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Atschlwi
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Solved!

adding units on a result

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hp prime

in the 1980s with the HP41 CV i habve been able to add units to the calc result like

speed=120 km/h

is there a way to do it the same way with the new PRIME calculator

thanks for any help

have a nice day

br 

may be i have to point out, that i try to solve this problem in  the DEFINE mode

so, i define a fromula and i would like to add units to the result

sorry & thanks

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Corruptissima
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 Hi Atschlwi,

Firstly, I need to know what your formula is and the units involved in the equation.

i.e 1MJ = 1000kJ = 1000kg*m^2/s^2(SI unit)=1MWatt.s(Watt-Seconds)=238.845896627kcal (Kilo Calories)

 

The equation
0.488*G^0.75*1_(MJ CE/kg) What is G? and MJ (Prime does not know that MJ is Megajoule)
so, it should be typed in as 1000000_J or 10^6_J.. CE(Conversion Energy?) does not stand for anything as Prime does not know What units (or SI units) the CE is comprised of.

Here are some of Energy units that Prime understands:

 Energyt.pngEnergy2.png

Here are picture of some conversions Prime can do

Energya1.pngEnergy21.png

If you can give me more information of the units that are involved in your equation e.g Acceleration =20_(m/s)/2_(s)=10_(m/s^2) .

View solution in original post

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Corruptissima
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Hi Atschlwi,


If you already have a formula defined as Function then you just need to modify the equation and add *1_(kph). i.e. Multiply the final result with 1 unit measure.

 

To modify the Defined function Select Shift toolbox(Mem) to get into Memory Manager -> Select User Variables then View Softkey -> then Scroll down until your Defined Function name appears -> Highlight your function then Select Edit Soft Key. At the end of your equation enter the following *1_(kph) the Ok to save the function then Esc to return to home mode.

 

My simple example here is Defined Function name: SPEED and Formula is A*B*1_(kph)

Speed1.png
To use my example function in Text mode or CAS mode is SPEED(67,23)

Speed2.png
Hope this helps.

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Atschlwi
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hi corruptissima

thank you very much for your hints

is it possible, that the calculater accepts only some how standard units.

will you pleasse figure it out for >>

mega joule convertible energy per kg weight

MJ CE / kg

0,488*G^0,75*1_(MJ CE/kg)

it doesn´t work

if i ad 1_(kph) >>> it works !!!!

 

appreciate your help

thanxs

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Corruptissima
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 Hi Atschlwi,

Firstly, I need to know what your formula is and the units involved in the equation.

i.e 1MJ = 1000kJ = 1000kg*m^2/s^2(SI unit)=1MWatt.s(Watt-Seconds)=238.845896627kcal (Kilo Calories)

 

The equation
0.488*G^0.75*1_(MJ CE/kg) What is G? and MJ (Prime does not know that MJ is Megajoule)
so, it should be typed in as 1000000_J or 10^6_J.. CE(Conversion Energy?) does not stand for anything as Prime does not know What units (or SI units) the CE is comprised of.

Here are some of Energy units that Prime understands:

 Energyt.pngEnergy2.png

Here are picture of some conversions Prime can do

Energya1.pngEnergy21.png

If you can give me more information of the units that are involved in your equation e.g Acceleration =20_(m/s)/2_(s)=10_(m/s^2) .

View solution in original post

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Atschlwi
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hi there

the formula calulates

convertible energy in mega joule  per kg weight (G) of a cow

      CE                                  MJ               /            kg

0,488 >> constant

CE =0,488*G^0,75*1_(MJ/kg)

 

so meanwhile i guess i could solve the issue

by simply removing then convertible energy CE out of the unit

because this should be clear some how anyways

 

again

thank you very much for your detailed explanation.

 

my be i´ll take a course on using the calculator one day

ha ha

 

have a nice one

 

BR

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Corruptissima
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HI Atschlwi,

 

Maybe this is the what you're looking for.

Only if Constant has value .488/Kg^2 the G is mass of a Cow in kg.

Then the equation CE is 0.488_(kg^(−2))*500^0.75*1_(kg)*1000000_(J)

so the Defined function looks like 
CEfor.png

And solution is shown in joule so you need to remove 1E6_(J) for megajoule answer.

CEfo2r.png

As there is no way to specify the MJ in prime other than programing it you have to convert the answer divide by 1E6.

CEfo3r.png

So the answer is 51.59971756559 MJ/kg. Alternatively, You can remove 1E6 in this equation since you know that answer is in terms of Megajoule/Kilo i'e 0.488_(kg^(−2))*500^0.75*1_(kg)*1_(J), But, the catch is that if you use the program later with the last equation, you might not remember if the answer should be in megajoules.

Here is a short program of calculation of CE value in MJ/kg. How ever the answer is returned in String so you cannot use it for direct number calculation.

CDPRO.pngCEProg.png

Program listing, So you can copy into the virtual Prime to test:

// Calculate Coversion Energy
// of MJ/kg of Per kilo of cow
EXPORT CE(G) // Function name
BEGIN
LOCAL ce; // Declaration of local value
ce:=0.488*G^.75; // Formular
RETURN (ce+"_MJ/kg");// Answer in MJ.kg
END;

 

Hope this help with your quest. 

 

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