you cannot divide m/s by m/s and have a result in anything
other than m/s.

1 m/s^2 / 10 m/s^2 = 0.01 m/s

you certainly cannot end up with a result that is simply
an amount of time.

the result would be given in m/s

you can drop the m (meters) and only use the seconds.

1 s^2 / 10 s^2 = 0.1 s^2

or

1 s / 100 s = .01 s

so we have dropped the velocity from the equation.

and we only kept the time.

we now subtract the result of .01 s from the number 1
so we must also drop the time from the equation in order to
subtract.

1 - .01 = .99

we then find the square root of the number .99 ?

0.99498743710662 ?

what could this number possibly represent?

we have dropped the units of distance
we have dropped the units of time

if we do associate this number with anything then
we must associate this number with a velocity.

we must give the result in m/s (distance / time)

so the time dilation equation is an equation that is
used to solve for velocity it is not used to solve
for time.

Time for a basic math refresher.

10 M/S / 10 M/S = (10/10) * (M/S / M/S) = 1 * 1 = 1

That is the correct math for the situation.

Bill Gill

1 what?

10 orange/peels / 10 orange/peels = (10/10) * (orange/peels / orange/peels) = 1 * 1 = 1

1 what?

should we give the answer in units of oranges or peels?
why would we only use 1 peel?

10 watts/second / 10 watts/second = (10/10) * (W/s/W/s) = 1 * 1 = 1

1 WHAT?
should we give the answer in units of watts or seconds?
why would we only use 1 second in the above?

the above is exactly what the equation is doing
except its using two separate quantities of
distance and time vs energy and time.


LOL

in the equation what is being divided is the distance
not the time.

the equation only uses the distance in the (m/s)

the equation never touches time so how can it be considered
that the equation delivers any amount of time as its result
or any number that can be associated with any amount of time?

so any number that can be a result from the equation is
a velocity.

ie .999999999 m/s

No, velocity is not what the equation solves for.

The number you are asking about is a ratio, I think, and as such it would be without units.

Once you have the ratio, you multiply it by t (in seconds) to get
the answer, t prime, which will also be in seconds--the dilated time.

~

just because the equation has a t' in front of it
does not mean that it solves for t'

the only element that actually undergoes any calculation
is distance ie... Meters

time is never used or calculated.

if time is never calculated then how could time be a result
of the calculation?

just because the inventor of the equation says so?

then lets try this equation out.

time = sqrt (1-(1 kgm^2)/(10 kgm^2))

there is no amount of time involved in the above equation.
nor is time calculated in the above calculation.
will the result of the above equation be an amount of time?

the result of the above equation can only be given in units
of kgm

or mass and distance

and the result will always be less than 1 kgm
because thats the intended purpose of the equation.

thats why the 1-(?/?) is there.

so no. the equation t'=tsqrt(1-(v2/c2) does not deliver
any number as its result that could possibly be associated
with any amount of time.

the only possible association that the number could be
associated with is units of distance (meters).

...just stunning.

"in front of it?!?"
Paul, by definition, an equation involves the stuff on both sides of the "equals sign" doesn't it? There is no place that is "in front of" the equation. The equation is about the relationship between t' and t, which are both "time" measured in seconds.

~ ?

yes , it is a ratio , it is a ratio of distance
since each expression v and c are expressed in units of
m/s (meters per second).

then the ratio is a ratio of meters per second.

50 m/s only has 1 second attached to it.
squaring that 50 m/s does not affect the number
that is associated with time (1 second)
it only affects the number that is associated with distance (50).

finding the square root of 2500 does not affect the time
either , it only affects the distance and 50 meter distance is
where the 2500 came from.

and the speed of light only has 1 second attached to it.

so the ratio cannot be time as the time units remain the
same throughout the equation so the ratio is a ratio
of distance.

solve for the ratio of time in the following equation
ratio t = 1m/sec / 10m/sec ?

now solve for the ratio of distance in the following equation
ratio d = 1m/sec / 10m/sec ?

if the equation is about the relationship between
t' and t then why is it that t' or t is never used in any
calculation that is performed while calculating distances
in the equation.

all calculations are performed on the distances of v and c

the only relationship that is being examined in the
equation is the relationship between
the v distance and the c distance in a single second.

and you can only have 1 second in 1 second.

...just because you've separated velocity into distance and time, in your mind,
it doesn't mean that time isn't included in the equation--in those terms, v and c.

...with "time" being the same for each.

I was taught, if "calculations are performed" on two different ratios--with numerators (such as v and c describe),
then the denominators (t in this case) should be the same.

How would you perform calculations on two ratios, if their denominators were different?
~

Now this is a classic thread

This has to go to the maths hall of fame, not quite as funny as asking what Obamas last name is, but getting close

I think you have just worked out the depth of problem to explain it to him Samwik. I think I would just ignore it and run away really not worth the effort. Remedial education is not something you can easily do on the internet ... run away now

Guys you know 't' is something you have instead of coffee ... right. Therefore I put it to you, 't' can't be in any equation but it has to be in a cup ... so you are all wrong. The answer to the equation is clearly a cup, otherwise you can't have 't'.

Sorry it was the best I could do without going down the fundamentalist education line, which might be a little bit close.

velocity is already separated into units of distance and time

and you can clearly see the separation when you look at
an amount of velocity such as 50 m/s

the m/s represents meters per second.
the m represents meters
the s represents a single second
the 50 represents the distance

so its not in my mind where the separation occurred but
in the written definition of velocity.






I never said that time was not "included" in the equation I said that the time that is in the equation is never calculated.

and "time" is never calculated in the equation , the only elements
in the entire process where calculations are performed
are the two distance elements (the two elements that can
experience a rate of change).

the time element does not experience a change.

so time does not change and cannot change according to the
equation.

and all the squaring and rooting around in the equation
has absolutely no effect on the single second time element.

the only things that change in the equation are the numbers
that are associated with distance.

at no time during the calculation is there a single change
that occurs or can occur to the single second that is included
into the equation.

orac.

here is my reply to you.

Thanks Orac!

Paul, if I understood Orac correctly, he was right about how my reply “worked out the depth of [the] problem,” relativity; unless he just meant the problem with algebraic skills.
That’s because I (in composing my reply) had noticed that
by confronting the logic (and meaning) of the equation’s calculations,
the manifestation of relativity operating became more obvious—since the process of “relativity,”
and its consequence, quite quickly become almost undeniably apparent—like a mathematical slap in the face.

Maybe that could be said more simply.
The implications of “relativity” become more obvious
when you observe the consequences (and meaning)
of performing those calculations ...for “time dilation” ...correctly.
But I suppose if one decided to deny relativity, then how those “calculations are performed” must also be denied.
===

And Paul, either way, with or without “t” in the denominators, of v and c,
there is still the t (time in seconds) in front of the square root sign.
Time is still used to calculate the dilated time.

It is just an equation relating time to dilated time,
by comparing one velocity with another—the speed of light—in a certain reference frame.
And one of those velocities is constant in all reference frames, so for comparing the effect of other different velocities,
the "constant velocity" makes a good common denominator.

So the idea that this might also be done by looking only at “distances” (without the time)
involved with those two different speeds, shouldn’t be surprising.
And the idea that you could simply compare the distances (without time),
to calculate the dilated time, might give you another insight into the nature of “relativity.”

Why again don’t you think the equation is true?
~

v2/c2 results in units of (distance) and (meters)/(seconds)

ie...

1 m/s ^2 / 10 m/s ^2 = .01 m/s ^2

you then subtract the result that is given in (distance)
and the rate of change over time (m/s) from the 1 that causes
the result to always be lower than 1

1 - .01 = .99

this leads us to your point below.



the square root of .99 is 0.99498743710662

so the current calculation is 0.99498743710662 m/s

it is a amount of velocity it has not been converted
into an amount of time.

its not simply a number nor does it only have units
of time attached to it.

remember this:
The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time.

in other words if you multiply 2 apples x 2 oranges

you still only have 2 apples and 2 oranges.

you dont have 4 apples or 4 oranges.

you cannot convert apples into oranges.

and you cannot convert velocity into time.

by multiplying the 1 second in the calculation
that you pointed to times the current
total in the calculation of 0.99498743710662 m/s

1 s * 0.99498743710662 m/s = 0.99498743710662 m/s

the above is not what the equation would have us expect
to be possible , the equation expects us to simply believe
that the result will be converted into time.
but you cant multiply an amount of time by an amount of velocity and have the result given in an amount of time.

1 second * 1000 m/s is not equal to 1000 seconds

so nothing changed !!! time has not yet been changed
nor has velocity been converted into time.

the result is not an amount of time.

the result of the equation is still a velocity.
and since 1 second * 1 second = 1 second the time
never changed it is still 1 second.

now if the 1 second were 2 seconds then thats another
story all together because the 2 seconds would then
be multiplied by the distance in the 0.99498743710662 m/s
and the result would become 1.98997487421324 m/s

once again you cant multiply an amount of time by an
amount of velocity and have the result given in an amount
of time.

or at least I cant.



because I can (think).

trying to work with this 1 inch x 2 inch reply box
is really hard to do.

so I make lots of mistakes that I have to correct.

it takes me awhile to make a decent reply.

but I think I get my point across most of the time
even if it never sinks in.

I can see nothing has changed, but I think I see several places where you are going awry.

First:Okay, this indicates a major misconception.
It (t) is not another "1 second" or "1 s" in front of the square root sign.
It (t) is whatever certain length of time you might want to calculate (the time dilation of),
such as the duration of a trip at near light speed.

And by using that "certain length of time" (say t = 2 years, for example)
and multiplying by the square root of that ratio [m/s divided by m/s],
you find the solution, getting the result for t' (t prime) ...the dilated time.
===

Assuming the equation is correct....
So, if the square root of all that stuff turned out to be one half,
then t' would be one year (of the traveler's perspective)
...for a (t equals) two year trip (from Earth's perspective).

Orac?!? Do I have those perspectives correct? Please feel free to correct anything, since I'm just figuring this out as I go along ...from a chemistry/biology science background. But I think I still recall algebra well enough.
===

But whatever, Paul, let's see if you can come up with a new 'exclusion principle' for this second point about the units, such as meter per second or (seconds * some distance in meters), which should just be basic algebra and was already pointed out, once or twice, in various ways. But humor me here.

But first one other point about the units: ...where you say:...that is correct! The answer wouldn't be (in units of) time:
It would be 1000 meters, a distance. Just look at the units, and at which units cancel and what is remaining.
If it were 2 seconds * 1000 m/s, you'd get 2000 meters, right?

So, about these units.... Here's where I'm hoping you'll humor me a bit.
Here is my "test" for you: Are these two lines below the same? In other words:
Is the second line equal to the first line, but just stated differently?

50 meters/second (divided by) 200 meters/second =

50 meters/second (multiplied by) 1 second/200 meters ?

~

p.s. "It's a trap!" ....No, not really; it's just basic algebra.
...or do you consider that fantasy science too?

this entire thread has been focused on an amount of 1 second of time.

so if you want to you can multiply the results x any amount of seconds you choose
but for now lets stick to 1 second for the time element (t).
we are examining the results in a time frame of 1 second.



exactly !

because you cannot multiply an amount of time (1 second) by a velocity (1000 m/s)
and have the result given only in units of time.

so when the equation divides v^2 by c^2 the results of that calculation are given in m/s

the equation then (for design purposes) subtracts the result given in m/s from the number 1
which always gives its result as a number lower than 1.

even so the result given after the calculation is still given in m/s

the number 1 in the equation is not a unit of time , its simply the number 1
and has no viable reason to be in the equation other than to falsely lower the result
of the previous calculation.

and finding the square root of the result given in m/s does not convert the result
into an amount of time either.
so the square root of the result would also be given in m/s

now back to your point...and mine
1 second * 1000 m/s is not equal to 1000 seconds

time * distance is not equal to time alone...

so when the equation multiplies t (1 second) by the result of all of the previous calculations
there is no change in the number or the units given or assigned to that number.
because any number multiplied by 1 remains the same.

1 * 1 orange = 1 orange
1s * 1 orange/s = 1 orange

the orange does not become time in either of the above
equations so the orange is still an orange.

and multiplying 1 by any velocity also does not convert
the units of velocity into units of time.

so after the final calculation (multiplication) the number is still in units of m/s
and should not be given or assigned as an amount or a unit of time alone.

the final calculation should be given in units of m/s

so the end product of the equation t' ( time dilation )
would be given in units of m/s

LOL

Paul, as usual, you logic seems to work for only about 1 second.
===

Do you think this is some sort of "designer math" question:

50 meters/second (divided by) 200 meters/second =
50 meters/second (multiplied by) 1 second/200 meters ?

Do you think it is true, or not? It should be easy, since it's all in "1 second" speak.

~

I admire your tenacity samwik .. good luck

Im not sure what you expect to gain from this but here we go.



the reason that the result is 1000 meters in the above equation
is because there is an amount of 1000 meters distance in 1000 m/s

meters PER second



right!

the purpose of the equation is to solve for the distance that would
be traveled over a time period of 2 seconds at a velcity of 1000 m/s

now for your equation and the line of text below it.

the equation itself



the equation has 3 elements on each side of the division symbol.

these are the number of meters (50 and 200)

these numbers are given in units of (meters for distance) and (seconds for time)
written as meters / second (meters per second)

the two units meters and seconds represent the two elements of distance and time.

both sides of the equation have these elements attached to the number that represents the amount of distance.
in order to show what the numbers represent.

(50 meters per second) and (200 meters per second)

the purpose of the equation is to solve for the difference in the rate of change
of position over an amount of time between the two sides of the equation and
by solving the equation you can find the difference in the rate of change of position over
an amount of time!!!

as in the below solved equation.

we know that in the equation we are discussing
t'=t*sqrt(1-(v2)/(c2)
the division that you are questioning and the result that I have given
is between v2 and c2 and by dividing v2 by c2 we get the ratio between
the two (v2/c2) so to stay closer to the OP I will
emphasize the ratio through an explaination of the result of the below equation.

50 meters/second (divided by) 200 meters/second = a ratio of .25 to 1 of 200 meters/second.

meaning that there is a .25 to 1 ratio between 50 m/s and 200 m/s
you can check this by multiplying 200 m/s times the .25 to 1 ratio between 50 m/s and 200 m/s.

so the difference in the rate of change or the ratio between the rate of change between 50 meters and 200 meters in a time period of 1 second is a ratio of .25 to 1 of 200 meters /second

the next line of text is a line of text and is not a proper equation.

50 meters/second (multiplied by) 1 second/200 meters ?

however were I to correct your error in assembling the equation by
inserting a (EQUAL) or (IS EQUAL TO) or (EQUALITY) symbol (=)

then the two sides of the equation have the same meaning
meters per second is the same as seconds per meter

ie ...

200 meters/second (meters per second) means that an objects rate of change
of position over a distance of 200 meters occurrs in a time peroid of 1 second.

200 meters per second = 1 second per 200 meters

1 second/200 meters (seconds per 200 meters) means that in a time period of 1 second
an object experiences a rate of change of position of 200 meters.

1 second per 200 meters = 200 meters per second

therefore to solve the line of text or the improper equation that you posted
through making it a proper equation by inserting the equals sign (=)

50 meters/second (multiplied by) 1 second/200 meters =

for example:
suppose you needed to know the total board length produced in meters and total production time in seconds for 250 machines 50 of the machines produce (50) 1 meter long boards each second and the other 200 machines produce (200) 1 meter long boards each second.

LOL^2

50 meters/second (multiplied by) 1 second/200 meters = 10,000 meters per second.

if we simply cancel out meters and seconds on both sides of the equation
we get 50 * 200 = 10,000 and thats it ... no units attached.
it has no value other than being a number that is derived through the
multiplication of 2 other numbers that have no units attached to them.

it is a number of unknown origin unless you associate it with the units
that it was formed from.


Im not certain why this seemed to be important to you however.

but your question was is it true...

if you were talking about the line of text then yes it is true
that the line of text is truly a line of text.

...here is where I can see you misinterpreted what I was asking about. My apologies for trying to draw this out longer, but I think the way I formatted the text (and the equation) may have been misleading. I can see it looks like I was aksing for the answer to two different equations, with one on each line of the reply; and with an 'equals sign' at the end of the first, but with a 'question mark' at the end of the second line.

But actually, there is only one equation, with one "side" of the equation being the first line (before the = sign), and the other "side" of the equation being the second line, following after the 'equal sign' and before the question mark, which is just the end of the sentence.

I'm not asking for a solution to the equation, but I just wanted to know if you thought the equation was true. I think this is an equality:

50 meters/second (divided by) 200 meters/second = 50 meters/second (multiplied by) 1 second/200 meters.

Do you think that single equation is a true equality, algebraically; is the single equation true, or do you think it is false?

Again, sorry for the confusion and repeating so much.
~

I didnt see that it was supposed to be in a single line.

what you have is

50 m/s / 200 m/s = 50 m/s * 1 s / 200 m

is the ratio between 50 m/s and 200 m/s or.25 to 1 equal to
the multiplication between 50 m/s and 1 s/200 meters or 10000 m/s



the two sides are not equal

the results of the equation would be false
but the equation itself is useable in a sense.

I dont concern myself with algebra
I never have , I only use math on spreadsheets and
in computer programs.

so to me the equation is false , it can only be used to
deliberately find a wrong number.

oh !!! I get it now.

you mean its like

t'=t*sqrt(1-(v2)/(c2)

is used to deliberately find a wrong number for a specific purpose.

"Ye shall know them by their fruits."

Before rejecting something outright, you might first seek some insight,
and then make a choice about your concerns.

[...ruminating on 'home schooling' possibilities and possible circumstances]
===

As I said in the other thread, "Paul, it is always nice to see what and how you think."
It's always nice to know where one stands, however frightening the perspective might be.

But now I think I can see why you can't conceive that this "time dilation formula" is true.

~

I remember taking algebra at some point in the past.

and that is why I dont concern myself with algebra.
to me and the way I use math its not necessary in fact in
would probably be a problem trying to code an application
and trying to use algebra to do it with.

a computer program could use algebra but only if the programmer
for some strange reason wanted to waste a lot of time in
doing so.

a computer performs math in much the same way as what we
were taught in grade school... but coding the math and
assigning values to variables so that the program will perform
the needed math routines is strictly the responsibility of the
computer programmer.

I take a working equation such as the one that we are discussing
and I start at the first calculation and work my way through to
the last calculation.

and I have to assign variables to the numbers that I use to
perform the calculations other wise all I end up with is
a number.

as in

v = 5
c = 100
t = 1
t1 = 0

when I press the calculate button

v=v*v
c=c*c
t=v/c
t=sqrt(t)
t=1-t
t1=1*t

t1 is then output to a textbox.

and thats it.

theres no need for me to get any further into it than
the simple math.

same for a spreadsheet.

of course I could have spent several days writing 1000 lines of
code to make it look technical , but thats just not the way
that I am.

getting technical can slow down the computing time of the equation.



I cant remember taking a computer programming course that
lasted longer than the time it took to boot me from the
class because I couldnt type fast enough however.

I even had the entire paragraph memorised that the speed typing test was given for thinking that that could at least
reduce the time it took to glance back and forth to the
page.

programming isnt about speed typing its about having the intelligence to know how to form code to perform a desired
result.

so I didnt feel bad about it , it informed me that the majority
of people who did complete the course were rushed through it
so fast that when they graduated they couldnt remember anything
that they had studied in the course.

which is a lot like possibly more than 99.999 percent of all algebra course graduates.

ask them !!!

I'm afraid I am confused with what you are doing. I see that you are reusing all of your variables, and that confuses me.
Why don't you define some intermediate variables to make sure that you aren't changing something that shouldn't be changed. For example:

v = 5
c = 100
t= 1
a = v*v
b = c*c
d= a/b
e = 1 - d
f = sqrt(e)
t1 = t * f

That is an accurate representation of the relativistic time contraction formula, setting aside that in the formula C is a constant equal to approximately 300 million meters/second.

And of course I would probably combine those all into one long formula, rather than doing it one step at a time. But if you aren't comfortable using shorthand then this way will give a correct answer.

Maybe you really should refresh your memory of algebra.

Bill Gill

I find its easier to do it the way I posted.

theres not as many variables to keep track of.

and its easier to do an initial build of a equation
one line at a time.

but sure you could do the equation like this.

dim v=5:c=100:t=1:t1=0:v1=0:v2=0
v1 = 1-(v*v)/(c*c):v2 = sqrt(v1):t1 = t * v2

or something like that I just dont do it that way
Id rather do 1 calculation on 1 line
it makes stepping through the calculations easier
and less confusing when you are dealing with hundreds of variables.

and unlike the above I would name the variables so that they
are more recognizable to me ... like the speed of light c would be sol and the speed of light C would be SOL

v1_m1
v1_m2
v1_m3

etc... having easier to find variables makes finding errors
in equation building much easier.

and I try to reuse the same variable as I go along in the calculations then I simply give the named variable the value of the variable that I used to work the equation.

also this way you dont have too many variables storing
unnecessary data.

and I try to never have duplicate equations if possible
by sending values of variables to a routine that handles
that specific equation.

as in

100 if a >= 100 then a = 0
101 if add_10=1 then goto 600
102 if a < 101 then goto 500
103 goto 100
500 a=a+1:add_10=1:goto 100
600 a=a+10:add_10=0:goto 100

the above loop will never stop.
it will add 1 to a
then it will add 10 to a
then 1
then 10

etc...etc...etc...
this way you can set up multiple routines
in the same loop.

if you dont already program you should try it
I think you would like it.
and its free.

visual studio express.

https://www.visualstudio.com/en-us/products/visual-studio-express-vs.aspx

And you can lose track of what you are doing much more easily. That way you can claim that you have disproved something when you have made major errors in your math.

I would probably write it as:

t1 = t*(sqrt(1-(v^2/C^2)))

That way I can see what the equation is and only have
3 variables, t, v, and t1. t and v of course are the input values and t1 is the output value, while C is a constant.

Bill Gill

c is a constant but the computer does not know
that c is a constant.


v , c , t and t1 = 4 variables.

C is a constant. You set it programmatically and then never change it. In many programming languages you can define C as a constant and you can't change it.

v and t are independent variables. They are the input values to the calculation and can be set to any value. Except of course that v cannot be greater than C.

t1 is a dependent variable. Its value depends on the outcome of the calculation performed using v and t.

You know, there is at least one site on the web where you can study math and physics for free. Khan Academy

If you studied that you might actually understand just what mistakes you are making.

Bill Gill

Im sorry bill

that is almost exactly what I did when I wrote

v=5 !!

like I said I could have spent several days in doing that
by writing 1000 lines of code , but I cant think of a real
reason to waste time on that type of filler.

I dont get paid to truncate or dilate code to make it look like
I have put a lot of work into it.

I only do what is needed for my own purposes.

but feel free to waste all the time you want to
while you are writing your own code.

some people even like to tell their life story in the
comments of code , perhaps you could write code and
then leave negative comments in the code about any thing
that you can find to be negative about.

if you could ever bring yourself to be negative about anything!

heres a page that you can start with as a beginer
to learn about constants and variables.

https://www.kidscodecs.com/variables-constants-data-types/

by the way you can also declare a value to the text
inside a textbox and disable the textbox at design time
this way the text inside the textbox is unchangeable but the value of the text inside the textbox has value is visible to the user and can be called at runtime
as in x = val(textbox1.text)

so you dont really need to declare any constants or variables
as long as you input the desired or required value into the
textbox at design time.

enabling or disabling the textbox allows the programmer
to either choose to allow users to control the value programmatically ( change the text inside the box ) or not.

and the text that is put into the textbox can be controlled
by only allowing numbers and decimals to be entered.

you can even program a string manipulation routine that will
handle the "^" symbol so that a user can enter a number
that is to be squared or cubed for instance.

and remember microsoft has developer forums that you can
become a valued member and contributor on.

you can also leave negative comments there also if you can
ever find anything to be negative about.

If I wanted to do something like that I wouldn't use a text box. When you disable a text box the text is grayed out. It is still readable, but it really looks odd. I would use a label. A label displays the text in the normal mode, but can't be edited by the user, so that makes it work just the same for your application, without being grayed out.

The fact is that I use labels quite a bit for displaying the results of calculations. You can change the text displayed in a label programmatically.
e. g. lblTprime.text = val(txtT.text) * sqrt(1 - val(txtV.text)^2/val(lblC.text)^2)

I do try give my text boxes and labels meaningful names. Starting them with lbl or txt gives me a quick clue as to what I am operating on and clarifies the operations I can perform on it.

This also assumes you are using Visual Basic, other languages use different conventions, but what you have been saying sounds to me like VB.

Of course I don't know what you are developing so I don't know why you would want your constant shown on the screen.

Bill Gill

Im not developing anything right now.

my dev computer isnt up and running again enough for
me to trust it.

Im waiting on a hard drive that I havent ordered yet.

because the one that was in the dev computer became infected
and I wont use it anymore to dev on or connect it to the internet.

yes I use VB .net

I didnt know that you programmed.

I wouldnt grey out a textbox for my own use just to store
a constant in it.

I would simply set it and forget it
in the declarations.

and using labels is the better choice as you say for
pretty much any text or numbers.

have you ever used a controls AccessibleDescription property
or AccessibleName property to make a control array with?

you can assign the control a numbered value at design time
and then control the many different captions on the buttons , labels or text in the textboxes of the controls
in your control array at run time.

you can also reassign numbers to the controls at run time
for sorting purposes etc...

by setting up routines.

its really good for using in any type of data situation
where data is stored accessed and updated frequently and you dont
want to use the microsoft databases for any particular reason.

you can make your own working excel like spreadsheet
or database application using textboxes , labels , buttons etc
and make them function exactly the way you want them to.

I hope you have good backups of your dev computer. It wouldn't be a good thing to lose all of your work.

I haven't used those particular properties. I do use text boxes and labels quite a bit. I don't really do a lot of programming. The last thing of any consequence I did was a couple of years ago I built my own calendar that works the way I want it to, instead of accepting what somebody else thinks is the right way. The same way you do your spread sheets. You can never find anything that really does what you want. It may take some work, but at least it does what you want it to.

One thing about my calendar it can show holidays in different calendars. The Hebrew calendar, the Muslim calendar, the Chinese calendar, and several different Hindu calendars. Fortunately I found a book with basic algorithms for all of those. I just had to interpret them into VB.

Bill Gill

I didnt have a backup of the OS or the programs but I did have
backups of my files that I thought I wanted to keep
such as my archive of sagg and health and nutrition papers
and info , pictures , etc...

also all of my programming projects were not affected because
I made a program that moves them off the hard drive to an external
drive and I can choose which projects I want to work on
that way and it will move them back to the hard drive as needed.

I made a program once upon a time that I named My Journal ...

it was basically a calendar database and you could add notes
to the journal every day.

it didnt use a database only flat files and each day had its own flat file so you never really could lose a day unless you lost
the folder the files were in.

you could click on any date on the calendar control and the
note for that day would pop up because the note was saved
as that date as in 01_07_2004 and it used the value of the date
that you clicked on through string manipulation.

then the program loaded the 20 notes possible for that date
into an array of rich text boxes.

and each box was editable.

it seemed like every time I started working on a idea
for a program before long it was available on the internet.

which is why I made the program to move my project files
off of my dev computer in the first place.

Good, it sounds as if you didn't lose anything of major importance. That is a big part of what you need backs up.

My calendar looks pretty much like a standard calendar except it spreads all over the computer screen. In addition to the holidays, which you can add to suit yourself, I have notes on events that will happen on that day. They are in a flat file that the program reads and displays the basic information on the screen. If you hover the pointer over one of them a pop-up shows up that gives all the detail I entered for it.

As far as a program to keep your notes that sounds like like a useful idea. Before I retired I kept notes on everything I talked to anybody about using notepad, one file per day. Then I wrote a routine that would search the notes to find where it was I talked about something. That was before Windows had a routine that would search files. That gave me a lot of help trying to come up with just what was said when if a question came up.

Bill Gill

well I would have done a backup.

but I only wanted to back up certain files and when I
got everything loaded in the full backup was going to be
around 90 gigabytes ... and microsoft server backup wont
let you do a partial backup the first time you do a backup.

I will remember that this time.

you know after all the updates are installed and all the hardware
is installed so that I will have a recovery backup.

I cant complain though because the windows server 08 r2 that
I installed in 2009 never did give me any real problems until
I started with the android studio and oracle vbox and such
as that.

Im not going to say that that is where the infection came from
however because I dont know.

the same OS install running for 9 years + is a really good record especially on a dev machine.

the self help sort of personal programs that I make for myself
are very useful as long as they arent dependent on any of the
ever changing internet like my netflix program that stopped
functioning every time they hired a new kid off the streets
and let him mess around with the code in order to make it
look as if he was doing something.

LOL

or thats the way it seems anyway.

It is pretty well accepted that you need to keep changing your website if you want people to keep coming. But you really need to be careful not to break it. And I expect it is probably not too good an idea to include links in your programs to specific parts of any website. The front page usually at least stays there, but lower level pages tend to change frequently. If you need the information on one of those lower level pages you can have a problem if they move the information to a different page, or format it differently. I have no idea how to handle that problem when you are developing a program that gets data from a given website. I expect if I came up with a case where I needed it I would hit Google and see if I could find a solution. There are a lot of places where you can find good ideas for doing special things in programming. I know I have gotten some good ideas from VBForums

Bill Gill

the problem was that the format of the internal links were changed.

I used string manipulation to build a flat file database
of folders and files.

and I also saved each titles image that was connected to
a title for display purposes when the title was displayed.

I had the images stored in the database.

there was a point where the description of the title was also
stored in the image through pixel manipulation.

I used pixel values to represent the stored data in the images.

so the program would load the image into a picturebox then
decrypt the data from the image and then display the data
within a separate picturebox along with the original image
loaded into it.

and this served as the users visual representation of the
title.

this made reading the data really difficult so I 80sixed it.

I notice that netflix has something similar to this these days
only the data is not embedded into the image itself.

it hovers over the image and is also really hard to read
unless you select the text and highlight it.

the benefits of the program to netflix would have been
that the images or descriptions would not need to be
downloaded every time that a user needed to view the image
and description using the program which would have saved
netflix a really large amount of bandwidth cost.

also the program had the ability to store titles in folders that represent the titles category such as drama , scifi , horror , etc or create his own category such as monday , tuesday , wednesday , thursday ... etc.

it was the inability of a user to store titles in folders
that caused me to build the program in the first place.

sifting through or trying to arrange the allowed 500 titles in you netflix list can be a real pain and the program I built did not have a limit as to the number of titles that you could store within the folders on your computer.

but the only thing I need to fix is the actual string manipulation of the URL of the title.

the URL is the only part of any netflix code that I use.

Im just waiting a few months to see if the URL's become stable
before I fix it.

I may have to add a start up routine that examines the netflix URL's and decides if an internal change to the program is needed and then makes the needed change to the program before the program tries to utilize the netflix data.

oh nooo I only use the microsoft help files when I need to
find a way to do something in VB.

ROLF

why would anyone want to spend only a few minutes looking in
the forums for needed programming information when they could
spend all day or even weeks reading the microsoft help files
before they find that they will most likely need to look in
the forums to actually find the needed information.

why fix your own lawnmower when it breaks down when theres a
guy at the repair shop who will keep it a whole month before he releases it.

if you fix it yourself , then you have to cut the grass that month.

Everything you could possibly need is in the Microsoft help files. Of course finding it is shall we say 'difficult?' Just finding the information on a key word can get to be frustrating. All you have to do is to high light the word and press F1. And it will give you information on everything that that word could possibly refer to. That's why I keep pointers to the the help forums.

Bill Gill