you should be able to calculate the volume of the unit tetrahedron without help, if you're a real mathemati/sci/an
Oooh, I remember this problem from my Applied Triple Integrals class.
>>7716821
The area of an equilateral triangle of side length x is (x^2)*sqrt(3)/4, so, integrating along the altitude gives you sqrt(3)/12.
>>7716856
SHAME
>>7716856
We are very sorry, but due to recent occurrences of gross error, we must request that you immediately relinquish your membership in the /sci/ community.
>>7716821
You can use put the points into a 4x4 matrix, solve for the determinate, then multiply it by 1/6.
>>7716821
I actually decided to do this just now, for shits and giggles.
Got it wrong the first time around, and wasted a good chunk of time trying to figure out where my reasoning was fucked, only to find that I accidentally wrote the square root of 9 as 9 somewhere along the way.
Man that pisses me off.
>>7716936
Can a tetrahedron be though of as one half of a parallelepiped? If so you could plug the vectors into a 3x3 matrix, compute the determinant, and then divide it by two to get the volume.
>>7717222
>Can a tetrahedron be though of as one half of a parallelepiped?
How many edges do you think a parallelepiped has?
>>7716856
this
except do it correctly
>>7716821
newfag here, tryna learn the ways.
I can figure out the area of a triangle. a triangular prism seems to me to be the next step up in a dimension, to find its volume you simply multiply the area of the triangle by its height.
for a tetrahedron, i'm lost as to where is logically goes. could someone break this down for me? I can look up the formulas but I don't know how to go from a triangle to a tetrahedron just by their formulas.
>>7716821
1/3 the base area times the altitude
Easy as fuck desu senpai
>>7717249
It's much easier to describe in abstract terms than in their linear relationship to each other desu
Basically if you want to integrate cleanly between space-relative volumes you integrate the average between the average of the side length and the altitude
I proved that in high school but fuck proving it again
[math]((1+3^{1/2})/2)^{3}/6[/math]
I believe but I'm doing this in my head
>>7717277
Oops I made a small error
[math]((1+3^{1/2}/2)/2)^{3}/6[/math]
>>7717213
>for shits and giggles
redneck detected
>>7717355
>2015
>Possessing a neck of tone comparable to the rest of the body
Get a load of this plebeian.
>>7717358
neck of tone detected
>>7717284
not a "small" error, m8
>>7716821
With eyes closed.
(1/3)(1/2)sqrt(3/4)sqrt(2/3)=(1/12)sqrt(2)
It's part of the "Geometry in the Dark" series of exercises to improve focus and visualisation.
>>7716856
>>7717284
>>7717428
We are very sorry, but the magnitude of recent errors has made it necessary to revoke your membership in the /sci/entific community.
>>7717440
But isn't sqrt(2)/12 correct? Or are you commenting on his solution (I have no idea how he got the left hand side)?
>>7717462
>isn't sqrt(2)/12 correct?
definitely not, sorry
>>7717468
http://mathworld.wolfram.com/RegularTetrahedron.html
?
>>7717476
equation 3 appears to be a horrible mistake
sqrt(3)/8
>>7717462
Of course it is correct. Calculation is standard procedure: (1/3) of the product of base area (1/2)sqrt(3/4) and height sqrt(2/3).
>>7718465
Shit, I didn't even realize there was a damned formula for this, and actually integrated!
This is what I got
>>7718589
Sorry was upside down and silly degrees
1/sqrt(6) desu senpai
>>7718465
>height sqrt(2/3)
this is the part that's wrong
>>7718591
Fucked up some bounds, but I rewrote and solved on my blackboard. Is this the concensus?
>>7719354
wiki confirms
just successfully integrated it
can I /sci/ now?
but can you derive the volume of the regular tetrahedron through manipulation of the stella octangula
you are allowed to use the formula the for volume of square pyramids
it's actually quite hard
>actually integrating it
I remember from my grade school geometry class that the formula for the volume defined by 2-d shape and a point is base*height/3. You get that formula from integrating since x^2 -> x^3/3 (because area increases with the square of the dilation). The area of the base is sqrt(3)/4 because 60 degree angles. The tricky bit is the height since the angle isn't anything nice. I think there's no easy way but to go all analytic geometry on it sadly. Angle bisectors won't help because it's not that nice.
(x1-x)^2 +(y1-y)^2 + h^2 = 1
so
(1-1/2)^2 + y^2 + h^2= 1
(1/2-1/2)^2 +(sqrt(3)/2-y)^2 + h^2= 1
Then subtract one equation from the other to cancel the y^2s, solve the resulting mess for y, then substitute back to get h.
Or you could be like archimedes and drop it in a bucket of water.
>>7718958
using the altitude of one face=sqrt(3/4) as hypotenuse:
h^2=(sqrt(3/4))^2-((1/3)sqrt(3/4))^2=(3/4)-(1/9)(3/4)=(8/9)(3/4)=(2/3)
alternatively, using edge=1 as hypotenuse:
h^2=(1)^2-((2/3)sqrt(3/4))^2=1-(4/9)(3/4)=1-(1/3)=(2/3)
in both cases: h=sqrt(2/3)
>>7718958
It's correct, I'll make a diagram for how to obtain it, and post it later.
Biochem major here, may not have a clue what I'm doing but I'm gonna take a stab at it. I've only taken calc I so bare with me...
Could you designate the back most vertice as point (0,0,0), solve the vertice at the tip using the side lengths, develop a function that describes the position of the first vertice as that point moves along the edge of the tetrahedron to the top vertice, and use an integral to calculate the surface area of an infinite number of triangles until you have the area?
Think of it like rotating the base of the tetrahedron 90 degrees to the right.
Just use the pyramid formula, Jesus Christ.
[math]V=\frac{1}{3}\cdot A_b \cdot h[/math]
[math]A_b=\frac{\sqrt{3}}{4}[/math]by simple geometry
[math]h=\frac{\sqrt{6}}{3}[/math] by pic related
[math]V=\frac{1}{3}\cdot \frac{\sqrt{3}}{4} \cdot \frac{\sqrt{6}}{3}=\frac{\sqrt{2}}{12}[/math] as someone said earlier.
altitude of one face
[math]\displaystyle a=\sqrt{1^2-\left ( \frac{1}{2} \right )^2}=\frac{\sqrt{3}}{2}[/math]
area of one face
[math][math]\displaystyle A_f= \frac{1}{2} \times 1 \times \frac{ \sqrt{3}}{2}= \frac{ \sqrt{3}}{4}[/math][/math]
>>7716821
I'm an engineer, though.
mathJax is being weird today
[math]\displaystyle A= \frac{1}{6}A_f= \frac{\sqrt{3}}{24} = \frac{1}{2} \times \frac{1}{2} \times d= \frac{1}{4}d[/math]
so [math]\displaystyle d= \frac{\sqrt{3}}{6}[/math]
height of the tetrahedron
[math]\displaystyle h= \sqrt{1^2- \left ( \frac{\sqrt{3}}{3} \right )^2}= \sqrt{\frac{2}{3}}[math]
mathJax is being very tedious today
height of the tetrahedron
[math]\displaystyle h= \sqrt{1^2- \left ( \frac{ \sqrt{3}}{3} \right )^2}= \sqrt{ \frac{2}{3}}[/math]
volume is just
[math]\displaystyle V= \frac{1}{3}hA_f = \frac{1}{3} \sqrt{ \frac{2}{3}} \times \frac{ \sqrt{3}}{4}= \frac{ \sqrt{2}}{12}[/math]
>>7716821
Simply cross section integral.
>>7720390
>Simply cross section integral.
That's what I figured.
Off the top of my head:
Given a side length of x, the cross-section triangle has an area of:
A = x * h / 2
A = x * (x * sqrt(3)/2) / 2
A = x^2 * sqrt(3) / 4
Now that we know the area A, we just integrate it over the height of the tetrahedron, which I'll temporarily assume is 1:
INT (from 0 to 1) (x^2 * sqrt(3) / 4) dx
But we know that the height of the tetrahedron is actually taller than 1, so we have to scale it up by whatever the height is, which I'll call "k" -- so the actual volume is:
V = INT (from 0 to 1) k * (x^2 * sqrt(3) / 4) dx
The only remaining thing to figure out is k, which is the height of the tetrahedron. This can be determined from the pythagorean of the internal vertical triangle, which I'll call ABC:
C = hypotenuse = slant height = sqrt(3) / 2
B = bottom leg = half of slant height = sqrt(3) / 4
A = height leg = sqrt(C^2 - B^2)
So k = A, whatever A turns out to be. Pop k into the above integral, and you've got the volume.
CBA to do all the math myself.
Let me know if this is wrong, please.
I found a video that shows a general solution to the answer using a simple cross section integral. I think if you just plug in the values for a and h you can obtain the volume of the unit tetrahedron.
>>7720816
>khan academy
>gives wrong answer
get the hell out
>>7720851
That's not khan academy you faggot.
>>7716821
High-school maths teacher here. This is a question I use at the start of the year for my seniors. Except I expect A-level students to do it mentally (no pen and paper).
The first three steps are "assumed knowledge" from middle school:
- Volume of a pyramid is one third the area of the base multiplied by height.
- Area of a triangle is half the altitude by base.
- 30/60 right triangles have sides in the ratio: 1, 2, √3.
Thus fairly quickly students should get:
V = ⅓ × (½ × ½ × √3) × H
V = (√3/12)H
By considering the altitudes that identify the orthocentre on the base face, the triangle is divided into six similar (30/60) right triangles.
Given the area of each is one third of the larger 30/60 triangle, each side length is reduced by a factor of 1/√3. Thus the hypotenuse is 1/√3.
This length, with the height and edge of the pyramid forms a right triangle.
Using Pythagoras, the height is obviously √(2/3).
Thus the volume is:
V = (√3/12)H
V = √3/12 × √2/√3
V = √2/12
Top students can do this in under 5 min, but most take between 5-10 min, at which point I put the others out of their misery.
Take the mass and divide by density. Or you could do a displacement measuring in a grad. Cylinder to find volume. There, proved my /sci/lls
https://www.google.com.mx/search?q=volume+tetrahedron&oq=volume+tetrahedron&aqs=chrome..69i57.4405j0j7&sourceid=chrome&es_sm=122&ie=UTF-8
>>7721179
What part of "without help" did you not understand, Anon?
>>7716856
I'm done with /sci/.
>>7720937
>no pen and paper
Very good. Here's another example of the exercises mentioned at >>7717428:
Close your eyes and visualise a regular octahedron inscribed in a sphere. Take your time to see it clearly. Applying what you have explored in the previous exercises, find the ratio of the volumes of the sphere and the octahedron (V_sphere/V_octahedron).
>>7716936
Would like to know if this works. Take 4 points, choose one as a base point, subtract to get 3 vectors, put in 3x3 matrix, get determinant, divide by 6... profit.
I seem to recall it does, but I'm not sure where the 1/6 comes from. Must be 1/6th of the parallelpiped defined by the 3 vectors, but I can't really visualize that you can divide the parallelpiped up into 6 tetrahedra that way. Nothing left in the center?
>>7721977
6=3!
>>7722299
But how is this relevant?
>>7721926
PIE
>>7722785
Anon delivers.
>>7717428
>"Geometry in the Dark" series...
would you mind kind anon tell me where is this from?
>>7721926
Nice. I'll need to get a copy of that series of problems.
I think this is somewhat easier in terms of mental calculation, because the 45 triangle values (1/√2 = √2/2) are easier to manipulate. However, I suspect many of my students will be less familiar with the octahedron.
I expect a similar proportion of success - but might try this one instead next term as a trial.
...
For the record, I'd let the radius be 1. This gives an edge length of √2.
Volume of the octohedron is double that of the corresponding square pyramid.
V = 2(⅓BH)
V = ⅔(2)(1)
V = 4/3
The volume of the unit sphere:
V = 4/3π
The ratio is π.
>>7722336
>how is this relevant?
becoz dubs-trips-dubs
>>7723338
>where is this from?
>>7723355
>need to get a copy
I'm afraid I will not be able to help. It's an unpublished manuscript circulated as teaching notes and it's not even in English. What I can offer though is an outline of the other exercises in this section. The one I just posted is an ad hoc translation of one of them, slightly appended for a certain gentleman (^.^)
r = sqrt(1/2)a -> a = sqrt(2)r -> a^3=2sqrt(2)r^3 [store]
V = (2/3)sqrt(1/2)a^3 = (2/3)sqrt(1/2) * 2sqrt(2)r^3 = (4/3)r^3
V/r^3 = (4/3) and thus V_sphere/V_octahedron = π
Now imagine a cube, an octahedron, a tetrahedron and a sphere. You can inscribe each into the sphere or let the sphere circumscribe it. The goal is to find the six values of the expression V/r^3. Remember, with eyes closed. Time is not an issue, you have all the time you need.
As already mentioned, the simple geometry is just a vehicle. The objective is to enhance focus, memory and visualisation, the ability to mentally manipulate equations and finally 'spool out' the result (what I did in the first post).
I'm currently working on the 'Intuitive Physics' section. Find the minimum h and µ with eyes closed..
Holy moly people are really going all in at solving this shit? I thought this board had higher standards.
>>7720937
If I were your student and solved it mentally in less than 30sec, would I get an insta-A? 5 minutes for TOP students... Sigh
(pi*sqrt(3))/48
octahedron with inscribed sphere is tricky
took me a while to see the similar triangles
>>7716821
sqrt(33) , divided by 48 .
[cube units cm^3 etc etc]
lol the volume of the cube is 1x1x1 - 1x1x1(30degrees)
fucking shape math what is this hiscghool lol
>>7727754
same anon here
my bad it's actually [sqrt(72)]/72 cubic units
>>7727792
That's a funny notation but the result is correct:
√(72) = √(2*4*9) = 6√(2) and 6√(2)/72 = √(2)/12
>>7716821
Volume of a unit cube is 1 u^3
rectangular prism is half that so 1/2 u^3
tetrahedron is half a rectangular prism so 1/4 u^3
>>7724686
2r for both if there's no resistance, and the stopping deceleration is 1g. Plus an infinitesimally small bit more, otherwise the object would "hang" forever at the top of the loop.
>>7720937
you know what? fuck Maths. I am glad I am doing my masters in Marketing.
>>7716852
>mfw this actually made me laugh
i've been here too long, guys
>>7729498
Interesting. Can you expand on it a bit further? I see it dropping down from the top of the loop if h is 2r. The two equal arrows are meant as a hint. And yes, assumption is standard g and friction ignored except in the red µ zone. Radius r and length s are constants, min h and µ should be found.
>>7729603
either that, or you need to reduce your sqrt(3) by a half again...
Q: Why is it that L and tan(a) have the same numerical value?
A: That is because (n)/√(n) = √(n)
Q: Does that count as a geometric reason?
A: No.
Oh
I know this
Umm
>1
times
>(3/4)^.5
times
>1/3 because conic solid
>Fug.... I'll get it
>Umm
>Where's a piece of paper
>Okay... (8/12)^.5... I think
So that's .08333... repeating, of course.
I probably fucked up somewhere.
>>7730201
geometry is about relations
integers, fractions and surds
not absurds like 0.1178511302
>>7729589
If h is exactly 2r, as the object approaches the top of the loop, its velocity approaches 0. It never attains the top, while its motion toward the top is never completely canceled by its rise up the ever gentler slope toward the top. That's only if there is no resistance at all. If you built the apparatus in the real world, it would hugely amplify differences in initial conditions around 2r plus whatever extra is needed to overcome resistance enough to get the object over the top in half of its drops.
>>7729589
One more thing: I am assuming the object's trajectory to be perfectly confined to the course, that it cannot fall away from it in any direction, that it is not like a marble on a ramp but more like an amusement park ride on monorail.
u can do it by integration
actually doing it is arithmetic, which an engineer or physics major would probably be better at
>>7730201
> irrational number
> repeating decimal fraction
>>7716936
I know what the determinant of a matrix is. I've never heard of the determinate of a matrix.
>>7730371
Thank you. I completed the 'eyes closed' exercise and again it was a real adventure. The mass/energy always vanished from the equations and I could reduce the problem to accelerations and potentials. To ensure that the sliding object stays in contact with the track I balanced the accelerations at the top of the loop and then added the resulting kinetic potential to the gravitational potential which then yields the required height above the loop top. I think the result is correct, calculation is in the picture.
The marble is a good idea. For the next exercise I will replace the sliding thing with a rolling ball to see how that changes the necessary height. After the kinetic potential I will probably invent the rotational potential..or something like that.
>>7730084
It resembles the popular 3 - 4 - 5 triangle, only in this case it's √(1)/2 - √(2)/2 - √(3)/2 and 1/4 + 2/4 = 3/4. It boils down to tan(alpha)=a/b=b/1 or b^2=a*1 (where the 1 is the unit of length to keep things dimensionally consistent) and (2/4)^2 is indeed 1/4. Is that a geometric explanation? Probably not, closer to geometric numerology, or something like 2*2=2+2.
>>7730441
The marble needs h=2.7r to loop the loop.
