How to write this simple program? converting to numbers???

[Ron]

I want to calculate the surface area of a sphere from an inputed radius
with option strict on. I guess I am not converting something
correctly.

Here is what I am doing:
I have a textbox txtradius that I enter a radius into.

I then have a lable that I want the surface area to be displayed in
called lblsurface

here is my logic.

lblsurface.text = (4 * 3.14 * txtradius * txtradius)

I am getting a conversion error.
I dim txtradius as double
dim lblsurface as double

Thanks for any help

[Spam Catcher]

"Ron" <pt*****@yahoo. comwrote in news:1169051101 .266111.300510@
51g2000cwl.goog legroups.com:

>
I then have a lable that I want the surface area to be displayed in
called lblsurface

here is my logic.

lblsurface.text = (4 * 3.14 * txtradius * txtradius)

I am getting a conversion error.
I dim txtradius as double
dim lblsurface as double

txtRadius is a textbox right?

In that case you should be doing:

Dim _Radius as Double = Ctype(txtRadiu. text, Double)

[Phill W.]

Ron wrote:

I want to calculate the surface area of a sphere from an inputed radius
with option strict on. I guess I am not converting something
correctly.

lblsurface.text = (4 * 3.14 * txtradius * txtradius)

You could boil all this down into a single statement, but here's the
long-hand way, so you can see all the conversions going on ...

Dim nRadius as Double = CDbl( txtRadius.Text )
Dim nArea as Double _
= 4 * Math.PI * Math.Pow( nRadius, 2 )

lblSurface.Text = nArea.ToString( "0.000" )

HTH,
Phill W.

[Ron]

here is what I have and it does not work.

Dim _Radius As Double = CType(txtRadius .Text, Double)
CDbl(lblsurface .Text, double) = 4 * 3.14 *
(CDbl(txtRadius .Text) * (CDbl(txtRadius .Text)))
lblVolume.Text = (4 / 3 * 3.14 * (CDbl(txtRadius .Text) *
(CDbl(txtRadius .Text) * (CDbl(txtRadius .Text)))))

still not working.

Phill W. wrote:

Ron wrote:

I want to calculate the surface area of a sphere from an inputed radius
with option strict on. I guess I am not converting something
correctly.

lblsurface.text = (4 * 3.14 * txtradius * txtradius)

You could boil all this down into a single statement, but here's the
long-hand way, so you can see all the conversions going on ...

Dim nRadius as Double = CDbl( txtRadius.Text )
Dim nArea as Double _
= 4 * Math.PI * Math.Pow( nRadius, 2 )

lblSurface.Text = nArea.ToString( "0.000" )

HTH,
Phill W.

[BillE]

In the second line you put
CDbl(lblsurface .Text, double)
it should be
CDbl(lblsurface .Text)
or
CType(lblsurfac e.Text, double)
"Ron" <pt*****@yahoo. comwrote in message
news:11******** *************@1 1g2000cwr.googl egroups.com...

here is what I have and it does not work.

Dim _Radius As Double = CType(txtRadius .Text, Double)
CDbl(lblsurface .Text, double) = 4 * 3.14 *
(CDbl(txtRadius .Text) * (CDbl(txtRadius .Text)))
lblVolume.Text = (4 / 3 * 3.14 * (CDbl(txtRadius .Text) *
(CDbl(txtRadius .Text) * (CDbl(txtRadius .Text)))))

still not working.

Phill W. wrote:

>Ron wrote:

I want to calculate the surface area of a sphere from an inputed radius
with option strict on. I guess I am not converting something
correctly.

lblsurface.text = (4 * 3.14 * txtradius * txtradius)

You could boil all this down into a single statement, but here's the
long-hand way, so you can see all the conversions going on ...

Dim nRadius as Double = CDbl( txtRadius.Text )
Dim nArea as Double _
= 4 * Math.PI * Math.Pow( nRadius, 2 )

lblSurface.Tex t = nArea.ToString( "0.000" )

HTH,
Phill W.

[AMDRIT]

First, I would use doubles, I would use decimals.

dim decSurface as decimal =0d
dim decRadius as decimal =0d
const decPi as decimal = 3.14d
dim decResult as decimal =0

if decimal.trypars e(lblsurface.te xt,decSurface) then
else
endif

and so on

decResult = (decRadius ^3) * decSurface * 4/3
"BillE" <be****@datamti .comwrote in message
news:%2******** ********@TK2MSF TNGP06.phx.gbl. ..

In the second line you put
CDbl(lblsurface .Text, double)
it should be
CDbl(lblsurface .Text)
or
CType(lblsurfac e.Text, double)
"Ron" <pt*****@yahoo. comwrote in message
news:11******** *************@1 1g2000cwr.googl egroups.com...

>here is what I have and it does not work.

Dim _Radius As Double = CType(txtRadius .Text, Double)
CDbl(lblsurface .Text, double) = 4 * 3.14 *
(CDbl(txtRadiu s.Text) * (CDbl(txtRadius .Text)))
lblVolume.Text = (4 / 3 * 3.14 * (CDbl(txtRadius .Text) *
(CDbl(txtRadiu s.Text) * (CDbl(txtRadius .Text)))))

still not working.

Phill W. wrote:

>>Ron wrote:
I want to calculate the surface area of a sphere from an inputed
radius
with option strict on. I guess I am not converting something
correctly.

lblsurface.tex t = (4 * 3.14 * txtradius * txtradius)

You could boil all this down into a single statement, but here's the
long-hand way, so you can see all the conversions going on ...

Dim nRadius as Double = CDbl( txtRadius.Text )
Dim nArea as Double _
= 4 * Math.PI * Math.Pow( nRadius, 2 )

lblSurface.Te xt = nArea.ToString( "0.000" )

HTH,
Phill W.

[Phill W.]

AMDRIT wrote:

First, I would use doubles, I would use decimals.

.. . .

const decPi as decimal = 3.14d

Why use Decimals when all the Math methods return Doubles?
Why avoid using the Math library? The last time I looked, PI <3.14!

Regards,
Phill W.

[AMDRIT]

Because

Dim dNumber1 As Decimal = Math.PI
Dim dNumber2 As Decimal = Math.PI ^ 2
Dim dNumber3 As Decimal = dNumber2 - dNumber1 + 5&
lblsurface.Text = dNumber3

is not the same as

Dim dNumber1 As Double = Math.PI
Dim dNumber2 As Double = Math.PI ^ 2
Dim dNumber3 As Double = dNumber2 - dNumber1
lblsurface.Text = dNumber3

and if you plan on reusing any of the numbers where they mean anything, then
I recommend decimals.

"Phill W." <p-.-a-.-w-a-r-d@o-p-e-n-.-a-c-.-u-kwrote in message
news:eo******** **@south.jnrs.j a.net...

AMDRIT wrote:

>First, I would use doubles, I would use decimals.

. . .

>const decPi as decimal = 3.14d

Why use Decimals when all the Math methods return Doubles?
Why avoid using the Math library? The last time I looked, PI <3.14!

Regards,
Phill W.

[AMDRIT]

My apologies, that second dNumber3 should have read Dim dNumber3 As Double =
dNumber2 - dNumber1 + 5&

"AMDRIT" <am****@hotmail .comwrote in message
news:Oq******** ******@TK2MSFTN GP06.phx.gbl...

>
Because

Dim dNumber1 As Decimal = Math.PI
Dim dNumber2 As Decimal = Math.PI ^ 2
Dim dNumber3 As Decimal = dNumber2 - dNumber1 + 5&
lblsurface.Text = dNumber3

is not the same as

Dim dNumber1 As Double = Math.PI
Dim dNumber2 As Double = Math.PI ^ 2
Dim dNumber3 As Double = dNumber2 - dNumber1
lblsurface.Text = dNumber3

and if you plan on reusing any of the numbers where they mean anything,
then I recommend decimals.

"Phill W." <p-.-a-.-w-a-r-d@o-p-e-n-.-a-c-.-u-kwrote in message
news:eo******** **@south.jnrs.j a.net...

>AMDRIT wrote:

>>First, I would use doubles, I would use decimals.

. . .

>>const decPi as decimal = 3.14d

Why use Decimals when all the Math methods return Doubles?
Why avoid using the Math library? The last time I looked, PI <3.14!

Regards,
Phill W.

[Phill W.]

AMDRIT wrote:

Because

Dim dNumber1 As Decimal = Math.PI
Dim dNumber2 As Decimal = Math.PI ^ 2
Dim dNumber3 As Decimal = dNumber2 - dNumber1 + 5&
lblsurface.Text = dNumber3

is not the same as

Dim dNumber1 As Double = Math.PI
Dim dNumber2 As Double = Math.PI ^ 2
Dim dNumber3 As Double = dNumber2 - dNumber1
lblsurface.Text = dNumber3

and if you plan on reusing any of the numbers where they mean anything, then
I recommend decimals.

Strange, then, that Our Friends in Redmond seem satified with mere
Doubles for the Math methods... :-)

Regards,
Phill W.