Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 Re: Help Me ! » 20mm conduit (round tubing) bend » 2020-04-11 09:03:40

Thanks for this!
What I was looking for is specific to the bender that I am using, the site of Dan explaining conduit bending is making use of an American bender which has a different radius to the one that I am using. I do however have an idea. I have some conduit and will mark every 10mm along the length from before the first bend to after the second bend. After bending to the required angle I can count the marks to get the info. I will make a video showing this and try to explain my problem.
Thank you for your help!
All the best
Stay safe!!

#2 Re: Help Me ! » 20mm conduit (round tubing) bend » 2020-04-10 09:31:09

Hello Bob,
Thanks for the response, this is the sort of thing I mean. I have found things like this but found it a little confusing. Like you mentioned in bending pipe, one side stretches and the other squashes (sort of). When bending a set the pipe shortens along the length, as in, measure a 1 metre length, bend it twice to create a set(two 45 angles) the pipe will not measure 1 metre in length along the original path. if this makes sense. That is what I want to figure out. I think it is called take up. American electricians use a multiplier to do with the angle height of the triangle. I want to find the multiplier.
Thanks for the help Bob, stay safe!

#3 Help Me ! » 20mm conduit (round tubing) bend » 2020-04-10 02:22:28

Replies: 5

i have some 20mm diameter round tubing (conduit) and a bender that has a former of i think 100mm radius. i set a mark on the former where the bend starts or just before the bend.
what i would like to know is how to get a multiplier for the length of conduit.
american benders use a multiplier to work out the length of conduit to get from one place to another and include a bend.
say i wanted to have a run of conduit along a worktop, straight for about 2 foot, then go up an angle of 45 degrees, to go over a 6 inch item. then continue straight for another foot. is there a way to work out the total length of that piece?
this is what i am stuck on. in the uk we use metric but i put this example in feet.
i am using uk equipment which is not the same as american.

#4 Re: Help Me ! » cable tray and trunking for electricians » 2019-08-08 22:23:06

Hey Bob,
WOW! that's a lot more math than I thought would be involved. Thank you for the response and the picture as well. I think that I have made a mistake somewhere. the 182mm is the hypotenuse, 170mm adjacent and 64mm opposite. The trunking that I have made will sit on top of the 182mm. So CA should be 182mm. I think in my description I may have asked for where I put the centre of that fold (EA). What I meant was that the point A is where I start my measurements from as it is the only thing I know so far. From there I work back to get the centre line and then 10 degrees out the other way to get the other angle. The point A is the first angle. So F should fold back to A and FH should sit on top of the object I want to get over. I don't know if that will change the math much? I was hoping for something simple like the rise over run thing. Something that I could do on site quickly. Do you know where I could find more of this sort of thing?
Thanks again, you have been a big help
All the best, Simon

#5 Re: Help Me ! » cable tray and trunking for electricians » 2019-08-08 00:44:35

i made a video that should help describe my problem. any help would be great thanks
if you know of any sites that show this sort of thing, let me know, that would be great

#6 Help Me ! » cable tray and trunking for electricians » 2019-07-14 03:46:28

Replies: 7

i am trying to learn how to accurately measure and cut cable tray and trunking to be able to fabricate my own angles.
both of these items come in 3 metre lengths and can be cut with a hacksaw.
i want to be able to measure accurately the starting point, the cuts for the angles and the end points for my length.
i want to be able to make up sweeping 90 degree and less bends and sets.
i dont know how to insert pictures and the img thing that i read doesnt make much sense to me.
i have googled this and looked on youtube but there is not enough information on it.
the cable tray is 3 metres in length, this doesnt matter but i think the width does. it is 150mm across. i know that for a sweeping 90 degree bend there are two cuts and two bends. each bend is a 45 degree angle. but the length of the part in the centre is the concern as i have seen different lengths. i have one formula which is square root of 2 * width squared. so the width is already squared, multiply this by 2 and find the square root.
a set is two 45 degree bends, imagine a bar going horizontal then wanting to go up a step. it would do this at a 45 degree angle, the first going up and the second would level to horizontal and sit on the step. the space underneath the set would be a triangle but not nessysaraleigh right angled. any help would be great thanks

#7 Re: Help Me ! » trig problem, law of cosines » 2016-08-04 02:39:48

thanks bob, unfortunately i dont understand the ruler/protracto method. the book i am using doesnt go into that sort of thing. thanks for your help.

#8 Help Me ! » trig problem, law of cosines » 2016-08-03 23:36:36

Replies: 5

i am having a problem with law of cosines. i am using the C^2 = A^2-2ABcosc+B^2
what do i do if i dont have a value for A?
i have tried to rearrange to find A but with no luck.
any help would be great.
i get to here usually,
2ABcosc = A^2+B^2-C^2,
i then work out 2Bcosc, as i would have those values but no A
sometimes i think square root both sides. this would remove the exponents on the right and square everything on the left but that does not work.
i am using a book called "trigonometry: essentials practice workbook with answers" by chris mcmullen. it uses only certain values that mean you do not require a calculator. the values you use are 0, 30, 45 and 90 degrees. sine, cosine and tangent for these values and only those values.
any help would be great thanks

#9 Help Me ! » drawing phasor diagram » 2016-04-17 00:15:45

Replies: 0


i have a parallel rlc circuit which seems to be inductive with a 1 amp magnitude(?)
the circuit is explained in a post below about rlc circuits in parallel with complex numbers.
i havent drawn a phasor diagram. the capacitor and resistor currents are really low, in the micro amps with the inductor at 1 amp which is the supply current. so i am guessing that the phasor diagram would be the voltage pointing to 3 oclock and the current pointing toward 12 oclock, the voltage would have a magnitude of 10(volts, the supply voltage) and the current would have a magnitude of 1 amp(the supply current).
would this be correct. i have read a few things on the internet but not to sure if i have grasped the idea.



#10 Help Me ! » RLC parallel circuit, complex numbers, done it but not sure » 2016-04-15 04:16:18

Replies: 1

hello all,

been a while!

i have just started my HND electronics course and i have to do some circuit analysis.
i have a parallel RLC circuit. three components, R = 12k ohms, L = 10mH, C = 10nF
with a 10sin(1000t) supply
i have to find the total impedance in polar form
supply current in polar form
current through each component in polar form
q factor
bandwidth and
resonant frequency

i written out what i have done and attached it (hopefully)

FluxBB bbcode test
FluxBB bbcode test

not sure about this. i think that i may have done something wrong with the angles.



#11 Re: Help Me ! » laplace problem » 2015-05-13 01:57:03

no, my tutor has told me that he thinks it is wrong, though he hasnt checked his answers?

he told me he would check with other students and get back to me. i posted this question on another forum and they told me this is correct. im not sure as i am terrible at this. the question is almost exactly the same as a question posted on the internet.

this is the web site, its the first question with the answer. the only difference is the capacitor, voltage and resistor values. the whole question is identical. i followed the example changing the values as i went. so i thought that it was accurate. i just wanted to check as i have no way of knowing if this is correct. i have changed the degree that i will be studying at uni because it has less maths. thats a change in career. thats how much i drag at maths.
i just wanted to know if this is correct answer with the correct work.


#12 Help Me ! » laplace problem » 2015-05-10 21:40:03

Replies: 3

my tutor says i got the wrong answer. could someone please help!

not sure here
any help!

#13 Re: Help Me ! » not sure of the answer » 2015-04-30 21:30:19

i have just looked at simplification and cant seem to find anything that looks like the problem i have.

i did 1/400 on the calculator and pressed the S<>D button and got 0.0025 which i can rewrite as 25x10^-3 which is what i have to show.

i still don't understand how to do this.

#14 Re: Help Me ! » not sure of the answer » 2015-04-30 07:46:44

thankyou for you help bobby, please could you explain how you got this. what is the process called so I can look it up.


#15 Re: Help Me ! » not sure of the answer » 2015-04-30 02:00:37


im not actually sure how to work it out. i have to rearrange it somehow. is there a name for that process.
the bottom part is supposed to be the top rearranged to be able to find I
i dont know how to do this. a little hint would be good


i just had another look and i think that it is 1/50 instead, not 1/200
otherwise the rest is the same

#16 Re: Help Me ! » not sure of the answer » 2015-04-30 00:43:38


thanks for that bobby

is this right?


#17 Help Me ! » not sure of the answer » 2015-04-30 00:01:05

Replies: 9

i have a problem  i have this

I =  5*10^-5/1+(2*10^-2) * s = ?

i dont know s, that shouldnt matter.

i think this equals (5*10^-3) 1/200+s

i have done this on latex but cant post it so this will probably look weird, its laplace transform.
i have to rearrange to find I
any help would be great



#18 Re: Help Me ! » j notation, equivalent circuit resistance » 2014-11-01 03:17:24


could someone please let me know if this is correct as I have been told that I have done this wrong though I didn't inform the person of how I did it. this is using division of conjugate complex numbers.

I have followed the method my tutor showed me and I think this is correct.

any help or a push in the write direction would be great!!!


#19 Help Me ! » j notation, equivalent circuit resistance » 2014-10-31 22:44:03

Replies: 2


I have a circuit with values
v1 = 20 + j0
r1 = 15 +j0
l1 = 0 +j10
c1 = 0 - j5

I have to use superposition and division of complex conjugate numbers.
the resistor, inductor and voltage source are in series. the capacitor is in parallel with the voltage source

I added the resistor and inductor, (15 + jo) + (0 + j10) = 15 + j10, find the conjugate of this 15 - j10
so then product over sum for the components that are in parallel. the inductor/resistor || capacitor

    (0 - j5)*(15 - j10)
(15 + j10)*(15 - j10)

for the numerator I got

0 x 15 = 0
0 x -j10 = 0
-j5 x 15 = -j75
-j5 x -j10 = (j^2)50 = -50   minus 50 because of (j^2) = -1

so the result is -50 -j75, which im not sure about

then the second line

15 x 15 = 225
15 x -j10 =-j150
j10 x 15 = j150
j10 x -j10 = -(j^2)100,    (j^2) = -1 x -100 = 100

225 -j150 +j150 +100, the middle terms cancel leaving 225 + 100 = 325

so I have,     -50 -j75/325

then I divide each of the numerator by the denominator so,

-50/325 = -0.153846

-75/325 = -0.230769

I think I can use this to find polar notation r and theta(the zero with a line through)

im not sure if any of this is right though, any help would be great thanks


#20 Re: Help Me ! » kinetic energy » 2014-05-20 09:18:09


sorry I didn't realize shivams had posted something. I go to a college in Hampshire, not teeside.
please could you point out the other forum or thread as I don't want to have to post the same thing if its already been answered with advice

#21 Re: Help Me ! » kinetic energy » 2014-05-20 06:47:09

thanks for that answer but im terrible with math.
the graph that the tutor showed me looked like a regular graph not a quadratic.
thank you for your help

all the best

#22 Re: Help Me ! » kinetic energy » 2014-05-20 06:10:42

I don't know, the question was given like that. the mass stays the same I imagine. I just think it may have been left in as a clue to finding the answer.

#23 Help Me ! » kinetic energy » 2014-05-20 02:59:55

Replies: 8


I have these two questions

c)    Plot a graph of the kinetic energy of the mass against time. Explain your calculations and state formulae used.

d)    Plot a graph of the kinetic energy of the mass against distance. Explain your calculations and state formulae used.

I have used this formula for the first one,
Ek = ½ mv2 = ½ ma2t2         
Ek is proportional to t2.

this for the second
Ek = ½ mv2 , but we can substitute, giving
Ek = mgy

in not too sure how to show the graphs, but is the above formula along the right lines?
also I have to explain the use of the formula. I think this is just explain why I used it and any transposing required. if anyone knows a better answer for this one please give some advice.

taken from here,

this is the first part of the question,

A horizontal force of 80 N acts on a mass of 6 kg resting on a horizontal surface.  The mass is initially at rest and covers a distance of 5 m in 0.92 s under the action of the force.  Assuming there are no energy losses due to air resistance and therefore that the acceleration is constant:

a)    Calculate the total energy expended in the acceleration, I got 11.81 acceleration

b)    Calculate the coefficient of kinetic friction between the mass and the surface.  Suggest materials from which the block and table might be made in order to give such results.  I got 0.155


#24 Help Me ! » normal distribution, find percentage » 2014-05-05 08:48:05

Replies: 1


I have a question

the lengths of pins produced by a machine follow a normal distribution with mean 2.54cm and standard deviation 0.04cm. a pin is rejected if the length is less than 2.44cm or more than 2.60cm

A/ find the percentage of pins that are accepted.

B/ if it is decided that 2.5% of the pins are to be rejected because they are too long and 2.5% because they are too short, determine the new range of acceptable lengths.

for the first part I have the formula, z = x-MU / sigma, I rearranged this

sigma is 0.04cm
MU is 2.54cm

so I used this z * sigma + MU = x, z * 2.58 = ? so here I thought I would take a value from a table and divide by 2.58 until I got an accurate z value. then use that z for the rest.
if anyone could confirm this method as im not to sure? the rest I think I can get.


#25 Re: Help Me ! » evaluate integration » 2014-03-20 10:54:41


I ended up with loge 1 * 1/3 e ^3*1 and, loge 0 * 1/3 e ^3*0
I also put those upper and lower limits in the x in the [] 2x/3 = 2*1/3 and 2*0/3
and e^3*1, e^3*0
any help

im fed up of this


Board footer

Powered by FluxBB