Distance between options in an af:selectOneRadio

Hi
Is it possible to control the distance between the options in an af:selectOneRadio component?
I am currently making an af:selectOneRadio with horizontal layout. The users complain say that they would prefere a longer distance between the radio options and I have tried doing that by putting spaces in the end of the labels but the spaces are ignored and the distance therefore stays the same. How I can make this distance longer?

try this between af:selectRadio
<af:spacer width="4"/>regards
srini

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Find driving distance between two points without using API by use of Lat & Long?

Using Google geocode API : http://maps.googleapis.com/maps/api/geocode/xml?address=thane&sensor=true
We performed get distance between search criteria entered by user and all related clubs by lat & long stored at db.
2. Two different points such as
(origin: Lat1 & Long1) and (destination: Lat2 & Long2)
We tried for to get distance between these two points,
(Lat2 & Long2) to (Lat1 & Long1)
But distance which we get by calculation is simple straight line distance
Origin Destination
(Lat1 & Long1) (Lat2 & Long2)
3. This is not driving distance as google shows in exact Km
4. For that Google provide another API (distancematrix API)
http://maps.googleapis.com/maps/api/distancematrix/xml?origins=Thane&sensor=true&destinations=khopat&mode=driving&language=en%20-%20EN
5. But there is limit for DistanceMatrix-Service without ClientID and client key
100 elements per query.
100 elements per 10 seconds.
2 500 elements per 24 hour period.
But as element request exceeds it shows : OVER_QUERY_LIMIT error
6. In case of Client ID and Client key
In Distance Matrix 100 000 elements per 24 hour period,a maximum of 625 elements per query and a maximum of 1 000 elements per 10 seconds.
As per this one there is option to get purchase these API but basic question is remain same for us if we are requesting single origin and multiple destination then how element calculation done by google?
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Elements
The information about each origin-destination pairing is returned in an element entry. An element contains the following fields:
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Duration: The duration of this route, expressed in seconds (the value field) and as text. The textual representation is localized according to
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Hey,
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If you manually deleted the separator and replaced it with a different
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Hyphenation settings are controlled by the settings in the paragraph
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Is there an inbuilt function in PL/SQL for calulating the distance between two GPS (lat/long) points?
I'm using Oracle 9i. There's a thing called SDO_GEOM available, but I'm not sure if this is what its used for or if it's the best option.

If the earth is a complete globe and its circumference is 40000km,
I make the function as follows.
create ore replace
function distance
(a_lat number,a_lon number,b_lat number, b_lon number)
return number is
circum number := 40000; -- kilometers
pai number := acos(-1);
a_nx number;
a_ny number;
a_nz number;
b_nx number;
b_ny number;
b_nz number;
inner_product number;
begin
if (a_lat=b_lat) and (a_lon=b_lon) then
return 0;
else
a_nx := cos(a_lat*pai/180) * cos(a_lon*pai/180);
a_ny := cos(a_lat*pai/180) * sin(a_lon*pai/180);
a_nz := sin(a_lat*pai/180);
b_nx := cos(b_lat*pai/180) * co s(b_lon*pai/180);
b_ny := cos(b_lat*pai/180) * sin(b_lon*pai/180);
b_nz := sin(b_lat*pai/180);
inner_product := a_nx*b_nx + a_ny*b_ny + a_nz*b_nz;
if inner_product > 1 then
return 0;
else
return (circum*acos(inner_product))/(2*pai);
end if;
end if;
end;
I rewrite it by using the factorization and the triangle function's sum and difference formulas:
cos(x-y) = cos(x)cos(y)+sin(x)sin(y)
As result, this function is same to the 1st method of Billy Verreynne.
create or replace
function distance
(a_lat number,a_lon number,b_lat number, b_lon number)
return number is
circum number := 40000; -- kilometers
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dz number;
dx2y2 number;
inner_product number;
begin
if (a_lat=b_lat) and (a_lon=b_lon) then
return 0;
else
dz := sin(a_lat*pai/180)*sin(b_lat*pai/180);
dx2y2 := cos(a_lat*pai/180)*cos(b_lat*pai/180)*cos((a_lon-b_lon)*pai/180);
inner_product := dz*dz + dx2y2;
if inner_product > 1 then
return 0;
else
return (circum*acos(inner_product))/(2*pai);
end if;
end if;
end;
Message was edited by:
ushitaki

Diversity - Distance between Antennas

Considering both antennas are similar-type and gain the two imp criterias i read in achiveing Diversity is
1. Antennas should be placed close enough to each other so that the RF coverage area is nearly identical.
2. The receiving antennas are spaced sufficiently apart to achieve independence (no coupling) between the received signals.
So the distance between both the antenna would be important.
In a cisco doc,
http://www.cisco.com/en/US/tech/tk722/tk809/technologies_tech_note09186a008019f646.shtml
For 2.4GHz wavelength = 12.5cm and for 5GHz wavelength = 6cm.
Hence for 2.4GHz wavelength,
Dist bet diversity Ant = 12.5cms Or 25cms or 37.5cms or 50cms
now from below POST:
http://forums.cisco.com/eforum/servlet/NetProf?page=netprof&forum=Wireless%20-%20Mobility&topic=WLAN%20Radio%20Standards&topicID=.ee6e8c2&fromOutline=&CommCmd=MB%3Fcmd%3Ddisplay_location%26location%3D.2cc18267
The spacing can be in multiples but is best in multiples of odd numbers. 1x or 3x. This allows for the phase to be always 100% out for the opposing antennae. 2x could cause phase shift overlay and could result in both antennae receiving the same phase shift differentiation.
Is this true? then my choise would narrow down to "Dist bet diversity Ant=12.5cms or 37.5cms'
But from below POST iam confused regarding point 1 and 2 (I have put them pointwise just for reference)
http://forums.cisco.com/eforum/servlet/NetProf?page=netprof&forum=Wireless%20-%20Mobility&topic=WLAN%20Radio%20Standards&topicID=.ee6e8c2&fromOutline=&CommCmd=MB%3Fcmd%3Ddisplay_location%26location%3D.2cc0332d
1. Never should you have the antennas exactly one wavelength away from each other. For the frequency of 2400, one wavelength is 4.92"... so any distance that is not a multiple of 4.92 and no more than a multiple of 4 is recommended.
2. You should use 1/2 wavelength distances and to be safe no more than multiple of 3.
My Questions are
1. Should the distance between my Antennas for diversity be any len more than 12.5 cms and less than 25cms or more than 37.5cms and less than 50cms?
2. Is it that the distance between my Antennas for diversity should not be exactly equal to the multiple of wavelengths? (12.5cms, 25cms, 37.5cms, 50cms)
3. Is it that i have to use only odd wavelengths multiples? (12.5cms, 37.5cms) and should not use even multiples?
4. The use 1/2 wavelength distances between the antennas is diversity. Is it ok? or is the minimum 1 wavelength as specfied in the cisco doc.
Also from the below POST iam confused regarding the MAX distance:
http://forums.cisco.com/eforum/servlet/NetProf?page=netprof&forum=Wireless%20-%20Mobility&topic=WLAN%20Radio%20Standards&topicID=.ee6e8c2&fromOutline=&CommCmd=MB%3Fcmd%3Ddisplay_location%26location%3D.1dd7905b
"Distance beyond 1-2 wavelengths is not critical. Since the antennas are not transmitting and/or receiving at the same time,there is no real chance of overloading the other antenna or radio front-end. "
There's not really a "max distance" as far as diversity is concerned. By placing the antennas some distance apart (the minimum, I believe, is ~one wavelength), you optimize the chance that a signal that is null (or interfered with) on one side, isn't on the other.
my question is
5. Is there a max distance between the antennas in diversity (like we read 4xwavelength) or there is no max distance? then wont there evolve a different RF coverage area for both the Antennas?
Another important question
> If my wireless card supports 5GHZ and 2.4GHZ and hence my antennas would be dual band. Then to achive antenna Diversity, what would be the distance between the two antennas? The minimum distance should be 1 wavelength, so For 2.4GHz wavelength = 12.5cm and for 5GHz wavelength = 6cm.
For a dual band antenna? how much should it be? Whats the min and max.
someone please can help me solve the huge confusion in my mind :)

First, let me say that this is certainly a well-done post, good job of back-tracking the links and expressing your questions.
The next thing is that there are more than a few nuances when dealing with RF and signal transmission and propagation, so to cover a specific topic fully would frequently cover more space than provided.
With all that dancing out of the way ;-), I'll take a shot at answering your questions...
"Is there a max distance ... ?"
Well, no, not really ... but there is a max *effective* distance and a fairly specific point of diminishing returns (as mentioned in the linked posts).
The design and implementation goal is to create a sort of "binocular view" with the antennas, such that the chance of a "null" is reduced, such that the signal / pattern developed from the diverse pair of antennas will create at least one clean and clear (reduced multipath, reduced nulls, possibly better line of sight)signal to (ideally) every client in the coverage area.
Even though only one antenna is active at any given time, having any conductor within the radiated field will distort the transmission envelope, either positively (in the case of a Yagi) or negatively (in the case of a grounded conductor, like a sprinkler head).
The amount of the distortion, and the shape of the distortion are functions of the size and shape of the parasitic element. The effect is so variable that it is one of the foundations for the contention that a comprehensive site survey is not really an option, it's mandatory; it's truely the only way to see what adaptations are necessary to make the "rules of thumb" function in the specific location.
Regarding antennas handling multiple bands:
The antennas (in this case) are close to resonant multiples. The differences are such that the logical / electrical length can be adjusted by adding inductive or capacitive loads at the transmitter or feed point.
Even though the physical shape of the antennas may be different from model to model, the electrical load characteristics for a given band / antenna remain similar, so electronically compensating to make an antenna primarily designed for 2.4GHz function well for 5.6GHz is not especially difficult. Going from the longer wavelength to the shorter is more effective and easier to design than trying to go the other way.
For example, a quarter-wave 2.4GHz antenna is easily compensated to create a good half wave (or loaded 5/8) 5.6GHz antenna.
So, (finally) to provide the proper spacing for a diverse pair of multi-band antennas, figure for the longer wavelength (2.4GHz in this case) and the (roughly double) distance is likely to be acceptable for the higher wavelength. The spacing doesn't have to be exact; personally, I used to shoot for ~18" (variable with the mounting location and antenna type)and I always had pretty good luck, according to the follow-up survey.
Because of the jillions of possible variables, you can't just calculate and hang the antennas and be done ... a survey should be done to verify the installations and to make educated adjustments that will optimize the system for that specific location and environment.
The initial calculations are ideal guidelines, then you make adjustments to cover things like metal wall studs, sprinkler heads, the boss' favorite "wall o' vines" ... you will rarely get a perfect environment, adjustments are almost always required.
It's late and I'm starting to babble ... ponder the above and let us know if it helped any. There are a great bunch of experienced folks here, and I think this will be the start of a great discussion.
Good Luck
Scott

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Make sure your customer knows that going to Gigabit may not be an option over MMF if the distance is too great. Standards specify support for 62.5/125 fiber out to 220m to 275m with SX optics, and out to 550m with LX optics and mode-conditioning patch cords. (LX has been pushed as far as 700m reliably, but exceeds the standards).
Their best option for faster-than-100-meg service is to use Fast EtherChannel: n x 100-meg links, if they have multiple strand-pairs available.

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Hello people!
i have a pattern with different sized objects. i would like to change the overall appearance by keeping the size of the objects, but changing the distances between the objects.
-> scaling the overall size of the pattern, but not the size of the objects

bobster,
In this case, you may create the pattern in a quite different way and just scale once according to need:
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2) Object>Path>Average, with Both ticked;
Now you have the first dot;
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I intended this post as a recommandation for HP.
I have an HP Envy Touchsmart 17t-j100 and everything is excelent up until now.
The only thing that is very annoying is the distance between USB ports. PLEASE increase the distance between USB ports for the future models. I have a lot of problems tryng to connect everything I need.
Even if I try to open the DVD drive, it will stuck in my USB device. I mean how much difficult is to put a distance of 1 cm between them?
If the USB cables are the standard size, there are no problems but most of the USB devices nowadays are much more wider than a standard USB cable and I keep having problems.
I mean it's a 17" laptop, there is plenty of space on the sides. In the end, it's better to drop the DVD drive and put a VGA slot instead, but with bigger distance between USB ports.
I am planning to buy another ENVY in the future but I really want my new laptop to have USB ports with a wider distance bewtween them.
Thank you very much and I hope HP will take care of this little problem on the future models.
Have a nice day.
This question was solved.
View Solution.

Hi @catapara89 ,
Thank you for visiting the HP Support Forums and Welcome. Thanks so much for taking the time to let us know of your suggestions on the USB slots. It is important and has been viewed.
Have a great week.
Thanks.
Please click “Accept as Solution ” if you feel my post solved your issue, it will help others find the solution.
Click the “Kudos, Thumbs Up" on the bottom to say “Thanks” for helping!

Calculating co-ordinate distances between specific atoms

Hi,
Below is some code to calculate distances between all pairs of atoms. However, i need to make it slightly more specific by only calculating the distance between certain pairs of atoms
input
ATOM 5 CA PHE 1 113.142 75.993 130.862
ATOM 119 CA LEU 7 113.101 72.808 140.110
ATOM 138 CA ASP 8 109.508 74.207 140.047
ATOM 150 CA LYS 9 108.132 70.857 141.312
ATOM 172 CA LEU 10 110.758 70.962 144.119
e.g distance between all pairs for atoms 5, 119, 150 and 172 (say), last three columns are x,y and z co-ordinates
code it self
import java.util.*;
import java.io.*;
public class Distance {
public static void main(String[] args) {
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System.out.println("\nResults printed to file DistanceCalculations" + "\n");
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public Distance() {
Vector atomArray = new Vector();
String line;
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System.out.println("Enter PDB file:" + "\n");
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String fileName = inputReader.readLine();
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Atom.printDistance (a,b);
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System.out.println("Input file problem");
} catch (Exception ex) {
System.out.println (ex);
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public double x, y, z;
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public Atom(String s) throws IllegalArgumentException {
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StringTokenizer t = new StringTokenizer (s, " ");
t.nextToken();
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catch (Exception ex) {
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ok, essentially
want to calculate distance between to ranges. Say
range 1 is the first three, range 2 the rest. THen
calculate distance between all possible pairs between
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Sylvia.

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What is the maximum distance between a time capsule and an airport express?

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If it is not possible to run the Ethernet cable, you might look at a pair of Ethernet powerline adapters to send a "psuedo" Ethernet signal over the AC wiring in your home.
This will not be anywhere near the performance of a regular Ethernet cable, but it will likely be much better than wireless.
Since Ethernet over powerline (EOP) is somewhat unpredictable, you would want to have a very clear understanding of the store's return policy in case things don't work out as hoped. Any computer / electronics superstore will have a selection of powerline adapters to choose from.

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How do I cancel the distance between the numbers?
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Calculate distance between Latitude and Longitude

Hi All,
I have one Latitude and Longitude points. I need to calculate distance between them.
Right now we are using HAVERSINE formula but the query takes longer time to complete.
Reference: http://www.movable-type.co.uk/scripts/latlong.html
Please assist on this.
Thanks in advance.

Check this link...
http://www.mrexcel.com/forum/excel-questions/202255-calculating-distance-between-two-latitude-longitude-points.html
I never did this before, but gave it a try using the formula mentioned in that link..
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Destination: 44.822075(B3)     -92.912498(C3)
Formula used:
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B: =2*ATAN2(SQRT(1-A),SQRT(A))
C: =R*B                  ---> DISTANCE!!!!
WITH t AS
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SELECT (2 * ATAN2 (SQRT ( (1 - E2)), SQRT (E2))) * 6371
FROM t;
Check if this gives correct values... (I did not verify this properly.. ) And this is faster in my opinion..
Please post your code for better suggestions from volunteers...
Cheers,
Manik.

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Distance between options in an af:selectOneRadio

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