Re: [Flashcoders] extending built-in classes and scope
if you are extending the xml classes to get search capability look at
www.xfactorstudio.com 's xpath implementation
On 9/16/06, dc [EMAIL PROTECTED] wrote:
hi list -
I am trying to add some functions to built in classes.
Q1) Using the prototype syntax. Is this an AS1 method, is there a
better way to do this with AS2?
Q2) this works within my main movie script.
however, when i try to use the extended XML object within my own
classes, the compiler fails with a no such method error.
is there a scoping issue, or somehow i have to redefine the XML object
prototype within my own class, Again?
main timeline this is OK:
XMLNode.prototype.findFirstNode = function (searchName) {
trace(searching for: + searchName);
}
xml = new XMLNode;
xml.findFirstNode(test); // fine up to here.
import pikkle.WireMenu; // fails here
--
inside the pikkle.WireMenu i have exactly the same code,
xml.findFirstNode(test);
and there it fails, as if it forgot that it had added extra methods to
the XMLNode class.
thanks for any tips!
/dc
---
David DC Collier
mailto:[EMAIL PROTECTED]
+81 (0)80 6521 9559
skype: callto://d3ntaku
---
Pikkle 株式会社
http://www.pikkle.com
---
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http://www.lennel.org
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Re: [Flashcoders] specifying type of an array contents?
On 9/17/06, dc [EMAIL PROTECTED] wrote: is there a way to tell flash what types an array contains? Not in ActionScript, unless you make some custom methods that access it and check for the type at runtime. In haXe, however, you can do just that. It can compile for FlashPlayer 6/7/8/9, into existing SWFs (or create them if you want): haxe.org The for..in loop also works like you described, and it has some other really nice features. Mark ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com
Re: [Flashcoders] extending built-in classes and scope
seems like your pikkle.WireMenu class got imported before the compiler
reached your timeline code to add findFirstNode method to the XML object.
Try adding your method, using a class ... Or, may be, try wrapping your code
to add that method to XML object's prototype in a initclip/ endinitclip
pair.
~Arul Prasad.
On 9/17/06, dc [EMAIL PROTECTED] wrote:
hi list -
I am trying to add some functions to built in classes.
Q1) Using the prototype syntax. Is this an AS1 method, is there a
better way to do this with AS2?
Q2) this works within my main movie script.
however, when i try to use the extended XML object within my own
classes, the compiler fails with a no such method error.
is there a scoping issue, or somehow i have to redefine the XML object
prototype within my own class, Again?
main timeline this is OK:
XMLNode.prototype.findFirstNode = function (searchName) {
trace(searching for: + searchName);
}
xml = new XMLNode;
xml.findFirstNode(test); // fine up to here.
import pikkle.WireMenu; // fails here
--
inside the pikkle.WireMenu i have exactly the same code,
xml.findFirstNode(test);
and there it fails, as if it forgot that it had added extra methods to
the XMLNode class.
thanks for any tips!
/dc
---
David DC Collier
mailto:[EMAIL PROTECTED]
+81 (0)80 6521 9559
skype: callto://d3ntaku
---
Pikkle 株式会社
http://www.pikkle.com
---
___
Flashcoders@chattyfig.figleaf.com
To change your subscription options or search the archive:
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Re: [Flashcoders] specifying type of an array contents?
Also, if you dont want to go all the way to haXe, MTASC has added the ability to indicate the type of an array. I forget the exact syntax, but you put the type in a comment. Regards Hank On 9/17/06, Mark Winterhalder [EMAIL PROTECTED] wrote: On 9/17/06, dc [EMAIL PROTECTED] wrote: is there a way to tell flash what types an array contains? Not in ActionScript, unless you make some custom methods that access it and check for the type at runtime. In haXe, however, you can do just that. It can compile for FlashPlayer 6/7/8/9, into existing SWFs (or create them if you want): haxe.org The for..in loop also works like you described, and it has some other really nice features. Mark ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com
[Flashcoders] Preloader or not
Hi there, I used a tutorial on Adobe.com to create a preloader from some tutorial files. Works great, but the preloader displays the animation ALWAYS. Even if there's nothing to preload, cause it has been loaden already. Is there any way to determin if the content already has been loaded and then go to content, skipping the animation? Kind regards, Ben ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com
Re: [Flashcoders] Preloader or not
Normally, it already checks for 'getBytesLoaded() == getBytsTotal()' (or a variation). Could you post your ActionScript? Benjamin van Gogh wrote: Hi there, I used a tutorial on Adobe.com to create a preloader from some tutorial files. Works great, but the preloader displays the animation ALWAYS. Even if there's nothing to preload, cause it has been loaden already. Is there any way to determin if the content already has been loaded and then go to content, skipping the animation? Kind regards, Ben ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com
[Flashcoders] a question of geometry
smart people, help a less smart one out: I'm faced with a problem similar to the seven circles theorem (http://mathworld.wolfram.com/SevenCirclesTheorem.html) in that i need to arrange x number of circles along a circular path, making sure they all touch their two neighbors. In this way, the path's radius is NOT given, but is rather made up from the sum of all the circles' diametres. I've boiled the problem down to this: I have a line of x length with y number of segments of nonuniform length. I know the final length of the line because i know the length of each individual segment. Now, i need to bend this line so that the end of the final segment touches on the beginning of the first one. As such, each segment must be given an angle somehow based on the overall amount of segments and their individual lengths. Beyond this, i'm stumped. I've been pouring over mathworld and google looking for such bendyness, and i've come up empty handed. Anyone have suggestions, possible solutions? Thanks, - A ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com
Re: [Flashcoders] a question of geometry
Ok, are the circles all the same size, or do they have the same length? Alias On 17/09/06, Andreas R [EMAIL PROTECTED] wrote: smart people, help a less smart one out: I'm faced with a problem similar to the seven circles theorem (http://mathworld.wolfram.com/SevenCirclesTheorem.html) in that i need to arrange x number of circles along a circular path, making sure they all touch their two neighbors. In this way, the path's radius is NOT given, but is rather made up from the sum of all the circles' diametres. I've boiled the problem down to this: I have a line of x length with y number of segments of nonuniform length. I know the final length of the line because i know the length of each individual segment. Now, i need to bend this line so that the end of the final segment touches on the beginning of the first one. As such, each segment must be given an angle somehow based on the overall amount of segments and their individual lengths. Beyond this, i'm stumped. I've been pouring over mathworld and google looking for such bendyness, and i've come up empty handed. Anyone have suggestions, possible solutions? Thanks, - A ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com
Re: [Flashcoders] a question of geometry
The circles can be any size. Alias™ wrote: Ok, are the circles all the same size, or do they have the same length? Alias On 17/09/06, Andreas R [EMAIL PROTECTED] wrote: smart people, help a less smart one out: I'm faced with a problem similar to the seven circles theorem (http://mathworld.wolfram.com/SevenCirclesTheorem.html) in that i need to arrange x number of circles along a circular path, making sure they all touch their two neighbors. In this way, the path's radius is NOT given, but is rather made up from the sum of all the circles' diametres. I've boiled the problem down to this: I have a line of x length with y number of segments of nonuniform length. I know the final length of the line because i know the length of each individual segment. Now, i need to bend this line so that the end of the final segment touches on the beginning of the first one. As such, each segment must be given an angle somehow based on the overall amount of segments and their individual lengths. Beyond this, i'm stumped. I've been pouring over mathworld and google looking for such bendyness, and i've come up empty handed. Anyone have suggestions, possible solutions? Thanks, - A ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com
Re: [Flashcoders] a question of geometry
And does the number of circles vary? Will it always be the same number? On 17/09/06, Andreas R [EMAIL PROTECTED] wrote: The circles can be any size. Alias™ wrote: Ok, are the circles all the same size, or do they have the same length? Alias On 17/09/06, Andreas R [EMAIL PROTECTED] wrote: smart people, help a less smart one out: I'm faced with a problem similar to the seven circles theorem (http://mathworld.wolfram.com/SevenCirclesTheorem.html) in that i need to arrange x number of circles along a circular path, making sure they all touch their two neighbors. In this way, the path's radius is NOT given, but is rather made up from the sum of all the circles' diametres. I've boiled the problem down to this: I have a line of x length with y number of segments of nonuniform length. I know the final length of the line because i know the length of each individual segment. Now, i need to bend this line so that the end of the final segment touches on the beginning of the first one. As such, each segment must be given an angle somehow based on the overall amount of segments and their individual lengths. Beyond this, i'm stumped. I've been pouring over mathworld and google looking for such bendyness, and i've come up empty handed. Anyone have suggestions, possible solutions? Thanks, - A ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com
Re: [Flashcoders] a question of geometry
To reiterate, there can be *any* number of circles of *any* size. The aim is to create a circular chain of circles where the overall diameter is a result of the circumference of the chain. Alias™ wrote: And does the number of circles vary? Will it always be the same number? On 17/09/06, Andreas R [EMAIL PROTECTED] wrote: The circles can be any size. Alias™ wrote: Ok, are the circles all the same size, or do they have the same length? Alias On 17/09/06, Andreas R [EMAIL PROTECTED] wrote: smart people, help a less smart one out: I'm faced with a problem similar to the seven circles theorem (http://mathworld.wolfram.com/SevenCirclesTheorem.html) in that i need to arrange x number of circles along a circular path, making sure they all touch their two neighbors. In this way, the path's radius is NOT given, but is rather made up from the sum of all the circles' diametres. I've boiled the problem down to this: I have a line of x length with y number of segments of nonuniform length. I know the final length of the line because i know the length of each individual segment. Now, i need to bend this line so that the end of the final segment touches on the beginning of the first one. As such, each segment must be given an angle somehow based on the overall amount of segments and their individual lengths. Beyond this, i'm stumped. I've been pouring over mathworld and google looking for such bendyness, and i've come up empty handed. Anyone have suggestions, possible solutions? Thanks, - A ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com
Re: [Flashcoders] a question of geometry
I think your problem is not possible. Here is why (hope I am able to articulate). Some basic background info: - The diameter of a circle is the distance from one edge of the circle to the furthest edge. - A line of circles edge to edge, so that their diameters are used, makes a strait line. - The radius of a circle is the distance from the center to any edge. Lets do some simple math... Diameter 1 = 10 Diameter 2 = 8 Diameter 3 = 7 Diameter 4 = 5 Thus, their sum = 30. Now lets use this as the radius of our big circle (BC). BCc = 2 * 3.14 * 30 = 188.4 (where BCc is the big circle circumference) BCd = 2 * 30 = 60 (where BCd is the big circle diameter) First lets look at placing the original circle centers on the big circle's path, so that their edges touch. It will quickly become clear that this is not going to work. 188.4 / 4 = 47.1 The sum of the original four diameters would have to equal 47.1. This is obviously not correct. Let see if we can do this by keeping all the circles inside the big circle. So, not only do we want to have all four of our original circles be touching, but their tangents need to touch the big circle as well. In addition to all of this the last circle needs to also touch the first circle. The most optimal form of this would assume that all four of our original circles have the same diameter. The sum of these circles would equal 2.4times the big circles diameter. Remember thought that this woulod be the optimal. So lets take a look at our numbers. Optimal diameter sum = 2.41 * 60 = 144.6 ** not needed info** Each unit circle's diameter = 144.6 / 4 = 36.15 So this means that the sum of our original circles has to equal at least 144.6 or greater. Our circles diameter sum = 30 which is far less than the required 144.6. See Circle Packing for optimal values. http://mathworld.wolfram.com/CirclePacking.html I hope that was somw what clear. Charles P On 9/17/06, Andreas R [EMAIL PROTECTED] wrote: i need to arrange x number of circles along a circular path, making sure they all touch their two neighbors. In this way, the path's radius is NOT given, but is rather made up from the sum of all the circles' diametres. ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com
Re: [Flashcoders] a question of geometry
From what is said, is it true that you want to take a large circle, divide it up into a random number of pie shaped slices and then use the outer edges of the pies as diameters of smaller circles so that when the smaller circles are drawn, they form a circle of circles. If so you need to decide how to break 360 degrees up into a number of angles. Start at the 0 point (wherever you want the start to be) and find the intersection a line drawn from the center of the circle to the edge and then do the same for the next radius. The 2 points form the diameter of the first small circle. The small circle's radius is half the distance between the points and the centre of the small circle is at the midpoint between these 2 points. Do this for each pair of radius lines and you will have a circle of circles. Ron Andreas R wrote: To reiterate, there can be *any* number of circles of *any* size. The aim is to create a circular chain of circles where the overall diameter is a result of the circumference of the chain. Alias™ wrote: And does the number of circles vary? Will it always be the same number? On 17/09/06, Andreas R [EMAIL PROTECTED] wrote: The circles can be any size. Alias™ wrote: Ok, are the circles all the same size, or do they have the same length? Alias On 17/09/06, Andreas R [EMAIL PROTECTED] wrote: smart people, help a less smart one out: I'm faced with a problem similar to the seven circles theorem (http://mathworld.wolfram.com/SevenCirclesTheorem.html) in that i need to arrange x number of circles along a circular path, making sure they all touch their two neighbors. In this way, the path's radius is NOT given, but is rather made up from the sum of all the circles' diametres. I've boiled the problem down to this: I have a line of x length with y number of segments of nonuniform length. I know the final length of the line because i know the length of each individual segment. Now, i need to bend this line so that the end of the final segment touches on the beginning of the first one. As such, each segment must be given an angle somehow based on the overall amount of segments and their individual lengths. Beyond this, i'm stumped. I've been pouring over mathworld and google looking for such bendyness, and i've come up empty handed. Anyone have suggestions, possible solutions? Thanks, - A ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com
Re: [Flashcoders] a question of geometry
These circles will overlap each other. If this is not what you want, then Charles is right. It is not possible to do what Andreas seems to be describing. Perhaps Andreas should draw a picture by hand of what he actually wants this to look like. Ron Ron Wheeler wrote: From what is said, is it true that you want to take a large circle, divide it up into a random number of pie shaped slices and then use the outer edges of the pies as diameters of smaller circles so that when the smaller circles are drawn, they form a circle of circles. If so you need to decide how to break 360 degrees up into a number of angles. Start at the 0 point (wherever you want the start to be) and find the intersection a line drawn from the center of the circle to the edge and then do the same for the next radius. The 2 points form the diameter of the first small circle. The small circle's radius is half the distance between the points and the centre of the small circle is at the midpoint between these 2 points. Do this for each pair of radius lines and you will have a circle of circles. Ron Andreas R wrote: To reiterate, there can be *any* number of circles of *any* size. The aim is to create a circular chain of circles where the overall diameter is a result of the circumference of the chain. Alias™ wrote: And does the number of circles vary? Will it always be the same number? On 17/09/06, Andreas R [EMAIL PROTECTED] wrote: The circles can be any size. Alias™ wrote: Ok, are the circles all the same size, or do they have the same length? Alias On 17/09/06, Andreas R [EMAIL PROTECTED] wrote: smart people, help a less smart one out: I'm faced with a problem similar to the seven circles theorem (http://mathworld.wolfram.com/SevenCirclesTheorem.html) in that i need to arrange x number of circles along a circular path, making sure they all touch their two neighbors. In this way, the path's radius is NOT given, but is rather made up from the sum of all the circles' diametres. I've boiled the problem down to this: I have a line of x length with y number of segments of nonuniform length. I know the final length of the line because i know the length of each individual segment. Now, i need to bend this line so that the end of the final segment touches on the beginning of the first one. As such, each segment must be given an angle somehow based on the overall amount of segments and their individual lengths. Beyond this, i'm stumped. I've been pouring over mathworld and google looking for such bendyness, and i've come up empty handed. Anyone have suggestions, possible solutions? Thanks, - A ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and
Re: [Flashcoders] a question of geometry
It's important to remember i don't expect a perfect circle. Perhaps my terminology is weak, i mean a ring of circles http://andreas.rayon.no/temp/circleOfCircles.png - A Ron Wheeler wrote: These circles will overlap each other. If this is not what you want, then Charles is right. It is not possible to do what Andreas seems to be describing. Perhaps Andreas should draw a picture by hand of what he actually wants this to look like. Ron Ron Wheeler wrote: From what is said, is it true that you want to take a large circle, divide it up into a random number of pie shaped slices and then use the outer edges of the pies as diameters of smaller circles so that when the smaller circles are drawn, they form a circle of circles. If so you need to decide how to break 360 degrees up into a number of angles. Start at the 0 point (wherever you want the start to be) and find the intersection a line drawn from the center of the circle to the edge and then do the same for the next radius. The 2 points form the diameter of the first small circle. The small circle's radius is half the distance between the points and the centre of the small circle is at the midpoint between these 2 points. Do this for each pair of radius lines and you will have a circle of circles. Ron Andreas R wrote: To reiterate, there can be *any* number of circles of *any* size. The aim is to create a circular chain of circles where the overall diameter is a result of the circumference of the chain. Alias™ wrote: And does the number of circles vary? Will it always be the same number? On 17/09/06, Andreas R [EMAIL PROTECTED] wrote: The circles can be any size. Alias™ wrote: Ok, are the circles all the same size, or do they have the same length? Alias On 17/09/06, Andreas R [EMAIL PROTECTED] wrote: smart people, help a less smart one out: I'm faced with a problem similar to the seven circles theorem (http://mathworld.wolfram.com/SevenCirclesTheorem.html) in that i need to arrange x number of circles along a circular path, making sure they all touch their two neighbors. In this way, the path's radius is NOT given, but is rather made up from the sum of all the circles' diametres. I've boiled the problem down to this: I have a line of x length with y number of segments of nonuniform length. I know the final length of the line because i know the length of each individual segment. Now, i need to bend this line so that the end of the final segment touches on the beginning of the first one. As such, each segment must be given an angle somehow based on the overall amount of segments and their individual lengths. Beyond this, i'm stumped. I've been pouring over mathworld and google looking for such bendyness, and i've come up empty handed. Anyone have suggestions, possible solutions? Thanks, - A ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com
Re: [Flashcoders] a question of geometry
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[Flashcoders] unscribe
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[Flashcoders] Flash Develop and XPath
Hi everyone, I´m trying to use XPathAPI from Flash Develop and I Can´t do I´t. Help me on it please ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com
Re: [Flashcoders] Save audio from webcam
On client it`s impossible 2006/9/18, David Clarke [EMAIL PROTECTED]: Hi Coders, Is there a way to save audio from a webcam (on the client) to the server for later use in flash (preferably mp3) without flash communication server? The scenario is that a user would record a message, which would be added to a larger library for all to browse and hear. Regards, David Clarke ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com
[Flashcoders] [Ann] Capivate 2.0 at Sydney Flash Platform Developers Group/NSW CFUG Joint meeting
Monday, 25th September, 6:30 for 7pm start. NSW Sports Club (www.flashdev.org.au/venue) Adobe Captivate 2 has just been released. Come and see how this favorite program has made creating engaging learning experience more fun and easier than ever before. Brian Chau (live and in person) from Adobe (Australia) will give us a demonstration on how simple it is to create software demonstration, simulation and scenario-based training all with simple point and click. This meeting will be a joint meeting with the Sydney Flash Platform Developers Group. Follow the Flash Developers signs. There'll be an open bar, complements of Adobe. Please rsvp on www.flashdev.org.au/rsvp. Chris -- Chris Velevitch Manager - Sydney Flash Platform Developers Group m: 0415 469 095 www.flashdev.org.au ___ Flashcoders@chattyfig.figleaf.com To change your subscription options or search the archive: http://chattyfig.figleaf.com/mailman/listinfo/flashcoders Brought to you by Fig Leaf Software Premier Authorized Adobe Consulting and Training http://www.figleaf.com http://training.figleaf.com
