Inspiron 1420 screen size?

Discussion in 'Dell' started by georgie, Oct 11, 2007.

  1. georgie

    georgie Guest

    I'm trying to find the dimensions of the Inspiron 1420 LCD display.
    Dell only gives the diagonal measurement. Could someone with that
    laptop please post the vertical and horizontal lengths of the viewable
    screen?

    Also, is the higher res 1440x900 suitable for "office work" or would
    1280x800 be better for reading documents?

    Thanks for your help in advance....
     
    georgie, Oct 11, 2007
    #1
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  2. georgie

    Pen Guest

    georgie wrote:
    > I'm trying to find the dimensions of the Inspiron 1420 LCD display.
    > Dell only gives the diagonal measurement. Could someone with that
    > laptop please post the vertical and horizontal lengths of the viewable
    > screen?
    >
    > Also, is the higher res 1440x900 suitable for "office work" or would
    > 1280x800 be better for reading documents?
    >
    > Thanks for your help in advance....
    >

    Dell docs are here.
    http://support.dell.com/support/edocs/systems/ins1420/en/OM/specs.htm#wp1102222
    gives all 3 dimensions.
    I've never used 144-0x900 so I can't help. 1280x800 is small enough for me.
     
    Pen, Oct 12, 2007
    #2
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  3. georgie

    Tom Scales Guest

    7.5" high
    12" wide

    I love the 1440x900 display. Even better than I expected.

    BTW, with the diagonal measurement, you could have calculated the screen
    size with a little geometry :)

    Tom

    > -----Original Message-----
    > From: georgie [mailto:]
    > Posted At: Thursday, October 11, 2007 6:34 PM
    > Posted To: alt.sys.pc-clone.dell
    > Conversation: Inspiron 1420 screen size?
    > Subject: Inspiron 1420 screen size?
    >
    > I'm trying to find the dimensions of the Inspiron 1420 LCD display.
    > Dell only gives the diagonal measurement. Could someone with that
    > laptop please post the vertical and horizontal lengths of the viewable
    > screen?
    >
    > Also, is the higher res 1440x900 suitable for "office work" or would
    > 1280x800 be better for reading documents?
    >
    > Thanks for your help in advance....
     
    Tom Scales, Oct 12, 2007
    #3
  4. georgie

    Journey Guest

    On Thu, 11 Oct 2007 19:53:11 -0400, "Tom Scales" <>
    wrote:

    >7.5" high
    >12" wide
    >
    >I love the 1440x900 display. Even better than I expected.
    >
    >BTW, with the diagonal measurement, you could have calculated the screen
    >size with a little geometry :)
    >
    >Tom


    Well, it looks like you have mixed reviews so far for resolution. I
    will join Tom and say that when I had a 14.1" laptop I liked the 1440
    x 900 resolution.
     
    Journey, Oct 12, 2007
    #4
  5. georgie <> wrote:

    >I'm trying to find the dimensions of the Inspiron 1420 LCD display.
    >Dell only gives the diagonal measurement. Could someone with that
    >laptop please post the vertical and horizontal lengths of the viewable
    >screen?


    I haven't checked the Dell website to see what the display in
    question looks like, but given the standard ratios of "legacy"
    screens compared with the display ratios of the new wide screens,
    anyone who bothered to pay attention in his classes in plane
    geometry in high school should find that calculation elementary.

    >Also, is the higher res 1440x900 suitable for "office work" or would
    >1280x800 be better for reading documents?


    That is a totally subjective judgement, depending on screen area,
    the viewer's age, the viewer's visual acuity, and the phases of
    the moon.

    >Thanks for your help in advance....


    FWIW, you're welcome.
    --
    OJ III
     
    Ogden Johnson III, Oct 12, 2007
    #5
  6. georgie

    Journey Guest

    On Thu, 11 Oct 2007 22:40:01 -0400, Ogden Johnson III
    <> wrote:

    >georgie <> wrote:
    >
    >>I'm trying to find the dimensions of the Inspiron 1420 LCD display.
    >>Dell only gives the diagonal measurement. Could someone with that
    >>laptop please post the vertical and horizontal lengths of the viewable
    >>screen?

    >
    >I haven't checked the Dell website to see what the display in
    >question looks like, but given the standard ratios of "legacy"
    >screens compared with the display ratios of the new wide screens,
    >anyone who bothered to pay attention in his classes in plane
    >geometry in high school should find that calculation elementary.
    >
    >>Also, is the higher res 1440x900 suitable for "office work" or would
    >>1280x800 be better for reading documents?

    >
    >That is a totally subjective judgement, depending on screen area,
    >the viewer's age, the viewer's visual acuity, and the phases of
    >the moon.
    >
    >>Thanks for your help in advance....

    >
    >FWIW, you're welcome.
    >--
    >OJ III


    Well then instead of talking about it why don't you give him a
    geometry lesson, professor OJ III.
     
    Journey, Oct 12, 2007
    #6
  7. Let's see ....

    [Diagonal Measurement]**2 = (width**2) + (height**2)

    width = (1440/800)*height

    System of two simultaneous equations; solve for Diagonal Measurement

    The rest is left as a exercise for the student.


    Journey wrote:
    > On Thu, 11 Oct 2007 22:40:01 -0400, Ogden Johnson III
    > <> wrote:
    >
    >> georgie <> wrote:
    >>
    >>> I'm trying to find the dimensions of the Inspiron 1420 LCD display.
    >>> Dell only gives the diagonal measurement. Could someone with that
    >>> laptop please post the vertical and horizontal lengths of the viewable
    >>> screen?

    >> I haven't checked the Dell website to see what the display in
    >> question looks like, but given the standard ratios of "legacy"
    >> screens compared with the display ratios of the new wide screens,
    >> anyone who bothered to pay attention in his classes in plane
    >> geometry in high school should find that calculation elementary.
    >>
    >>> Also, is the higher res 1440x900 suitable for "office work" or would
    >>> 1280x800 be better for reading documents?

    >> That is a totally subjective judgement, depending on screen area,
    >> the viewer's age, the viewer's visual acuity, and the phases of
    >> the moon.
    >>
    >>> Thanks for your help in advance....

    >> FWIW, you're welcome.
    >> --
    >> OJ III

    >
    > Well then instead of talking about it why don't you give him a
    > geometry lesson, professor OJ III.
     
    Barry Watzman, Oct 12, 2007
    #7
  8. georgie

    Journey Guest

    On Fri, 12 Oct 2007 01:50:17 -0400, Barry Watzman
    <> wrote:

    >Let's see ....
    >
    >[Diagonal Measurement]**2 = (width**2) + (height**2)
    >
    >width = (1440/800)*height
    >
    >System of two simultaneous equations; solve for Diagonal Measurement
    >
    >The rest is left as a exercise for the student.


    Barry, what do you say we cut to the chase?:

    http://tinyurl.com/2jlqdx

    Follow the simple instructions on that page.
     
    Journey, Oct 12, 2007
    #8
  9. georgie

    alien Guest

    "Journey" <> wrote in message
    news:...
    > On Fri, 12 Oct 2007 01:50:17 -0400, Barry Watzman
    > <> wrote:
    >
    >>Let's see ....
    >>
    >>[Diagonal Measurement]**2 = (width**2) + (height**2)
    >>
    >>width = (1440/800)*height
    >>
    >>System of two simultaneous equations; solve for Diagonal Measurement
    >>
    >>The rest is left as a exercise for the student.

    >
    > Barry, what do you say we cut to the chase?:
    >
    > http://tinyurl.com/2jlqdx
    >
    > Follow the simple instructions on that page.


    I came up with 3.3 Is that right?

    alien
     
    alien, Oct 12, 2007
    #9
  10. georgie

    alien Guest

    "Journey" <> wrote in message
    news:...
    > On Fri, 12 Oct 2007 01:50:17 -0400, Barry Watzman
    > <> wrote:
    >
    >>Let's see ....
    >>
    >>[Diagonal Measurement]**2 = (width**2) + (height**2)
    >>
    >>width = (1440/800)*height
    >>
    >>System of two simultaneous equations; solve for Diagonal Measurement
    >>
    >>The rest is left as a exercise for the student.

    >
    > Barry, what do you say we cut to the chase?:
    >
    > http://tinyurl.com/2jlqdx
    >
    > Follow the simple instructions on that page.


    BTW, a cut and paste from the last line on that page:

    (Can you think of an important physical mechanism that this model
    neglects?)

    Yes, my brain. :)

    alien
     
    alien, Oct 12, 2007
    #10
  11. georgie

    Journey Guest

    On Fri, 12 Oct 2007 15:17:45 GMT, "alien" <>
    wrote:

    >
    >"Journey" <> wrote in message
    >news:...
    >> On Fri, 12 Oct 2007 01:50:17 -0400, Barry Watzman
    >> <> wrote:
    >>
    >>>Let's see ....
    >>>
    >>>[Diagonal Measurement]**2 = (width**2) + (height**2)
    >>>
    >>>width = (1440/800)*height
    >>>
    >>>System of two simultaneous equations; solve for Diagonal Measurement
    >>>
    >>>The rest is left as a exercise for the student.

    >>
    >> Barry, what do you say we cut to the chase?:
    >>
    >> http://tinyurl.com/2jlqdx
    >>
    >> Follow the simple instructions on that page.

    >
    >BTW, a cut and paste from the last line on that page:
    >
    > (Can you think of an important physical mechanism that this model
    >neglects?)
    >
    >Yes, my brain. :)
    >
    >alien


    It's interesting to see people talking about it being a simple high
    school problem, yet I haven't seen anyone solve the problem yet --
    funny.

    I have a math minor so it shouldn't be too hard. Today though I am
    preparing a computer for sale (selling the M1210 that I got for $630
    for $1,000 -- nice profit), and taking someone to a recovery-based
    meeting for the first time so don't have time at the moment.

    When I have time I'll give it a shot, and if I can't, I'll chalk it up
    to middle age and "if you don't use it you loose it".
     
    Journey, Oct 12, 2007
    #11
  12. I gave you the formula. It's simple algebra, just substitute the 2nd
    formula into the first.


    Journey wrote:
    > On Fri, 12 Oct 2007 15:17:45 GMT, "alien" <>
    > wrote:
    >
    >> "Journey" <> wrote in message
    >> news:...
    >>> On Fri, 12 Oct 2007 01:50:17 -0400, Barry Watzman
    >>> <> wrote:
    >>>
    >>>> Let's see ....
    >>>>
    >>>> [Diagonal Measurement]**2 = (width**2) + (height**2)
    >>>>
    >>>> width = (1440/800)*height
    >>>>
    >>>> System of two simultaneous equations; solve for Diagonal Measurement
    >>>>
    >>>> The rest is left as a exercise for the student.
    >>> Barry, what do you say we cut to the chase?:
    >>>
    >>> http://tinyurl.com/2jlqdx
    >>>
    >>> Follow the simple instructions on that page.

    >> BTW, a cut and paste from the last line on that page:
    >>
    >> (Can you think of an important physical mechanism that this model
    >> neglects?)
    >>
    >> Yes, my brain. :)
    >>
    >> alien

    >
    > It's interesting to see people talking about it being a simple high
    > school problem, yet I haven't seen anyone solve the problem yet --
    > funny.
    >
    > I have a math minor so it shouldn't be too hard. Today though I am
    > preparing a computer for sale (selling the M1210 that I got for $630
    > for $1,000 -- nice profit), and taking someone to a recovery-based
    > meeting for the first time so don't have time at the moment.
    >
    > When I have time I'll give it a shot, and if I can't, I'll chalk it up
    > to middle age and "if you don't use it you loose it".
     
    Barry Watzman, Oct 12, 2007
    #12
  13. georgie

    georgie Guest

    Okay, I'm up for a challenge. I had to add a little algebra to that
    geometry, so first needed a quick refresher on solving equations with
    square roots.
    7.47" x 11.95" (HxW)
    Q.E.D.

    Thanks for making the answer to my question both easy and interesting
    Tom. ;)



    On Oct 11, 7:53 pm, "Tom Scales" <> wrote:
    > 7.5" high
    > 12" wide
    >
    > I love the 1440x900 display. Even better than I expected.
    >
    > BTW, with the diagonal measurement, you could have calculated the screen
    > size with a little geometry :)
    >
    > Tom
    >
    > > -----Original Message-----
    > > From: georgie [mailto:]
    > > Posted At: Thursday, October 11, 2007 6:34 PM
    > > Posted To: alt.sys.pc-clone.dell
    > > Conversation: Inspiron 1420 screen size?
    > > Subject: Inspiron 1420 screen size?

    >
    > > I'm trying to find the dimensions of the Inspiron 1420 LCD display.
    > > Dell only gives the diagonal measurement. Could someone with that
    > > laptop please post the vertical and horizontal lengths of the viewable
    > > screen?

    >
    > > Also, is the higher res 1440x900 suitable for "office work" or would
    > > 1280x800 be better for reading documents?

    >
    > > Thanks for your help in advance....
     
    georgie, Oct 13, 2007
    #13
  14. georgie

    Journey Guest

    On Fri, 12 Oct 2007 19:35:30 -0700, georgie <>
    wrote:

    >Okay, I'm up for a challenge. I had to add a little algebra to that
    >geometry, so first needed a quick refresher on solving equations with
    >square roots.
    >7.47" x 11.95" (HxW)
    >Q.E.D.
    >
    >Thanks for making the answer to my question both easy and interesting
    >Tom. ;)


    I cannot even begin to explain how intensely I was restraining myself
    from providing this answer to you since the instant you posted. Thank
    you -- now I can rest.
     
    Journey, Oct 13, 2007
    #14
  15. georgie

    Boris Guest

    Barry Watzman <> wrote in news:470f0b93$0$32516
    $:

    > Let's see ....
    >
    > [Diagonal Measurement]**2 = (width**2) + (height**2)
    >
    > width = (1440/800)*height
    >
    > System of two simultaneous equations; solve for Diagonal Measurement
    >


    Easy. Done this many times to determine the height lost when considering
    widescreen vs standard screen with same diagnonal measurement. (Without
    the aspect ratio, though, there's an infinite number of H X W
    possibilities).

    Let a=height
    Let b=width
    Let c=diagonal

    Given:
    a=900
    b=a(1440/900); or b=1.6a
    c=14.1
    a**+b**=c**

    Solution:
    a** + (1.6a)** = 14.1**
    etc.
    ..
    ..
    ..
    a = sqrt of 55.84550561
    a = 7.47298505
    b = 1.6a
    b = 11.95677608

    height = 7.47298505"
    width = 11.95677608"
    diagonal = 14.1"
     
    Boris, Oct 13, 2007
    #15
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