Hey /diy/ I would like your opinions on a tools bag. I'

>>1084940
thinking about getting a backpack. i believe its more handy and better to carry. The one on the picture seems alright, google makita tool backpack, ive seen on of those IRL and it looks like there is more room in he middle, for example to a laptop.
it also has a hard plastc shell covering the bottom part, which is pretty good. saves your eq. from any water on the ground and protects against impact( a little)

Also; could you help me with a calculation? Im studying electrics and i cant get this
equation to work on my calculator
And i cant really start a new thread.

46200/230*√3*cosϕ=

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>>1084940
It is better. I'm a gas mechanic and have pic related as a test run. You have all your stuff with you, without having this one-sided load on one shoulder. You also have both hands free (which helped me not falling down stairs). I'd recommend it.

>>1084951
not op
thats right! didn't think about the whole two hands free thing

>>1084946
Are you trying to solve for theta? What is it for? If the equation is cos(46200/230*√3) that's approximately -0.6964 radians.

>>1084959
tnx, its for calculating total amps needed in an apartment iafter i know the total wattage; Its current =wattage/voltage*v3 (since its 3 lined cable) and cosϕ dont remember why cos :9

>>1084968
the example in my book is; Current=wattage(32000W) divided by 230V*v3*cos

the answer in the book example here is 72 A.
does this make sense? I'm not sure if im explaining it right

>>1084946
Can't find the exact reference for your Makita, can you link it please ?

>>1084951
This one look almost perfect for what I want, I'll try to find one IRL to compare, thx !

>>1084985
here, try this
http://www.ebay.co.uk/itm/Makita-Professional-Tool-Rucksack-Toolbag-Backpack-Tool-Bag-Organiser-P-72017-/310410112649

>>1084972
May I see the problem in the book? Last I checked, as a retarded EE student, p=iv or power (watts) = current * voltage, so to find current with a given wattage and voltage should be i=p/v.

I'll admit I'm not familiar with AC (first year student), but I'm guessing at 60 hz, the standard NA standard, the cosine will equal 0.5 because cos(60°)=0.5

So i=(32kW/230V)*0.5
i=69.5652 A
It must come to the the cable, that v3? Or I'm doing something wrong with the AC. Sorry I'm not really helping that much.

>>1084996
thank you for looking into it, will have the example in a sec. Im in scandinavia, so we have a regular of 230 volt in the outlet at 50hz

(I) is the icon for current -amps
(P) is the icon for power -Watts
(U) is the icon for pressure- Volt

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>>1084996

>>1085001
here it is, after summing upp all effect needed in the apartment in the example(on the prev. page) it basically say:

we sum up all the power and end up with 32KW, if everything is turned on at the same time the aprt. would need 72 amps inn.
this is found/calculated by: I=P/U*√3*cosϕ
which equals 72 Amps,

my example uses the same formula but with 46200 W so im guessing i would get something above 80A

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>>1085001
I'm not sure I'm doing this right, but from what I've found.

>>1085017
>>1084996
hope this makes sense, and tnx again for looking into it?

>>1085020
that is the textbook example yes. tnx but im not able to calculate it. whenever I try putting it in the calculator i can never get the right answer, im probably doing something wrong with either v3 or cos.

what id you use there?
how did you get the
pf=cos(50*pi/180)
and what does pf stand for?

>>1085026
auming:
cos has to do with the frequency? so since we use 50hz you get >>1085020
50*pi? is that why you have 50 there?

>>1085026
I got pf from Power Factor found:
http://www.rapidtables.com/electric/Power_Factor.htm
and the formula found:
http://www.rapidtables.com/calc/electric/Watt_to_Amp_Calculator.htm

PF = |cos(a)|
50 hz for the angle (in degrees)
pi/180 for converting it into radians

Are we solving for phi?

>>1085032
what do you mean by "are we solving for pi?"

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>>1085034
I'm retarded, confusing the example with the problem.

If we are solving for current, i:
We know:
P = 46200 W
U = 230 V

but, what is cosφ?
From the example, I found that:
φ = 0.0000 + 0.4764i, some imaginary number, wut?
But when I plug it back into the example, I get 72 A, as I should, but why is this?

>>1085034
sorry for derailing the thread btw.

I think tht one of the vital parts of any toolbag is that solid plastic shell you find underneath, some has it streching quite high on the sides of the bag too. Like stanley Fatmax.

It protects your tools but could also protect the ground as you put it down. As a joiner/cabinet fitter, i usually get customers with new and really expensive flooring, which sometimes get dents just by looking at it. then its quite handy to have that protection under your bag, which prevents any of your tools from hitting the floor

>>1085044
tnx, could it make sense that φ means radient angle/degree?
back when i first was introduced to cos i learned it can mean two things, and in EE we only use it as one of them- radians

>>1085044
just read up that the PF here in Norway with 230V@50Hz is around 0.44. which is why we probably get that number.

Thank you for all the help. I gotta read up on PF and try to stick to my main language. sadly most of these things are better explained in English

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>>1085059
I'm still trying to figure out what φ is equivalent to.
From what I have read, it is the apparent power phase angle, which sadly doesn't mean much to me right now. How do you find that? The included picture means nothing, but is what I have found, given that φ is a constant, which I sincerely doubt is.

>>1085063
i see. i was hoping it would be a constant. would make things easier. what type of program do you use there?

>>1085065
I'm currently using MATLAB.

It was a requirement for one of my classes, and I got a deal for being with my university, so I might as well use it when I can. Without the license, it's ridiculously priced. It's essentially a glorified calculator. There's other programs that work similar, and many of them are pretty good.

Hope you can figure the problem out, sorry I wasn't able to help much, and for derailing a thread.

>>1085087
no, thank you for all the help

>>1085061
Power factor depends on the circuit you're running, it's not determined by your country's voltage.

https://www.youtube.com/watch?v=81n7HrLn3Ng

After some search I've found this for 110€

https://www.milwaukeetool.com/hand-tools/storage/48-22-8200

We don't use Milwaukee tools in France, any info on this brand ? The bag looks great.

>>1084940
>Printer bitch
>IT

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>>1084951
the Stanley looks nice

>>1084946
>46200/230*√3*cosϕ=
make sure your calculator is set for radians/degrees before you choose your angle for phi

>>1084968
oh god, lol

OK OP cos(phi) in this case is actually a symbol for something called "Power factor" in essence a power factor is just a variable that dictates the lag of the sin wave

things like motors and anything involving coils/inductance/transformers all form a power factor, which shifts the sin wave slightly and ends up "holding" power (this is why some companies can claim that their motor outputs over 3Hp despite the fact at 15A and 120W only being able to pull 2.4HP from the wall is because of the momentary power over 3HP from the power factor

when your book says "cos(Phi)" they are referring to the power factor of the circuit (which is different for all circuits

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Not exactly a bag but I'm a biomedical equipment tech and have to transport tools and parts through hospitals and labs. I have a main hardshell "technician's case" and several ancillary toolboxes depending on the job I'm doing that I roll around on one of these carts. I'd imagine that if you could fit one in your vehicle, it ought to work well in an office environment.