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#1 2006-02-03 06:37:10

johny
Member
Registered: 2005-11-19
Posts: 34

A problem!!!!

Heres the question:

A lorry has a downwards wiping windscreen wiper on a 40cm long metal arm. The rubber wiper part is at the end of the arm and is an articulate double module that alters in length during motion. The rubber alters between 30cm and 20cm, during a wiping sweep of the glass.

The full angle of sweep is 120 degrees and motion is symmetrical about the y-axis.

The bottom end of the rubber follows a path according to the equation y= 1/50x^2

The upper end of the rubber follows a path according to the equation y= 1/50x^2 + 30.

We need to calculate the area of the glass cleared in one full sweep of the arm.

I couldnt really do the question, but i believe it does involve integration. Can anyone please helo me out here??

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#2 2006-02-03 09:17:50

irspow
Member
Registered: 2005-11-24
Posts: 455

Re: A problem!!!!

The problem seems simple enough, however I am having trouble visualizing what is happening here.  The description of the wiper has left me baffled.  It probably isn't your fault though as I am often baffled by thy simplest things.

  In general integrating a function will give the area below this curve and the x axis.  You however have two functions of y.  You will need to subtract the integration of the lower y function from the integration of the upper one.

  If I could visualize this arm-blade relationship better the equation would be quite simple, really.  Your limits of integration would be related to the range of x for both equations dictated by the total angle of rotation.  The functions that you already have are also easily integratable.

  Could you post a picture of the mechanism in question?

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#3 2006-02-03 22:46:22

johny
Member
Registered: 2005-11-19
Posts: 34

Re: A problem!!!!

Yeah i am really sorry i forgot to post the pic!!!! i will upload it!!!!

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#4 2006-02-03 22:49:45

johny
Member
Registered: 2005-11-19
Posts: 34

Re: A problem!!!!

Heres the pic!

View Image: Maths100.JPG

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#5 2006-02-04 04:54:26

irspow
Member
Registered: 2005-11-24
Posts: 455

Re: A problem!!!!

Good picture johny.  Also there are more complications coming up.  We will need to know the rate at which the rubber is changing in length.  Is this a constant rate of change?  In other words we a radius that is changing here and it needs to be defined.

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#6 2006-02-04 05:22:10

johny
Member
Registered: 2005-11-19
Posts: 34

Re: A problem!!!!

Hey irspow, well it doesnt say any rate so it might aswell be constant. The only think it says is the rubber alters between 30cm and 20cm, thats about it.....so i would think its constant rate of change. If any more questions, not a prb smile!!!!

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#7 2006-02-04 05:57:02

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: A problem!!!!

The upper end of the rubber follows a path according to the equation y= 1/50x^2 + 30.

But then the lenght of the rubber is 30.  In your diagram, it is only 20.

Assuming that's supposed to be 20, what you have are:

y1= 1/50x^2 + 20
y2= 1/50x^2
y3= √(2)x + 30
y4= -√(2)x + 30

a = intersection between y2 and y3
b = intersection between y1 and y3
c = intersection between y1 and y4
d = intersection between y2 and y4

Last edited by Ricky (2006-02-04 05:57:28)


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#8 2006-02-04 06:31:45

johny
Member
Registered: 2005-11-19
Posts: 34

Re: A problem!!!!

Hey ricky.....i know it does seem confusing but i posted the exact question and diagram from the book!!!

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#9 2006-02-04 06:38:50

irspow
Member
Registered: 2005-11-24
Posts: 455

Re: A problem!!!!

I can't tell what part of the wiper is fixed to the blade.  Top, bottom, or center?  I tried resolving the y functions given with trigonomic functions and couldn't figure it out.

This information is critical to determine the upper and lower bounds for x in the integration.

Last edited by irspow (2006-02-04 06:40:22)

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#10 2006-02-04 06:51:58

johny
Member
Registered: 2005-11-19
Posts: 34

Re: A problem!!!!

I guess the wiper is fixed to the blade below 10cm from the top, meaning at the center.

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#11 2006-02-04 07:05:32

irspow
Member
Registered: 2005-11-24
Posts: 455

Re: A problem!!!!

That can't be true. Then the distance from the top of the arm to the top of the blade at 0° would be 5 and not the 10 shown in the diagram.  It is now seeming a little funny how these little details can be huge obstacles in what would otherwise be a simple calculation.

Either way, Ricky's general solution holds true when you do figure out the correct limits of integration.

Last edited by irspow (2006-02-04 07:07:33)

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#12 2006-02-04 07:32:27

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: A problem!!!!

is an articulate double module that alters in length during motion

If that line is directly out of your book, then I suggest you get another book...

Seriously, what kind of text book uses that language without explaining it?  I haven't got a clue what it means.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#13 2006-02-04 07:37:10

irspow
Member
Registered: 2005-11-24
Posts: 455

Re: A problem!!!!

That is kind of where my thinking has gone.  The book probably made some strange assumption and that the readers would make the same also without specifically stating what it was.  I am from the U.S. and I don't know of any wipers that change in length here, so the entire premise is new to me.

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#14 2006-02-04 07:39:45

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: A problem!!!!

is an articulate double module that alters in length during motion

If that line is directly out of your book, then I suggest you get another book...

Seriously, what kind of text book uses that language without explaining it?  I haven't got a clue what it means.

Edit:

Whoops, missed this line:

A lorry has a downwards wiping windscreen wiper on a 40cm long metal arm.

Ok, so we start at +40, not +30.  Then everything comes together.

y1= 1/50x^2 + 30
y2= 1/50x^2
y3= √(2)x + 40
y4= -√(2)x + 40

Everything else is the same.

is an articulate double module that alters in length during motion

What this means is that the distance from the vertex and the start of the rubber is constantly changing.  The equations above show this distance changing, it's nothing extra.  Just an explanation.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#15 2006-02-04 07:46:37

johny
Member
Registered: 2005-11-19
Posts: 34

Re: A problem!!!!

Ok so when i put the values in the equation u gave me before would give me the area of glass cleared???

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#16 2006-02-04 07:50:20

irspow
Member
Registered: 2005-11-24
Posts: 455

Re: A problem!!!!

It is nice that you now understand all of this Ricky, but how do you know the limits of integration for y1 and y2 taking into account that the blade's length is changing by 10 centimeters.

  Oh, and we really only need y1, y2, and y3.  You would just have to double the integration of these because both halves of the sweep are symmetrical.

  No, johny, they would be the four equation that you would need to integrate over the proper limits of integration.  The sum of these integrations would give you the area that you are seeking.

Last edited by irspow (2006-02-04 07:52:56)

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#17 2006-02-04 07:58:23

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: A problem!!!!

It is nice that you now understand all of this Ricky, but how do you know the limits of integration for y1 and y2 taking into account that the blade's length is changing by 10 centimeters.

The equations describe that change of 10 centimeters.

p0 = the vertex
p1 = point of intersection between y2 and y3
p2 = (0,0) (the end of the blade passes through this point)

||p1 - p0|| is the distance between p1 and p0:

||p1 - p0|| - 10 = ||p2 - p0||

But the wiper passes through both p1 and p2.  So the wiper must be changing size.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#18 2006-02-04 08:02:26

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: A problem!!!!

To sum up for Jonny, the answer (so far) is:


y1= 1/50x^2 + 30
y2= 1/50x^2
y3= √(2)x + 40
y4= -√(2)x + 40

a = intersection between y2 and y3
b = intersection between y1 and y3
c = intersection between y1 and y4
d = intersection between y2 and y4


You can use what irspow said to simplify the integration, if you wish.

Last edited by Ricky (2006-02-04 08:02:37)


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#19 2006-02-04 08:06:52

irspow
Member
Registered: 2005-11-24
Posts: 455

Re: A problem!!!!

Sooo, what are the proper limits of integration for y1 and y2?  I am thick-headed and am still confused.  If you know them then you could post the area for johny.

edit*

   I am not trying to be a jerk here,  I really want to understand the solution.

Last edited by irspow (2006-02-04 08:08:17)

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#20 2006-02-04 08:09:00

johny
Member
Registered: 2005-11-19
Posts: 34

Re: A problem!!!!

i doubt u guys are thick heads but i sure am, so if u have time then it would be awesome if u post the area and i would be really greatful smile

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#21 2006-02-04 08:20:55

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: A problem!!!!

Oh, ok, I misunderstood you before irspow.  But I'm still not sure what you mean now:

Sooo, what are the proper limits of integration for y1 and y2?

What do you mean limits for y1 and y2?  My solution never integrates y1 and y2.  It integrates y1 - y2, is that what you mean?

If so, the limits are b to c.

b = intersection between y1 and y3
c = intersection between y1 and y4

√(2)x + 40 = 1/50x^2 + 30
b=-5(√(35)-5)*√(2) or -6.47766

1/50x^2 + 30= -√(2)x + 40
c=5(√(35)-5)*√(2) or 6.47766

so you integrate y1 - y2 from ~-6.5 to ~6.5


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#22 2006-02-04 08:39:05

irspow
Member
Registered: 2005-11-24
Posts: 455

Re: A problem!!!!

I think you need something like;

  {∫f(1) - f(2) dx] from 0 to b + ∫f(3) - f(2) dx] from b to max x for f(2)} times 2

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#23 2006-02-04 09:25:22

irspow
Member
Registered: 2005-11-24
Posts: 455

Re: A problem!!!!

My problem is in post #14 it was stated that the top of the wiper blade is where the change in length occurred.  If that is true then the blade is fixed at the bottom of the arm.  Thus the radius of y2 would always be 40 from the vertex.

From this;

40 - 40cos60° would equal y2 at the end of its travel.

40 - 40cos60° = 20

Thus max x = 40√(1 - cos²60°) ≈ 34.641...

But plugging this value of x into y1 = 24

24 ≠ 20,  so there is a problem with one of the assumptions being made.

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#24 2006-02-04 09:39:48

irspow
Member
Registered: 2005-11-24
Posts: 455

Re: A problem!!!!

Actually, I just ran through the trigonomic relations regarding the situations with the blade fixed at the top and y1 and they also didn't match.  I also revisited the situation for fixed bottom and it too failed to create a match for y1.  I cannot see what is going on here.

If we use the 6.47766 value as the upper bound for y1 then;

sin 60° = 6.47766 / L ;  where L is the length from the vertex to the top of wiper

L = 7.4797...

Where does a variation of (10 - 7.4797...) = 2.5202... for the top of the blade come from at the extreme of travel?

Last edited by irspow (2006-02-04 09:48:57)

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#25 2006-02-04 17:24:32

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,552

Re: A problem!!!!

Is the wiper blade centered on the end of the wiper arm?


igloo myrtilles fourmis

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