Literally just finished doing this out the back garden. The wife has been on to me now for a while to finish the path. For some reason I was just completely stumped as how to make the curve work. Can't thank you enough for finally giving me the answer. Fantastic video.
+TheHidiho Tickled pink that I was able to help. Actually learned this formula from having to build archways for sets from drawings that gave no dimensions, except maybe the width of the arch.
+donnydoors Well Sir, you have been a life saver. Only problem is now that it has been done so well, she will recommend me to do her own friends paths. Which, with what I have learned from you, I'll be more than able to do. So maybe a blessing in disguise? Once again, I can't begin to thank you enough.
+TheHidiho I admit, it gives me great satisfaction to know that one of my videos, which was simply an off the cuff choice to demonstrate some geometric math, has been used to practical effect. My thanks back to you.
+brian vanderbrug I made use of this formula as a set builder, but I wasn't sure how to illustrate in in a video. And here you've proven that my idea wasn't so off the wall after all. You're more than welcome.
I am in the process of doing my curved walkway, and I have been stumped on how to lay out the curves to as accurately as possible. I cannot begin to tell you how helpful this video is going to be for me in doing so. Thanks so much for this.
Terrific. BTW, I have an ebook (PDF) that covers the garden layout and more. Check out my web site (credited in door videos) and send me an email. I'll set you up.
Your formula is derived from a more basic and very useful circle formula called the law of intersecting chords. I don't know who passed it, whether it be Solon, Polonius, Hipparchus, Aristotle, Ptolmey, Phythogoras or Zorba, but it holds for all circles among men. It goes like this, When in a Traffic Circle in Los Angles The two intersecting streets lay along intersecting chords. At the end of the traffic circle where the two chords end they form two parts bisected in this case by the centre of the circle. The law holds for most towns whether the streets cross in the exact center. The length of the two parts of one street, (contained within the traffic circle) when multiplied together equals the lengths of the two bisected streets coming the other way similarly multiplied together. And this is on government holidays too -and- the unions have agreed to be so bound. So any two chords have this property that intersect at any angle. The diameter luckily is such a chord too. So the height of our chord (the distance to the circumference from its so bisected center) multiplied by its other part that line -- which goes from the chord center thru the circle's center, to the other side of the circle -- is the same QUANTITY as the two halves of the LAID-OUT chord-in-question WHEN multiplied by each other. Neither chord has to go thru the center FOR THE multiplication to apply. That in our case we make one chord do so (by law of Hipparchus) is our luck.. a trick we gardeners employ) It does not invalidate the law that we use a diameter as a chord. In Los Angles and most other bisectoral towns diameter is just another citizen chord. Lets try it.. to find a radius. Not hard. From your example we have a chord length of 7 feet. Two halves are 3.5 3.5 X 3.5 = 12.25 Now our height is mentioned at 2.5 feet. 12.25 /2.5 = 4.9 feet. We divided the product we obtained from the two halves of the one chord by the "other half" of the other chord, the diameter, to obtain the answer to its left over part of its total length. The chord product-pairs are equal if we take their lengths to the circle's circumference from their mutual points of bisection. (Here I am saying the "other half" is the chord height. So I am using the parlance that there are two "halfs" of the chord on either side of its bisecting chord that are not necessarily equal...) Now 4.9 feet is the "another half", so to speak, of our crossing-chord. i.e. The "diameter chord". i.e. The one that contains our chord height.. which, since it intersects the middle of the other chord at right angles must pass thru the center of the circle. Add 4.9 to 2.6, the height of our chord, and we have 7.4 feet. Half of that, since we have just summed the circles diameter -- is our radius. 3.7 feet.
I'm sure this method is useful for other situations as well, but it seems like you could have skipped a few steps toward the end. Once you knew the radius and measured the string, you could have planted a stake at the apex, brought the other stake down directly across the mid point between the length, and staked it there. It should give the exact same results. But like I said, the ending steps you described are probably more versatile and accurate. Great video(s).
+Andrew Morales A square line down from the apex the length of the radius? But how do you keep it square. The arcs from each corner guarantee the placement.Thanks for the compliment.
donnydoors Well, the same way you made your apex point. It'd be easier to show than describe, but it doesn't really matter. Your method is foolproof, which is more important. Also, I'm glad to see you're still active on RUclips. I'd love to see more videos from you.
+donnydoors Well I'm very happy to hear that. I love informative videos, and especially informative videos that are very clear and direct, and even more so, informative videos that are very clear and direct and that also have a distinct personality. Your videos check all three boxes.
The arc tangent of the difference between the radius and the chord height divided by half the chord length is half of half the chord angle. So if the chord is 60 degrees, half of that is 30, and the small triangle at the top of the chord formed by the half the chord "base" and the height of the chord (in the orthogonal direction of the radius to the circumference of the circle from the chord middle.) has its acute angle of half the half angle of the angle containing the entire chord. In this case 15 degrees. This allows you to layout chords.. How do you get a reasonable chord? Simple, layout a suitable chord length and a chord height which is equal to 0.268 of half the chord length. Make the half chord length 5 feet lets say. The height is 1 foot 4 inches. The chord is ten feet. The radius of our arc is 10 feet also as it happens. The arc is 60 degrees, or 1.6 of a circle. The half arc is 30 degrees.
I took this from surveyor's chord formulae. Layout a circle. Draw two radii in a pie shape with an acute angle between them. Draw in a flat line joining the ends where they intersect the circumference.. natch .. as the other ends meet at the center. Draw in a another radius half way between them, splitting or bisecting the angle. We have two equal triangles. The "chord angle" is the full angle so to speak, between the two outer radii. The half angles are the angles of the two right triangles formed by the center radius and the two outer radii respectively. Since they are right triangles we can find the length of the chord quickly by Sin formula. We know the radii's length, and we either select a convenient chord length, -or- we know the included angle between the two outer radii, thus half of it, the angle we want the Sin of, can be calculated by division of the chord angle by two. We assume we know the chord angle here. Our chord angle is let's say 45 degrees. 'our chosen radius is 10 feet. Our half angle, the angle of the right triangles is 22.5 degrees. Our Sin is 0.382683. So half our chord length is 3.83 feet. Our chord is 7.654 feet. Now we construct another two triangles by drawing two lines -- from the end of the chord line or line which joins the two outer radii ends at the circumference -- to the center radius where it intersects the circumference. These two equal triangles sit on top of the other two between the chord and the circumference forming a "hat" on the chord. It happens that these two triangles forming a hat on the top of the chord have their acute angles -- (formed by the chord and their line of hypotenuse joining with, or intersecting the far end of the middle radius) -- being HALF the half-angle of our original two triangles -- which we remember were formed by our first construction (with two radii, the middle radius and the chord line.) We may need to know the depth of the chord. Simply we may that take this by example from a suggested half-angle of 22.5 degrees. Our half chord is therfore by calculation with a SIN of the half angle = to 3.83 feet. ( Sin of 22.5 degrees times the known radius of 10 feet ) Our "half of the half angle" for our top two triangles (forming the hat of our chord) is 11.25 degrees. Tangent of this is 0.198912 Times 3.83 feet is 0.761 feet. So our distance from the chord to the circumference is 0.761 feet, Layout a line 7 feet 8 inches, then from the middle layout a line 9 inches at right angles and layout what is called a fair curve (by eye) between the ends of the lines, the 7 foot 8 inch line and our 9 inch line. *********************************** Where we don't know the chord angle.. but we can eyeball the chord length and our offset from an obstruction so to speak. You can obviously work in reverse. Take a selected chord. Take a selected offset, which we define as the "tangent distance" from the chord line to the circumference at half way along the chord line.. we then calculate the radius. Here is how: Let's take 15 feet as our full chord, Take 2 feet as our offset to the circumference at the middle of the chord. This in your case is the distance from to our wall. This creates 2/7.5 as a "half the half angle tangent." The arc or inverse tangent of 2/15 = 14.931 degrees which is: "half the (chord) half-angle. " The half-angle of our chord is therefore 29.863 degrees. The Sin of this angle is 0.497925. Divide half the chord, or 7.5 feet by this SIN and we get 15.0625 feet for our radius. If you are confused please accept my congratulations. You are normal. IF you understood it you would be mathematically schizoid. What I need is a better understanding of sketch up so I can draw illustrations.
Well, you could start by drawing to scale what you have in mind . Then measure the position of the apex: 40' over from where the path hits the wall/property line. Then from there to the whatever distance to the apex the scaled drawing tells you. Five feet over measure out the distance from the wall to the point where the circle intersects that line Do this three more times to the right and five times to the left. When you have enough points you simply hand draw the curve of the circle point to point to point. If five points on either side isn't enough for a decent curve, add extra points in between. To simplify, find the radius point using the formula, Then draw out the curve to scale. Measure from the wall multiple points on the curve. When there are enough points, hand draw the curve from point to point. Hope this helps.
Just my kind of maths, Mr Doors. All prepped now to lay out my new curved patio. Many thanks for sharing and making me laugh, too!
Boy, is it rare I get a laugh. Thanks. Hope you make good use of this.
Literally just finished doing this out the back garden. The wife has been on to me now for a while to finish the path. For some reason I was just completely stumped as how to make the curve work. Can't thank you enough for finally giving me the answer. Fantastic video.
+TheHidiho Tickled pink that I was able to help. Actually learned this formula from having to build archways for sets from drawings that gave no dimensions, except maybe the width of the arch.
+donnydoors Well Sir, you have been a life saver. Only problem is now that it has been done so well, she will recommend me to do her own friends paths. Which, with what I have learned from you, I'll be more than able to do. So maybe a blessing in disguise? Once again, I can't begin to thank you enough.
+TheHidiho I admit, it gives me great satisfaction to know that one of my videos, which was simply an off the cuff choice to demonstrate some geometric math, has been used to practical effect. My thanks back to you.
I just laid out the curves for my backyard and it looks awesome! Thanks so much for posting the formula!
+brian vanderbrug I made use of this formula as a set builder, but I wasn't sure how to illustrate in in a video. And here you've proven that my idea wasn't so off the wall after all. You're more than welcome.
I am in the process of doing my curved walkway, and I have been stumped on how to lay out the curves to as accurately as possible. I cannot begin to tell you how helpful this video is going to be for me in doing so. Thanks so much for this.
Terrific. BTW, I have an ebook (PDF) that covers the garden layout and more. Check out my web site (credited in door videos) and send me an email. I'll set you up.
Thank you Donny, you helped me solve a tricky planning task.
Terrific. Hope I didn't over complicate. Put that one and more in a book bound to be out some day soon.
donnydoors you should, I was searching Maths sites and couldn’t find the answer, until I found you. Thanks again.
Great videos ! Very helpful
Glad you enjoyed them.
this was extremely interesting to watch
+AdiusOmega I'm so glad to hear that. So it's not just a movie of an old duff stumbling around in a patch of dirt.
You remind me of napoleon hill,great video. Your a funny guy!
I had to look up Napoleon. Were I only as prolific. Thanks for the compliment. Hopefully, your viewing was beneficial,
Amazing. I love it!
Your formula is derived from a more basic and very useful circle formula called the law of intersecting chords. I don't know who passed it, whether it be Solon, Polonius, Hipparchus, Aristotle, Ptolmey, Phythogoras or Zorba, but it holds for all circles among men.
It goes like this, When in a Traffic Circle in Los Angles The two intersecting streets lay along intersecting chords. At the end of the traffic circle where the two chords end they form two parts bisected in this case by the centre of the circle.
The law holds for most towns whether the streets cross in the exact center. The length of the two parts of one street, (contained within the traffic circle) when multiplied together equals the lengths of the two bisected streets coming the other way similarly multiplied together. And this is on government holidays too -and- the unions have agreed to be so bound.
So any two chords have this property that intersect at any angle. The diameter luckily is such a chord too. So the height of our chord (the distance to the circumference from its so bisected center) multiplied by its other part that line -- which goes from the chord center thru the circle's center, to the other side of the circle -- is the same QUANTITY as the two halves of the LAID-OUT chord-in-question WHEN multiplied by each other.
Neither chord has to go thru the center FOR THE multiplication to apply. That in our case we make one chord do so (by law of Hipparchus) is our luck.. a trick we gardeners employ) It does not invalidate the law that we use a diameter as a chord. In Los Angles and most other bisectoral towns diameter is just another citizen chord.
Lets try it.. to find a radius. Not hard. From your example we have a chord length of 7 feet. Two halves are 3.5 3.5 X 3.5 = 12.25 Now our height is mentioned at 2.5 feet. 12.25 /2.5 = 4.9 feet.
We divided the product we obtained from the two halves of the one chord by the "other half" of the other chord, the diameter, to obtain the answer to its left over part of its total length.
The chord product-pairs are equal if we take their lengths to the circle's circumference from their mutual points of bisection.
(Here I am saying the "other half" is the chord height. So I am using the parlance that there are two "halfs" of the chord on either side of its bisecting chord that are not necessarily equal...)
Now 4.9 feet is the "another half", so to speak, of our crossing-chord. i.e. The "diameter chord". i.e. The one that contains our chord height.. which, since it intersects the middle of the other chord at right angles must pass thru the center of the circle. Add 4.9 to 2.6, the height of our chord, and we have 7.4 feet. Half of that, since we have just summed the circles diameter -- is our radius. 3.7 feet.
I'm sure this method is useful for other situations as well, but it seems like you could have skipped a few steps toward the end. Once you knew the radius and measured the string, you could have planted a stake at the apex, brought the other stake down directly across the mid point between the length, and staked it there. It should give the exact same results. But like I said, the ending steps you described are probably more versatile and accurate. Great video(s).
+Andrew Morales A square line down from the apex the length of the radius? But how do you keep it square. The arcs from each corner guarantee the placement.Thanks for the compliment.
donnydoors Well, the same way you made your apex point. It'd be easier to show than describe, but it doesn't really matter. Your method is foolproof, which is more important.
Also, I'm glad to see you're still active on RUclips. I'd love to see more videos from you.
+Andrew Morales Keep an eye out. All this Reddit and Digg activity (which has greatly subsided) forces me to get out another video ASAP.
+donnydoors Well I'm very happy to hear that. I love informative videos, and especially informative videos that are very clear and direct, and even more so, informative videos that are very clear and direct and that also have a distinct personality. Your videos check all three boxes.
+Andrew Morales Well, I must express my gratitude for such a compliment.
Ha, ha, ha, ha, ha :) Thanks for sharing that info. You made me laugh.
The arc tangent of the difference between the radius and the chord height divided by half the chord length is half of half the chord angle. So if the chord is 60 degrees, half of that is 30, and the small triangle at the top of the chord formed by the half the chord "base" and the height of the chord (in the orthogonal direction of the radius to the circumference of the circle from the chord middle.) has its acute angle of half the half angle of the angle containing the entire chord. In this case 15 degrees.
This allows you to layout chords.. How do you get a reasonable chord? Simple, layout a suitable chord length and a chord height which is equal to 0.268 of half the chord length. Make the half chord length 5 feet lets say. The height is 1 foot 4 inches. The chord is ten feet. The radius of our arc is 10 feet also as it happens. The arc is 60 degrees, or 1.6 of a circle. The half arc is 30 degrees.
Gotta re-watch the video and go over your text to fully comprehend this.
I took this from surveyor's chord formulae.
Layout a circle. Draw two radii in a pie shape with an acute angle between them. Draw in a flat line joining the ends where they intersect the circumference.. natch .. as the other ends meet at the center.
Draw in a another radius half way between them, splitting or bisecting the angle. We have two equal triangles.
The "chord angle" is the full angle so to speak, between the two outer radii. The half angles are the angles of the two right triangles formed by the center radius and the two outer radii respectively.
Since they are right triangles we can find the length of the chord quickly by Sin formula. We know the radii's length, and we either select a convenient chord length, -or- we know the included angle between the two outer radii, thus half of it, the angle we want the Sin of, can be calculated by division of the chord angle by two. We assume we know the chord angle here.
Our chord angle is let's say 45 degrees. 'our chosen radius is 10 feet. Our half angle, the angle of the right triangles is 22.5 degrees. Our Sin is 0.382683. So half our chord length is 3.83 feet. Our chord is 7.654 feet.
Now we construct another two triangles by drawing two lines -- from the end of the chord line or line which joins the two outer radii ends at the circumference -- to the center radius where it intersects the circumference.
These two equal triangles sit on top of the other two between the chord and the circumference forming a "hat" on the chord.
It happens that these two triangles forming a hat on the top of the chord have their acute angles -- (formed by the chord and their line of hypotenuse joining with, or intersecting the far end of the middle radius) -- being HALF the half-angle of our original two triangles -- which we remember were formed by our first construction (with two radii, the middle radius and the chord line.)
We may need to know the depth of the chord. Simply we may that take this by example from a suggested half-angle of 22.5 degrees. Our half chord is therfore by calculation with a SIN of the half angle = to 3.83 feet. ( Sin of 22.5 degrees times the known radius of 10 feet )
Our "half of the half angle" for our top two triangles (forming the hat of our chord) is 11.25 degrees. Tangent of this is 0.198912 Times 3.83 feet is 0.761 feet.
So our distance from the chord to the circumference is 0.761 feet, Layout a line 7 feet 8 inches, then from the middle layout a line 9 inches at right angles and layout what is called a fair curve (by eye) between the ends of the lines, the 7 foot 8 inch line and our 9 inch line.
***********************************
Where we don't know the chord angle.. but we can eyeball the chord length and our offset from an obstruction so to speak.
You can obviously work in reverse. Take a selected chord. Take a selected offset, which we define as the "tangent distance" from the chord line to the circumference at half way along the chord line.. we then calculate the radius.
Here is how:
Let's take 15 feet as our full chord, Take 2 feet as our offset to the circumference at the middle of the chord. This in your case is the distance from to our wall.
This creates 2/7.5 as a "half the half angle tangent." The arc or inverse tangent of 2/15 = 14.931 degrees which is: "half the (chord) half-angle. "
The half-angle of our chord is therefore 29.863 degrees. The Sin of this angle is 0.497925. Divide half the chord, or 7.5 feet by this SIN and we get 15.0625 feet for our radius.
If you are confused please accept my congratulations. You are normal. IF you understood it you would be mathematically schizoid. What I need is a better understanding of sketch up so I can draw illustrations.
Rock star!
LOL!
“Out of my way, son!”
Not much use if your path radius is 80 foot and that centre point is in your neighbours back garden 😀
Well, you could start by drawing to scale what you have in mind . Then measure the position of the apex: 40' over from where the path hits the wall/property line. Then from there to the whatever distance to the apex the scaled drawing tells you. Five feet over measure out the distance from the wall to the point where the circle intersects that line Do this three more times to the right and five times to the left. When you have enough points you simply hand draw the curve of the circle point to point to point. If five points on either side isn't enough for a decent curve, add extra points in between.
To simplify, find the radius point using the formula, Then draw out the curve to scale. Measure from the wall multiple points on the curve. When there are enough points, hand draw the curve from point to point.
Hope this helps.
I prefer to build with woodglut plans.