VB6 to Basic4 android

Erel

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• Jul 18, 2011

• #2

B4X:

P1 = (0.0005831 + power(cE, -16.1992104 + (0.2892432 * age)))

Math.Exp (x) => power(cE, x)

 

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ciginfo

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• Jul 18, 2011

• #3

OK,thank you
B4A is very different of VB
I can 't convert also this code:

RR = 10 ^ ((0.2718 * Math.Log(Terme) * Math.Log(Terme) * 0.434294482 * 0.434294482) - (1.023 * Math.Log(Terme) * 0.434294482) + 0.9425)

Thank you if you can help me, and is there any tuto to calculate with B4A ?

François

 

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Erel

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• Jul 18, 2011

• #4

See the first post here: Basic4android Search: vb6

 

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ciginfo

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• Jul 18, 2011

• #5

I've seen but not found how to convert : Log, Round, ^

With VB
A = Log(B) / 2

A = (B ^ 2)

A = Round(B)

And with B4A ????

 

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Erel

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• Jul 18, 2011

• #6

Log => Logarithm.
^ => Power
Round => Round

 

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ciginfo

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• Jul 18, 2011

• #7

I have tried like this

Dim Zd, Zu, MoM As Double
MoM = 1
Zd = ((0.434294481903252 * (Logarithm(MoM))) - 0.305) / 0.235

ERROR

Compiling code. Error
Error compiling program.
Error description: Missing parameter.
Occurred on line: 119
Zd = ((0.434294481903252 * (Logarithm(MoM))) - 0.305) / 0.235
Word: )

I don't undersatand

 

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K

kickaha

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• Jul 18, 2011

• #8

Logarithm requires two parameters

Number (as double), and base (as double)
presuming base 10, your line would be

B4X:

Zd = ((0.434294481903252 * (Logarithm(MoM, 10))) - 0.305) / 0.235

 

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ciginfo

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• Jul 18, 2011

• #9

OK, I understand now
Thank you very much

 

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ciginfo

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• Jul 20, 2011

• #10

Logarithm requires two parameters

Number (as double), and base (as double)
If I use base = 10 it returns result very different vith VisualBasic

With VB
MoM = 15
Zd = ((0.434294481903252 * (Log(MoM))) - 0.305) / 0.235
Zd = 3.706

With Java
MoM = 15
Zd = ((0.434294481903252 * (Math.Log(MoM))) - 0.305) / 0.235
Zd = 3.706

Vith B4A
MoM = 15
Zd = ((0.434294481903252 * (Logarithm(MoM, 10))) - 0.305) / 0.235
Zd = 0.875

How choose the base, vith Vb and with Java there is not Base???
What is base?

Thank you

 

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K

kickaha

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• Jul 20, 2011

• #11

VB uses natural logs, so use the constant cE as the base.

 

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