Extra distance travelled along a sine wave path
$begingroup$
Let's say I have a blue line which is 10 metres long. I then draw a single cycle of a sine curve along this line, in red which has a maximum distance of 0.1 meter from the line (so an amplitude of 0.1m).
If I walk along the sine curve, I will walk further than 10 metres. But how do I calculate this new distance travelled?
What about if I increase the frequency so that there are five full cycles along my 10 metre blue line? How does the frequency affect it, in other words?
geometry curves
asked Nov 3 '17 at 16:33
Max WilliamsMax Williams
1162
$endgroup$
add a comment |
$begingroup$
Let's say I have a blue line which is 10 metres long. I then draw a single cycle of a sine curve along this line, in red which has a maximum distance of 0.1 meter from the line (so an amplitude of 0.1m).
If I walk along the sine curve, I will walk further than 10 metres. But how do I calculate this new distance travelled?
What about if I increase the frequency so that there are five full cycles along my 10 metre blue line? How does the frequency affect it, in other words?
geometry curves
asked Nov 3 '17 at 16:33
Max WilliamsMax Williams
1162
$endgroup$
add a comment |
$begingroup$
Let's say I have a blue line which is 10 metres long. I then draw a single cycle of a sine curve along this line, in red which has a maximum distance of 0.1 meter from the line (so an amplitude of 0.1m).
If I walk along the sine curve, I will walk further than 10 metres. But how do I calculate this new distance travelled?
What about if I increase the frequency so that there are five full cycles along my 10 metre blue line? How does the frequency affect it, in other words?
geometry curves
asked Nov 3 '17 at 16:33
Max WilliamsMax Williams
1162
$endgroup$
Let's say I have a blue line which is 10 metres long. I then draw a single cycle of a sine curve along this line, in red which has a maximum distance of 0.1 meter from the line (so an amplitude of 0.1m).
If I walk along the sine curve, I will walk further than 10 metres. But how do I calculate this new distance travelled?
What about if I increase the frequency so that there are five full cycles along my 10 metre blue line? How does the frequency affect it, in other words?
geometry curves
geometry curves
asked Nov 3 '17 at 16:33
Max WilliamsMax Williams
1162
asked Nov 3 '17 at 16:33
Max WilliamsMax Williams
1162
asked Nov 3 '17 at 16:33
Max WilliamsMax Williams
1162
asked Nov 3 '17 at 16:33
Max WilliamsMax Williams
1162
asked Nov 3 '17 at 16:33
Max WilliamsMax Williams
1162
1162
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
HINT
This answer may not help directly for higher frequencies/waves but stating it nevertheless.., the following requires some imagination.
Sinusoidal waves are development of slantly cut cylinder intersections seen as surface development.
If a cylinder radius $a$ is cut at an angle $alpha$ to radial plane the ellipse axes are $ (a, a sec alpha) $ then the ellipse has an eccentricity $ e= sin alpha$.
$$ 2 pi a = 10 m, quad a tan alpha = dfrac{.01}{2} ; $$
Solve for $ (a,alpha, e)$.
The perimeter is $ 4 a , E(e)$ which a standard result using Elliptic integrals; it can be found by integration or google it for a start.
EDIT1:
Useful to remember the waveform equations of same amplitude $A$ for single and multiple frequencies $n$:
$$ y_1= A sin dfrac{2 pi x}{lambda}$$
$$ y_n= A sin dfrac{2 pi xcdot n}{lambda}$$
answered Nov 3 '17 at 17:14
NarasimhamNarasimham
20.8k52158
$endgroup$
add a comment |
$begingroup$
You need the arc length formula for this: $intsqrt{1+f'(x)^2}dx$ on interval $[0,10]$. What you need to do is establish the formula of your curve, which is not hard to do. After all, you know the period (then use $b=frac{2pi}{period}$) and the amplitude as you stated. Anyway, the real problem is that the arc length formula is not going to come out "nice", in the sense that I greatly expect that you are going to deal with an integral that cannot be evaluated in terms of elementary functions. You are then left with an approximation using an TI for example
answered Nov 3 '17 at 16:38
imranfatimranfat
8,03141532
$endgroup$
$begingroup$
Thanks - is there a non-mathematicians' version? If it approximates the distance then that's fine.
$endgroup$
– Max Williams
Nov 3 '17 at 16:40
$begingroup$
Of course there is a less "stringent" mathematics version. How much less mathematically you want to go?. You can "avoid" calculus in the following manner: Use increments of $0.1$ for $x$ and calculate the corresponding $y's$ for each increasing $x$. Then use Pythagorean theorem $100$ times $(10/0.1)$ and add up distances. That would be a good approximation. It's a little work but not difficult. You still need to set up your sine formula, also not difficult. (This answers the first part of your question)
$endgroup$
– imranfat
Nov 3 '17 at 16:43
$begingroup$
I was hoping for a formula which doesn't require calculus (by which i mean, for example, integrating functions - sorry if that's not calculus, i'm not a mathematician), and into which I can just plug some numbers and carry out only simple operations. Imagine I'm trying to do it in a computer program where I can only do basic operations, raising to a power, sin, cos and tan and their converses.
$endgroup$
– Max Williams
Nov 3 '17 at 16:47
$begingroup$
My second approach in my previous comment does not require calculus. The Pythagorean theorem is what you need (remember that famous $a^2+b^2=c^2$ on a right triangle from high school?). I think that is as basic as you can go. To calculate a distance between two points, the formula translates to $c=sqrt{(0.1)^2+(y_2-y_1)^2}$. You need to do that 100 times as you have 100 increments. Then add up. Having something like Excel available would be tremendously helpful
$endgroup$
– imranfat
Nov 3 '17 at 16:49
$begingroup$
Ok, thanks. I guess I was too optimistic for a simple one-hit equation then!
$endgroup$
– Max Williams
Nov 3 '17 at 16:53
|
show 2 more comments
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT
This answer may not help directly for higher frequencies/waves but stating it nevertheless.., the following requires some imagination.
Sinusoidal waves are development of slantly cut cylinder intersections seen as surface development.
If a cylinder radius $a$ is cut at an angle $alpha$ to radial plane the ellipse axes are $ (a, a sec alpha) $ then the ellipse has an eccentricity $ e= sin alpha$.
$$ 2 pi a = 10 m, quad a tan alpha = dfrac{.01}{2} ; $$
Solve for $ (a,alpha, e)$.
The perimeter is $ 4 a , E(e)$ which a standard result using Elliptic integrals; it can be found by integration or google it for a start.
EDIT1:
Useful to remember the waveform equations of same amplitude $A$ for single and multiple frequencies $n$:
$$ y_1= A sin dfrac{2 pi x}{lambda}$$
$$ y_n= A sin dfrac{2 pi xcdot n}{lambda}$$
answered Nov 3 '17 at 17:14
NarasimhamNarasimham
20.8k52158
$endgroup$
add a comment |
$begingroup$
HINT
This answer may not help directly for higher frequencies/waves but stating it nevertheless.., the following requires some imagination.
Sinusoidal waves are development of slantly cut cylinder intersections seen as surface development.
If a cylinder radius $a$ is cut at an angle $alpha$ to radial plane the ellipse axes are $ (a, a sec alpha) $ then the ellipse has an eccentricity $ e= sin alpha$.
$$ 2 pi a = 10 m, quad a tan alpha = dfrac{.01}{2} ; $$
Solve for $ (a,alpha, e)$.
The perimeter is $ 4 a , E(e)$ which a standard result using Elliptic integrals; it can be found by integration or google it for a start.
EDIT1:
Useful to remember the waveform equations of same amplitude $A$ for single and multiple frequencies $n$:
$$ y_1= A sin dfrac{2 pi x}{lambda}$$
$$ y_n= A sin dfrac{2 pi xcdot n}{lambda}$$
answered Nov 3 '17 at 17:14
NarasimhamNarasimham
20.8k52158
$endgroup$
add a comment |
$begingroup$
HINT
This answer may not help directly for higher frequencies/waves but stating it nevertheless.., the following requires some imagination.
Sinusoidal waves are development of slantly cut cylinder intersections seen as surface development.
If a cylinder radius $a$ is cut at an angle $alpha$ to radial plane the ellipse axes are $ (a, a sec alpha) $ then the ellipse has an eccentricity $ e= sin alpha$.
$$ 2 pi a = 10 m, quad a tan alpha = dfrac{.01}{2} ; $$
Solve for $ (a,alpha, e)$.
The perimeter is $ 4 a , E(e)$ which a standard result using Elliptic integrals; it can be found by integration or google it for a start.
EDIT1:
Useful to remember the waveform equations of same amplitude $A$ for single and multiple frequencies $n$:
$$ y_1= A sin dfrac{2 pi x}{lambda}$$
$$ y_n= A sin dfrac{2 pi xcdot n}{lambda}$$
answered Nov 3 '17 at 17:14
NarasimhamNarasimham
20.8k52158
$endgroup$
HINT
This answer may not help directly for higher frequencies/waves but stating it nevertheless.., the following requires some imagination.
Sinusoidal waves are development of slantly cut cylinder intersections seen as surface development.
If a cylinder radius $a$ is cut at an angle $alpha$ to radial plane the ellipse axes are $ (a, a sec alpha) $ then the ellipse has an eccentricity $ e= sin alpha$.
$$ 2 pi a = 10 m, quad a tan alpha = dfrac{.01}{2} ; $$
Solve for $ (a,alpha, e)$.
The perimeter is $ 4 a , E(e)$ which a standard result using Elliptic integrals; it can be found by integration or google it for a start.
EDIT1:
Useful to remember the waveform equations of same amplitude $A$ for single and multiple frequencies $n$:
$$ y_1= A sin dfrac{2 pi x}{lambda}$$
$$ y_n= A sin dfrac{2 pi xcdot n}{lambda}$$
answered Nov 3 '17 at 17:14
NarasimhamNarasimham
20.8k52158
edited Nov 7 '17 at 7:09
answered Nov 3 '17 at 17:14
NarasimhamNarasimham
20.8k52158
answered Nov 3 '17 at 17:14
NarasimhamNarasimham
20.8k52158
answered Nov 3 '17 at 17:14
NarasimhamNarasimham
20.8k52158
20.8k52158
add a comment |
add a comment |
$begingroup$
You need the arc length formula for this: $intsqrt{1+f'(x)^2}dx$ on interval $[0,10]$. What you need to do is establish the formula of your curve, which is not hard to do. After all, you know the period (then use $b=frac{2pi}{period}$) and the amplitude as you stated. Anyway, the real problem is that the arc length formula is not going to come out "nice", in the sense that I greatly expect that you are going to deal with an integral that cannot be evaluated in terms of elementary functions. You are then left with an approximation using an TI for example
answered Nov 3 '17 at 16:38
imranfatimranfat
8,03141532
$endgroup$
$begingroup$
Thanks - is there a non-mathematicians' version? If it approximates the distance then that's fine.
$endgroup$
– Max Williams
Nov 3 '17 at 16:40
$begingroup$
Of course there is a less "stringent" mathematics version. How much less mathematically you want to go?. You can "avoid" calculus in the following manner: Use increments of $0.1$ for $x$ and calculate the corresponding $y's$ for each increasing $x$. Then use Pythagorean theorem $100$ times $(10/0.1)$ and add up distances. That would be a good approximation. It's a little work but not difficult. You still need to set up your sine formula, also not difficult. (This answers the first part of your question)
$endgroup$
– imranfat
Nov 3 '17 at 16:43
$begingroup$
I was hoping for a formula which doesn't require calculus (by which i mean, for example, integrating functions - sorry if that's not calculus, i'm not a mathematician), and into which I can just plug some numbers and carry out only simple operations. Imagine I'm trying to do it in a computer program where I can only do basic operations, raising to a power, sin, cos and tan and their converses.
$endgroup$
– Max Williams
Nov 3 '17 at 16:47
$begingroup$
My second approach in my previous comment does not require calculus. The Pythagorean theorem is what you need (remember that famous $a^2+b^2=c^2$ on a right triangle from high school?). I think that is as basic as you can go. To calculate a distance between two points, the formula translates to $c=sqrt{(0.1)^2+(y_2-y_1)^2}$. You need to do that 100 times as you have 100 increments. Then add up. Having something like Excel available would be tremendously helpful
$endgroup$
– imranfat
Nov 3 '17 at 16:49
$begingroup$
Ok, thanks. I guess I was too optimistic for a simple one-hit equation then!
$endgroup$
– Max Williams
Nov 3 '17 at 16:53
|
show 2 more comments
$begingroup$
You need the arc length formula for this: $intsqrt{1+f'(x)^2}dx$ on interval $[0,10]$. What you need to do is establish the formula of your curve, which is not hard to do. After all, you know the period (then use $b=frac{2pi}{period}$) and the amplitude as you stated. Anyway, the real problem is that the arc length formula is not going to come out "nice", in the sense that I greatly expect that you are going to deal with an integral that cannot be evaluated in terms of elementary functions. You are then left with an approximation using an TI for example
answered Nov 3 '17 at 16:38
imranfatimranfat
8,03141532
$endgroup$
$begingroup$
Thanks - is there a non-mathematicians' version? If it approximates the distance then that's fine.
$endgroup$
– Max Williams
Nov 3 '17 at 16:40
$begingroup$
Of course there is a less "stringent" mathematics version. How much less mathematically you want to go?. You can "avoid" calculus in the following manner: Use increments of $0.1$ for $x$ and calculate the corresponding $y's$ for each increasing $x$. Then use Pythagorean theorem $100$ times $(10/0.1)$ and add up distances. That would be a good approximation. It's a little work but not difficult. You still need to set up your sine formula, also not difficult. (This answers the first part of your question)
$endgroup$
– imranfat
Nov 3 '17 at 16:43
$begingroup$
I was hoping for a formula which doesn't require calculus (by which i mean, for example, integrating functions - sorry if that's not calculus, i'm not a mathematician), and into which I can just plug some numbers and carry out only simple operations. Imagine I'm trying to do it in a computer program where I can only do basic operations, raising to a power, sin, cos and tan and their converses.
$endgroup$
– Max Williams
Nov 3 '17 at 16:47
$begingroup$
My second approach in my previous comment does not require calculus. The Pythagorean theorem is what you need (remember that famous $a^2+b^2=c^2$ on a right triangle from high school?). I think that is as basic as you can go. To calculate a distance between two points, the formula translates to $c=sqrt{(0.1)^2+(y_2-y_1)^2}$. You need to do that 100 times as you have 100 increments. Then add up. Having something like Excel available would be tremendously helpful
$endgroup$
– imranfat
Nov 3 '17 at 16:49
$begingroup$
Ok, thanks. I guess I was too optimistic for a simple one-hit equation then!
$endgroup$
– Max Williams
Nov 3 '17 at 16:53
|
show 2 more comments
$begingroup$
You need the arc length formula for this: $intsqrt{1+f'(x)^2}dx$ on interval $[0,10]$. What you need to do is establish the formula of your curve, which is not hard to do. After all, you know the period (then use $b=frac{2pi}{period}$) and the amplitude as you stated. Anyway, the real problem is that the arc length formula is not going to come out "nice", in the sense that I greatly expect that you are going to deal with an integral that cannot be evaluated in terms of elementary functions. You are then left with an approximation using an TI for example
answered Nov 3 '17 at 16:38
imranfatimranfat
8,03141532
$endgroup$
You need the arc length formula for this: $intsqrt{1+f'(x)^2}dx$ on interval $[0,10]$. What you need to do is establish the formula of your curve, which is not hard to do. After all, you know the period (then use $b=frac{2pi}{period}$) and the amplitude as you stated. Anyway, the real problem is that the arc length formula is not going to come out "nice", in the sense that I greatly expect that you are going to deal with an integral that cannot be evaluated in terms of elementary functions. You are then left with an approximation using an TI for example
answered Nov 3 '17 at 16:38
imranfatimranfat
8,03141532
answered Nov 3 '17 at 16:38
imranfatimranfat
8,03141532
answered Nov 3 '17 at 16:38
imranfatimranfat
8,03141532
answered Nov 3 '17 at 16:38
imranfatimranfat
8,03141532
8,03141532
$begingroup$
Thanks - is there a non-mathematicians' version? If it approximates the distance then that's fine.
$endgroup$
– Max Williams
Nov 3 '17 at 16:40
$begingroup$
Of course there is a less "stringent" mathematics version. How much less mathematically you want to go?. You can "avoid" calculus in the following manner: Use increments of $0.1$ for $x$ and calculate the corresponding $y's$ for each increasing $x$. Then use Pythagorean theorem $100$ times $(10/0.1)$ and add up distances. That would be a good approximation. It's a little work but not difficult. You still need to set up your sine formula, also not difficult. (This answers the first part of your question)
$endgroup$
– imranfat
Nov 3 '17 at 16:43
$begingroup$
I was hoping for a formula which doesn't require calculus (by which i mean, for example, integrating functions - sorry if that's not calculus, i'm not a mathematician), and into which I can just plug some numbers and carry out only simple operations. Imagine I'm trying to do it in a computer program where I can only do basic operations, raising to a power, sin, cos and tan and their converses.
$endgroup$
– Max Williams
Nov 3 '17 at 16:47
$begingroup$
My second approach in my previous comment does not require calculus. The Pythagorean theorem is what you need (remember that famous $a^2+b^2=c^2$ on a right triangle from high school?). I think that is as basic as you can go. To calculate a distance between two points, the formula translates to $c=sqrt{(0.1)^2+(y_2-y_1)^2}$. You need to do that 100 times as you have 100 increments. Then add up. Having something like Excel available would be tremendously helpful
$endgroup$
– imranfat
Nov 3 '17 at 16:49
$begingroup$
Ok, thanks. I guess I was too optimistic for a simple one-hit equation then!
$endgroup$
– Max Williams
Nov 3 '17 at 16:53
|
show 2 more comments
$begingroup$
Thanks - is there a non-mathematicians' version? If it approximates the distance then that's fine.
$endgroup$
– Max Williams
Nov 3 '17 at 16:40
$begingroup$
Of course there is a less "stringent" mathematics version. How much less mathematically you want to go?. You can "avoid" calculus in the following manner: Use increments of $0.1$ for $x$ and calculate the corresponding $y's$ for each increasing $x$. Then use Pythagorean theorem $100$ times $(10/0.1)$ and add up distances. That would be a good approximation. It's a little work but not difficult. You still need to set up your sine formula, also not difficult. (This answers the first part of your question)
$endgroup$
– imranfat
Nov 3 '17 at 16:43
$begingroup$
I was hoping for a formula which doesn't require calculus (by which i mean, for example, integrating functions - sorry if that's not calculus, i'm not a mathematician), and into which I can just plug some numbers and carry out only simple operations. Imagine I'm trying to do it in a computer program where I can only do basic operations, raising to a power, sin, cos and tan and their converses.
$endgroup$
– Max Williams
Nov 3 '17 at 16:47
$begingroup$
My second approach in my previous comment does not require calculus. The Pythagorean theorem is what you need (remember that famous $a^2+b^2=c^2$ on a right triangle from high school?). I think that is as basic as you can go. To calculate a distance between two points, the formula translates to $c=sqrt{(0.1)^2+(y_2-y_1)^2}$. You need to do that 100 times as you have 100 increments. Then add up. Having something like Excel available would be tremendously helpful
$endgroup$
– imranfat
Nov 3 '17 at 16:49
$begingroup$
Ok, thanks. I guess I was too optimistic for a simple one-hit equation then!
$endgroup$
– Max Williams
Nov 3 '17 at 16:53
$begingroup$
Thanks - is there a non-mathematicians' version? If it approximates the distance then that's fine.
$endgroup$
– Max Williams
Nov 3 '17 at 16:40
$begingroup$
Thanks - is there a non-mathematicians' version? If it approximates the distance then that's fine.
$endgroup$
– Max Williams
Nov 3 '17 at 16:40
$begingroup$
Of course there is a less "stringent" mathematics version. How much less mathematically you want to go?. You can "avoid" calculus in the following manner: Use increments of $0.1$ for $x$ and calculate the corresponding $y's$ for each increasing $x$. Then use Pythagorean theorem $100$ times $(10/0.1)$ and add up distances. That would be a good approximation. It's a little work but not difficult. You still need to set up your sine formula, also not difficult. (This answers the first part of your question)
$endgroup$
– imranfat
Nov 3 '17 at 16:43
$begingroup$
Of course there is a less "stringent" mathematics version. How much less mathematically you want to go?. You can "avoid" calculus in the following manner: Use increments of $0.1$ for $x$ and calculate the corresponding $y's$ for each increasing $x$. Then use Pythagorean theorem $100$ times $(10/0.1)$ and add up distances. That would be a good approximation. It's a little work but not difficult. You still need to set up your sine formula, also not difficult. (This answers the first part of your question)
$endgroup$
– imranfat
Nov 3 '17 at 16:43
$begingroup$
I was hoping for a formula which doesn't require calculus (by which i mean, for example, integrating functions - sorry if that's not calculus, i'm not a mathematician), and into which I can just plug some numbers and carry out only simple operations. Imagine I'm trying to do it in a computer program where I can only do basic operations, raising to a power, sin, cos and tan and their converses.
$endgroup$
– Max Williams
Nov 3 '17 at 16:47
$begingroup$
I was hoping for a formula which doesn't require calculus (by which i mean, for example, integrating functions - sorry if that's not calculus, i'm not a mathematician), and into which I can just plug some numbers and carry out only simple operations. Imagine I'm trying to do it in a computer program where I can only do basic operations, raising to a power, sin, cos and tan and their converses.
$endgroup$
– Max Williams
Nov 3 '17 at 16:47
$begingroup$
My second approach in my previous comment does not require calculus. The Pythagorean theorem is what you need (remember that famous $a^2+b^2=c^2$ on a right triangle from high school?). I think that is as basic as you can go. To calculate a distance between two points, the formula translates to $c=sqrt{(0.1)^2+(y_2-y_1)^2}$. You need to do that 100 times as you have 100 increments. Then add up. Having something like Excel available would be tremendously helpful
$endgroup$
– imranfat
Nov 3 '17 at 16:49
$begingroup$
My second approach in my previous comment does not require calculus. The Pythagorean theorem is what you need (remember that famous $a^2+b^2=c^2$ on a right triangle from high school?). I think that is as basic as you can go. To calculate a distance between two points, the formula translates to $c=sqrt{(0.1)^2+(y_2-y_1)^2}$. You need to do that 100 times as you have 100 increments. Then add up. Having something like Excel available would be tremendously helpful
$endgroup$
– imranfat
Nov 3 '17 at 16:49
$begingroup$
Ok, thanks. I guess I was too optimistic for a simple one-hit equation then!
$endgroup$
– Max Williams
Nov 3 '17 at 16:53
$begingroup$
Ok, thanks. I guess I was too optimistic for a simple one-hit equation then!
$endgroup$
– Max Williams
Nov 3 '17 at 16:53
|
show 2 more comments
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