vertical gain would be .3 miles.

Observation: When the grade is low (say 10% or less) the horizontal gain is very close to the actual running distance.

How does this connect to triathlons?

Consider the chaos at the beginning of the swim. At the start everyone is headed to the first buoy. People swimming on top of people. Let's say it's 1/2 mile to the buoy, 880 yards. Now let's say you moved 10% of that distance, approximately 90 yards to the side and swam from that point to the buoy. Not nearly the choas. And, as demonstrated above, you've added very little to your swim distance. In the example above, moving 10% of the distance to the side increased the travel distance from 3 to 3.015. In the half mile example, if you moved 90 yards to the side, your distance to the buoy would be approximately 884 yards instead of 880 yards.

Therefore... It's worth it to move way to the side to avoid the battle field of swimming chaos.

Steve

Bobby, this isn't a fourth grade math problem, and here's why.


While you are correct that 10 feet of rise for every 100 feet traveled horizontally is how % grade is calculated, that does not tell us the length of the hypotenuse (which is what the treadmill distance measurement shows).

To illustrate this, let's think about everyone's favorite triangle, the 3-4-5 right triangle.



For every 4 feet traveled forward on this triangle, you go up 3 feet.

3/4=75% grade.

Using (almost) everyone's math, traveling 5ft on a 75% grade=5*.75=3.75 of vertical traveled. This is clearly not the correct answer.

To get the correct answer, as klehner earlier mentioned, you need to multiple sin(arctan(3/4)*5 to get 3.

Therefore, the correct answer is indeed sin(arctan(10/100))*3 miles*5280 feet/mile=1576.

My question to you is, which college did you go to? I want to know so I can avoid sending my (future) kids there.

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Ignorance is bliss until they take your bliss away.

Hmmmm... the treadmill I use has that figured out no thinking involved, and who cares you are on a TREADMILL!

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"You do what you have to do , so can do what you want to do."

The rigorous solution may not be 4th grade math, but as I wrote above the problem can be approximated with 99.5% accuracy as a simple multiplication problem. Plus we are talking about a treadmill here and I ask, what do you think is the accuracy of the incline measure? Half a degree? Tenth of a degree? ATan of 10% is 5.711 degrees and ASin of 10% is 5.739 degrees. So the difference is less than 3 hundredths of a degree which results in a difference of 8 ft over 1576 ft.
As to your question about my schoolin': Who are you calling a college boy?
Bobby

Good point regarding swimming from the side. Another advantage to that can be that you can head toward that buoy without having to sight for it just by keeping yourself properly offset from the swimmers to your side. Just take a peek as you breath that way and keep a gradually decreasing distance from the 'pack'.

OK, disclaimer: I'm an Engineer with a Masters in Education, and I teach math at BOCES on a hit and miss basis.

The right answer is clearly not 4th grade math. Probably somewhere up in the area of Trig, which in NY is taugh in 8th or 9th grade, depending on the school and the student. That is the ability to solve the math problem once laid out. The ability to lay the problem out (ie, understanding that the treadmill is not the "run" in the rise over run) is probably over the head of most high school physics students to firure out on their own. A few would see it right away, and most of the others would have the "ah-ha" moment after it was explained, but few adults that do not work in a related field would see it.

I'm not convinced that the people using the rule of thumb (or estimate) actually understood it was an estimate when they posted it. It looks a lot lilke back pedaling to me, but I will get you the benefit of the doubt. To argue that the treadmill accuracy makes the estimate valid is actually contrary. The less accurate the measurement the more important the math (but the less important the precision).

Now, if you read the OP's original question and assume it was a realitivity question and not a math or physics question, the answer is 0. The belt is moving down, but the runner is staying at the same height, he doesn't move in relation to the room (or the treadmill frame). He does move in relation to the belt, but that is caused by the belts movement, not the runners.

Treadmills are a great tool for teaching Realitivity to laymen. Lots of fun discussions can come out of some simple questions.

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"...the street finds its own uses for things"

Talk about a pointless post.

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Steve Johnson
DARK HORSE TRIATHLON |

Here's a easy to use online calculator for your question. In your case, 3 miles is the length of the Hypotenuse (the slant side in a triangle, the other two sides share a 90 degree angle), and the angle is 10 degrees. The calculator then spit out the lengths of the Opposite Side (vertical distance) and the Adjacent Side (hortizontal distance). The vertical distance for your question is 0.5209 miles.

http://www.easycalculation.com/.../triangle-angles.php

10 deg does not equal 10% grade, ie 45 deg =100% grade
10% grade = ~5.71deg
10 deg =~17.63% grade

FAIL...

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Sorry I didn't realize that 10 degree incline on a treadmill does not equal to a 10% angle. However, this can easily be adjusted in the calculator - just input the actual angle corresponding to the treadmill measure (see previous post).

Use a little brain and don't just whine "Fail".

Interesting, and thanks for the correction. One observation though, wouldn't a 100% grade be past being vertical (vertical is 90 degrees)?

Oh brother...

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that just shows that you don't understand what has been discussed in this thread

(rise/run)*100%

90deg=infinite% div by zero

Ok mr. math teacher, now that you say it, I think I got it. Never really bothered to read the entire thread as it seems too much info for a relatively simple question, and so I missed your context. Was just trying to help the fellow efficently figure out the answer to his question without having to seemingly take a mental journal back to 11th grade as I remember how painlful that was - even for someone like me who was a math major in university.

Which one?

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Am I mis-reading this statement, or is it a joke? Because if you are serious...

(Or is English not your native language and you are having problems the comunication part?)

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"...the street finds its own uses for things"

maybe

our lady of the dark ages

AutomaticJack, bring it on! What did I say that was incorrect, other than making the initial error of assuming the treadmill measure of grade was in angle degrees (I'm no engineer). Or maybe you just like to shoot your mouth off in an automatic and racist fashion bypassing your prefrontal cortex altogether? Am I communicating well enough for you there?

Now, all your righteous and dumb sarcasm aside, let me get back to the point of this thread in the first place. There are two easy steps, both using the online calculator.

1. To convert 10% grade to a value in angle degree.
Input "1" for the "Opposite Side", and "10" for the "Adjacent Side". This corresponds to a 10% grade, and you're solving for the angle and the "Hypotenuse".
The calculator will tell you that the angle is 5.71 degrees. The value of Hypotenuse is unimportant and you can ignore.

2. To figure out how much elevation gain after running 3 miles at 10% grade (or 5.71 degrees)
Input "5.71" for "Angle", and "3" for "Hypotenuse Side". The calculator will tell you that the "Opposide Side" is 0.2985 (miles). That is your elevation gain.

Problem solved - using only some elementary geometry knowledge and no trignometry jargon required.

This statement alone...

"Interesting, and thanks for the correction. One observation though, wouldn't a 100% grade be past being vertical (vertical is 90 degrees)? "

Supports my post. If I insulted you, then you diserved it. Racist? Is that because I re-read your post and thought that maybe you were a non-native speaker and mis-understood the question or mis-typed your response? Because if that is where you are going, then beside a math problem you have a reading comprehension problem.

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"...the street finds its own uses for things"

I'd have kept quiet about the Maths degree if I were you.....

We all agree that the answer is 1576 feet. Ken Lehner provided a trigonometry solution, Jolly Rogers showed an elegant Pythagorian approach. Now we have an online calculator for playing with triangles. Where is that picture of the dead horse?

It's a zombie horse. We must continue to beat it so that it doesn't come back to life!

Ok, seems like the consensus is that i'm the bad guy here - imprehensible to me but i accept it anyway. I agree others may have already provided the solutions, and I was simply trying to provide a simpler way, one that doesn't involve scientific calculators. The reception to my honest effort? Well, be that as it may, I rest my case and for me, the horse is now buried ;)

Nice to meet!

They teach trig in 8th and 9th grade now??

When I was in HS, the normal progression was 9th (alg), 10th (geometry), 11th (adv. alg), 12th (trig). If you were accelerated all that simply moved up 1 year and you ended with calculus as a senior.

This is a problem that someone would learn in chemistry when you do "factor labeling" or "cancellation of units" --- or in trig (which is junior or senior year for an average student).

I'm getting a masters in teaching math - and if an adult isn't in a math/sci career and hasn't done math in awhile, it doesn't mean he didn't know it or he's dumb. It means he hasn't done math in awhile and under different circumstances, maybe could've figured out the problem. Sometimes it's hard to see where to start.

So stop being mean to bmanners :-)

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maybe she's born with it, maybe it's chlorine
If you're injured and need some sympathy, PM me and I'm very happy to write back.
disclaimer: PhD not MD

Me too :D

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maybe she's born with it, maybe it's chlorine
If you're injured and need some sympathy, PM me and I'm very happy to write back.
disclaimer: PhD not MD

NYS kind of "bastardizes" the math progression that I (and apparently you) went though in the 70's (and I went to HS in Indiana). Back then the progression was just as you listed. Now there isn't a Geometry (proofs, etc) and Trig (Triangle Math) by themselves. You take Algebra in 9th (and it includes basic Trig), then Geometry (with more Trig), then Pre-Calc (which is what we learned as algebra), and then Calc (which is really basic, and optional). If you are in an honors program then everything accelerates a year and as a senior take real calc (which includes basic Diffy-Q).

My education degree is not at the level that a real teacher with a BS would have, so I am a little weak on why it is done that way now. I believe the idea is that all of the math courses are tied together, so one builds on the other (or teaches what is needed from a previous course but was left out). I hope that makes sense. I'm not really the right person to explain it. I know that everything requires writing, and all teachers are expected to grade and correct grammar and spelling in all classes, which is a change from my HS days.

I teach a course call "Manufacturing Mathematics" at BOCES, which is a Trig class for machinist. A very strange course to a normal math teacher, since you start with long division and fraction reduction and end with calculating chamfer angles and manually indexing a BCD on a plate, without really going into the why, just the how.

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"...the street finds its own uses for things"

I agree. This is one of the most entertaining threads yet.

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Hans Bielat
TorHans LLC Co-Founder, Owner, Chief Innovation Officer
http://www.torhans.com

So when the treadmill says 10% grade and 3 miles, we are assuming those numbers are 100% accurate??! ;-)

Convert it to kilometres (3 miles = 4.8km), 4.8km at 10% is 480m (so much easier this metric stuff), 480m =1,575ft

Much simpler than maths (note the British s on the end here)!

Disclaimer: The 3m conversion to 4.8km is an estimate not an exact figure but close enough to get your answer within 10ft

I always just simplify it to 50 feet per mile per %. So 3 miles at 10 percent is 50x3x10=1500 feet. Close enough and easy math.


I'm probably 5% off the real answer, but I doubt most treadmills are close to that accurace on the grade.