Return Styles: Pseud0ch, Terminal, Valhalla, NES, Geocities, Blue Moon.

Pages: 1-

I'm short and I want that to change.

Name: Anonymous 2006-01-14 19:59

I'm short (5' 7") and almost 18 years old.

At this point in life people don't really get much taller than whatever height they currently sustain. I exercise and weight lift on a fairly regular basis and my face ain't half bad, but everytime I look in the mirror I can't help but feel absolutely retarded because of how short I am. Girls like a boy keeps himself in shape but one of the key things that can make or break the way a girl thinks about you... is height.

Most women aren't very tall and having a towering presence, becuase most straight women like to be "swept off their feet" or have a big strong man to "take them", over them is something most find very sexy.

So, finally, to my point, would taking human growth hormones be a safe and sensible idea in order to increase my stature? What are the advantages and drawbacks to using it? 4chan, please, endow me with your knowledge!

Name: Anonymous 2006-01-14 21:24

Some people do grow taller when they are 18-22 years. So growth hormones might help you grow vertically instead of horizontally.

Name: Anonymous 2006-01-14 21:28

Wow.  And I thought my roommate in college was insecure about his height.  And he's 5"3'.

5'7" isn't short, so unless you can find a 7" penis pump, deal with it.

Name: Anonymous 2006-01-14 21:30

7'.  7'.  Though 7" is almost as funny.  Damnable typos.

Name: guy who started the thread 2006-01-14 22:11

5' 7" is when almost all the girls you know (or want) are your same height or taller. Women don't like it when a guy is shorter than them, even by one inch. A girl told me that... as one of the reasons she didn't like me anymore. If I get taller using growth hormones, maybe I won't develop a complex over this later in life.

Are there any dangers to using this stuff? What's this I hear about penises shrinking? Is that something that actually happens or is it just more urban myth bullshit (which I know it probably is but in the back of my head I can't help but keep the smallest shred of doubt)?

Name: Mel !iKGMr61IHM 2006-01-14 23:22

Go to China, pay them a few thousand dollars, and they'll break your legs and set the bones apart from each other so once they've gradually healed and reconnected, you'll be taller.

Name: Anonymous 2006-01-15 0:58

5' 7" isn't too short at all.  It's true women prefer guys who are as tall or taller than them, but there are lots of girls shorter than you are.  And since you're working out, you've got an advantage, since the muscle mass will give you more presence.  You don't have a problem worth risking your health on with growth hormones.

Name: Anonymous 2006-01-15 7:06

Nature usually doesn't deal you low cards unless you are seriously fucked. If you are tall you will be good at running and if you are short you will be good at weight lifting. All you have to do is work out 5 hours a day.

Name: Anonymous 2006-01-15 11:28

HGH and insulin like growth factor are both worthless when taken in pill form.  They need to be injected, which can cost hundreds.  The Anti-Aging Institute can do this for you.

Name: Anonymous 2006-01-15 11:41

>>5

"Women don't like it when a guy is shorter than them, even by one inch. A girl told me that... "

And you actually believed her?  Bwaahahaaa!

A) Never believe what they say when they tell you why they don't like you.  It's rarely, if ever, the *real* reason.

B) Even if it *is* the reason, your evidence is *purely* anecdotal.  And I meet your anecdotal evidence and raise you by one:  (1) My kid brother is dating an absolute *babe* who's about 6 inches taller, and they're in it for the long-haul.  (2) My *mother* married my *father* when she was about 4 inches taller than him.

The *real* reason she didn't like you is because you're a whiny, insecure little bitch.

Name: Anonymous 2006-01-15 15:21

Is 180 cm. considered tall?

Name: Anonymous 2006-01-16 7:02

You do know that taking HGH increases your risk for cancer?

Name: Anonymous 2006-01-16 9:37

WOMEN LIE.  DON'T BELIEVE ANYTHING THEY SAY. IN FACT, TAKE WHATEVER THEY SAY, AND THE OPPOSITE IS TRUE.

Name: Anonymous 2006-01-16 10:38

>>13 is a woman

Name: Anonymous 2006-01-16 16:50

>>10
truth

Name: Anonymous 2006-01-16 17:00

>>10
untrue, i find that i tend to look down on short guys (ho ho ho)

it's not that i make any huge decisions based around this or anything and i could well wind up marrying a short guy, i simply prefer men my height or taller (186cm+)

and besides, if you're going to think of a fake reason to ditch a guy his height is a seriously bad excuse

Name: Anonymous 2006-01-16 17:22

>>10

You obviously haven't dealt with to many women these days. A guy under 6ft gets no repect unless the girl is either;fat, ugly, will do it with anyone, or is insecure enough about her own looks to not care.

Name: Anonymous 2006-01-16 17:25

>>17

You've just described about 70-75% of the female population.

Name: Anonymous 2006-01-17 13:30

Fuck...
I think I'll go gay.

Name: Anonymous 2006-01-17 17:54

me 2

Name: Anonymous 2006-01-17 18:50

>>6

I recall reading about a girl who wanted to be an air hostess, but was too short, so she had her legs broken and re-set in that manner. Unfortunately, after two years of this, she was still not tall enough .. hahaaha

Name: Anonymous 2006-01-17 23:25

air hostess? such a low goal

Name: Anonymous 2006-01-18 19:11

so nobody really knows anything about HGHs?

Name: Anonymous 2006-01-18 19:54

>>1
You're not short stupid
Stop being emo and go have fun
Oh and if you take hormones you're really, really retarded
Even worse so that you ask Anonymous instead of your doctor

Name: Anonymous 2006-01-21 0:39

HGH is fine in injected form.  It increases growth of muscle and orgrans.  The body naturally produces it until you get older, so the procedures merely supplements it.

Name: Anonymous 2006-01-21 3:55

>>1
You consider that short??? I've got frens that are like 5' 4". And they STILL manage to find girls shorter than them. In fact one of them even had a girl thats like taller than him.

I think 5'3" or above is an ok height for any guy. Majority of girls I have seen are either around or shorter. Though that is asian girls :D If we are talking about western girls, well then u r screwed :D

On the flip side, I got a female friend who is like only 5' (and 22 years old). You have to pity her bcos every guy to her are like huge and that she can't really use the cheap height excuse if she wanted to.

Name: Anonymous 2006-01-21 15:59

I'm 6'1"

Will you admit you are inferior to me to boost my ego?

Name: Anonymous 2006-01-21 20:46

I'm 6'4", being tall has it's downsides too, trying to find a decent pair of shoes size 13 for example... and growth hormones won't help if you stopped growing already, at least that's what I thought...

Name: Anonymous 2006-01-22 12:31

Some people have growth spurts at 18-22 years.

Name: Anonymous 2006-01-22 18:23

I'm 25, motherfucker.

Name: guy who started the thread 2006-01-23 1:22

ok, well i guess i dont really have much to worry about then. im 17 and i have almost no facial hair, which means my growth spurt should come up pretty soon i guess

Name: Anonymous 2009-04-02 17:25

fds

Name: Anonymous 2009-04-02 17:26


2L \int_0^h \sqrt{R^2-(R-y)^2} dy

Substitute R-y = R \sin \theta, dy = -\cos \theta d\theta, so that \theta = \sin^{-1} \frac{R-h}{R}

2L \int_0^{\sin^{-1} \frac{R-h}{R}} \sqrt{R^2-(R \sin \theta)^2} (-R\cos \theta) d\theta = -2LR^2 \int_0^{\sin^{-1} \frac{R-h}{R} \cos^2 \theta d\theta
= -2LR^2 \int_0^{\sin^{-1} \frac{R-h}{R} \frac{1+\cos 2\theta}{2} \theta d\theta
= -2LR^2 \left[\frac{\theta}{2} + \frac{\sin 2\theta}{2}\right]_0^{\sin^{-1} \frac{R-h}{R}}

Name: Anonymous 2009-04-02 17:26

1) Refer to the following picture: http://www.imagedump.com/index.cgi?pick=get&tp=547398

The volume is the length of the tank times the area of the yellow shaded area of the circular cross section.  To find the area, we're going to integrate the length of a horizontal slice running vertically from the bottom of the tank to the top of the water.  The length marked as "dist" is \sqrt{R^2-(R-y)^2}, so we have to integrate twice this distance from 0 to h, and multiply by L.  The formula for the volume is:

2L \int_0^h \sqrt{R^2-(R-y)^2} dy

Substitute R-y = R \sin \theta, dy = -\cos \theta d\theta, so that \theta = \sin^{-1} \frac{R-h}{R}

2L \int_0^{\sin^{-1} \frac{R-h}{R}} \sqrt{R^2-(R \sin \theta)^2} (-R\cos \theta) d\theta = -2LR^2 \int_0^{\sin^{-1} \frac{R-h}{R}} \cos^2 \theta d\theta
= -2LR^2 \int_0^{\sin^{-1} \frac{R-h}{R}} \frac{1+\cos 2\theta}{2} \theta d\theta
= -2LR^2 \left[\frac{\theta}{2} + \frac{\sin 2\theta}{2}\right]_0^{\sin^{-1} \frac{R-h}{R}}
[math] = -2LR^2 \sin^{-1} \frac{R-h}{R} +

Name: Anonymous 2009-04-02 17:28

1) Refer to the following picture: http://www.imagedump.com/index.cgi?pick=get&tp=547398

The volume is the length of the tank times the area of the yellow shaded area of the circular cross section.  To find the area, we're going to integrate the length of a horizontal slice running vertically from the bottom of the tank to the top of the water.  The length marked as "dist" is \sqrt{R^2-(R-y)^2}, so we have to integrate twice this distance from 0 to h, and multiply by L.  The formula for the volume is:

2L \int_0^h \sqrt{R^2-(R-y)^2} dy

Substitute R-y = R \sin \theta, dy = -\cos \theta d\theta, so that \theta = \sin^{-1} \frac{R-h}{R}

2L \int_0^{\sin^{-1} \frac{R-h}{R}} \sqrt{R^2-(R \sin \theta)^2} (-R\cos \theta) d\theta = -2LR^2 \int_0^{\sin^{-1} \frac{R-h}{R}} \cos^2 \theta d\theta
= -2LR^2 \int_0^{\sin^{-1} \frac{R-h}{R}} \frac{1+\cos 2\theta}{2} \theta d\theta
= -2LR^2 \left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_0^{\sin^{-1} \frac{R-h}{R}}
= -LR^2 \sin^{-1} \frac{R-h}{R} - \frac{LR^2}\sin \sin^{-1} \frac{R-h}{R} \cos \sin^{-1} \frac{R-h}{R}

Name: Anonymous 2009-04-02 17:30

1) Refer to the following picture: http://www.imagedump.com/index.cgi?pick=get&tp=547398

The volume is the length of the tank times the area of the yellow shaded area of the circular cross section.  To find the area, we're going to integrate the length of a horizontal slice running vertically from the bottom of the tank to the top of the water.  The length marked as "dist" is \sqrt{R^2-(R-y)^2}, so we have to integrate twice this distance from 0 to h, and multiply by L.  The formula for the volume is:

2L \int_0^h \sqrt{R^2-(R-y)^2} dy

Substitute R-y = R \sin \theta, dy = -\cos \theta d\theta, so that \theta = \sin^{-1} \frac{R-h}{R}

2L \int_0^{\sin^{-1} \frac{R-h}{R}} \sqrt{R^2-(R \sin \theta)^2} (-R\cos \theta) d\theta = -2LR^2 \int_0^{\sin^{-1} \frac{R-h}{R}} \cos^2 \theta d\theta
= -2LR^2 \int_0^{\sin^{-1} \frac{R-h}{R}} \frac{1+\cos 2\theta}{2} \theta d\theta
= -2LR^2 \left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_0^{\sin^{-1} \frac{R-h}{R}}
= -LR^2 \sin^{-1} \frac{R-h}{R} - LR^2 \sin \sin^{-1} \frac{R-h}{R} \cos \sin^{-1} \frac{R-h}{R}
= -LR^2 \sin^{-1} \frac{R-h}{R} - LR^2 \frac{R-h}{R} \sqrt{1-\left(\frac{R-h}{R}\right)^2}

Name: Anonymous 2009-04-02 17:31

1) Refer to the following picture: http://www.imagedump.com/index.cgi?pick=get&tp=547398

The volume is the length of the tank times the area of the yellow shaded area of the circular cross section.  To find the area, we're going to integrate the length of a horizontal slice running vertically from the bottom of the tank to the top of the water.  The length marked as "dist" is \sqrt{R^2-(R-y)^2}, so we have to integrate twice this distance from 0 to h, and multiply by L.  The formula for the volume is:

2L \int_0^h \sqrt{R^2-(R-y)^2} dy

Substitute R-y = R \sin \theta, dy = -\cos \theta d\theta, so that \theta = \sin^{-1} \frac{R-h}{R}

2L \int_0^{\sin^{-1} \frac{R-h}{R}} \sqrt{R^2-(R \sin \theta)^2} (-R\cos \theta) d\theta = -2LR^2 \int_0^{\sin^{-1} \frac{R-h}{R}} \cos^2 \theta d\theta
= -2LR^2 \int_0^{\sin^{-1} \frac{R-h}{R}} \frac{1+\cos 2\theta}{2} \theta d\theta
= -2LR^2 \left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_0^{\sin^{-1} \frac{R-h}{R}}
= -LR^2 \sin^{-1} \frac{R-h}{R} - LR^2 \sin \sin^{-1} \frac{R-h}{R} \cos \sin^{-1} \frac{R-h}{R}
= -LR^2 \sin^{-1} \frac{R-h}{R} - LR^2 \frac{R-h}{R} \sqrt{1-\left(\frac{R-h}{R}\right)^2}
= -LR^2 \sin^{-1} \frac{R-h}{R} - L (R-h) \sqrt{R^2-(R-h)^2}

Name: Anonymous 2009-04-02 17:39

1) Refer to the following picture: http://www.imagedump.com/index.cgi?pick=get&tp=547398

The volume is the length of the tank times the area of the yellow shaded area of the circular cross section.  To find the area, we're going to integrate the length of a horizontal slice running vertically from the bottom of the tank to the top of the water.  The length marked as "dist" is \sqrt{R^2-(R-y)^2}, so we have to integrate twice this distance from 0 to h, and multiply by L.  The formula for the volume is:

2L \int_0^h \sqrt{R^2-(R-y)^2} dy

Substitute R-y = R \sin \theta, dy = -R\cos \theta d\theta, so that \theta = \sin^{-1} \frac{R-y}{R}

2L \int_{\pi/2}^{\sin^{-1} \frac{R-h}{R}} \sqrt{R^2-(R \sin \theta)^2} (-R\cos \theta) d\theta = -2LR^2 \int_{\pi/2}^{\sin^{-1} \frac{R-h}{R}} \cos^2 \theta d\theta
= -2LR^2 \int_{\pi/2}^{\sin^{-1} \frac{R-h}{R}} \frac{1+\cos 2\theta}{2} d\theta
= -2LR^2 \left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_{\pi/2}^{\sin^{-1} \frac{R-h}{R}}
= -LR^2 \sin^{-1} \frac{R-h}{R} - LR^2 \sin \sin^{-1} \frac{R-h}{R} \cos \sin^{-1} \frac{R-h}{R}
= -LR^2 \sin^{-1} \frac{R-h}{R} - LR^2 \frac{R-h}{R} \sqrt{1-\left(\frac{R-h}{R}\right)^2}
= -LR^2 \sin^{-1} \frac{R-h}{R} - L (R-h) \sqrt{R^2-(R-h)^2}

Name: Anonymous 2009-04-02 17:40

1) Refer to the following picture: http://www.imagedump.com/index.cgi?pick=get&tp=547398

The volume is the length of the tank times the area of the yellow shaded area of the circular cross section.  To find the area, we're going to integrate the length of a horizontal slice running vertically from the bottom of the tank to the top of the water.  The length marked as "dist" is \sqrt{R^2-(R-y)^2}, so we have to integrate twice this distance from 0 to h, and multiply by L.  The formula for the volume is:

2L \int_0^h \sqrt{R^2-(R-y)^2} dy

Substitute R-y = R \sin \theta, dy = -R\cos \theta d\theta, so that \theta = \sin^{-1} \frac{R-y}{R}

2L \int_{\pi/2}^{\sin^{-1} \frac{R-h}{R}} \sqrt{R^2-(R \sin \theta)^2} (-R\cos \theta) d\theta = -2LR^2 \int_{\pi/2}^{\sin^{-1} \frac{R-h}{R}} \cos^2 \theta d\theta
= -2LR^2 \int_{\pi/2}^{\sin^{-1} \frac{R-h}{R}} \frac{1+\cos 2\theta}{2} d\theta
= -2LR^2 \left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_{\pi/2}^{\sin^{-1} \frac{R-h}{R}}
= -LR^2 \sin^{-1} \frac{R-h}{R} - LR^2 \sin \sin^{-1} \frac{R-h}{R} \cos \sin^{-1} \frac{R-h}{R}
= -LR^2 \sin^{-1} \frac{R-h}{R} - LR^2 \frac{R-h}{R} \sqrt{1-\left(\frac{R-h}{R}\right)^2}
= LR^2\frac{\pi}{2}-LR^2 \sin^{-1} \frac{R-h}{R} - L (R-h) \sqrt{R^2-(R-h)^2}

Name: Anonymous 2009-04-02 17:42

Anon delivers:

1) Refer to the following picture: http://www.imagedump.com/index.cgi?pick=get&tp=547398

The volume is the length of the tank times the area of the yellow shaded area of the circular cross section.  To find the area, we're going to integrate the length of a horizontal slice running vertically from the bottom of the tank to the top of the water.  The length marked as "dist" is \sqrt{R^2-(R-y)^2}, so we have to integrate twice this distance from 0 to h, and multiply by L.  The formula for the volume is:

2L \int_0^h \sqrt{R^2-(R-y)^2} dy

Substitute R-y = R \sin \theta, dy = -R\cos \theta d\theta, so that \theta = \sin^{-1} \frac{R-y}{R}

2L \int_{\pi/2}^{\sin^{-1} \frac{R-h}{R}} \sqrt{R^2-(R \sin \theta)^2} (-R\cos \theta) d\theta = -2LR^2 \int_{\pi/2}^{\sin^{-1} \frac{R-h}{R}} \cos^2 \theta d\theta
= -2LR^2 \int_{\pi/2}^{\sin^{-1} \frac{R-h}{R}} \frac{1+\cos 2\theta}{2} d\theta
= -2LR^2 \left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_{\pi/2}^{\sin^{-1} \frac{R-h}{R}}
= -LR^2 \sin^{-1} \frac{R-h}{R} - LR^2 \sin \sin^{-1} \frac{R-h}{R} \cos \sin^{-1} \frac{R-h}{R}
= -LR^2 \sin^{-1} \frac{R-h}{R} - LR^2 \frac{R-h}{R} \sqrt{1-\left(\frac{R-h}{R}\right)^2}
= LR^2\frac{\pi}{2}-LR^2 \sin^{-1} \frac{R-h}{R} - L (R-h) \sqrt{R^2-(R-h)^2}

Don't change these.
Name: Email:
Entire Thread Thread List