discoverynews:

Sea Urchin-Inspired House Captures Tidal Energy

If you’ve ever dreamed of having a seafront home shaped like a sea urchin — who hasn’t? — then hold onto your swim fins.

The Hydroelectric Tidal House, envisioned by architectural designer Margot Krasojević, draws inspiration from some of nature’s weirdest sea creatures — echinoderms like starfish and sea urchins whose symmetrical shapes have long fascinated biologists. Learn more

discoverynews:

Sea Urchin-Inspired House Captures Tidal Energy

If you’ve ever dreamed of having a seafront home shaped like a sea urchin — who hasn’t? — then hold onto your swim fins.

The Hydroelectric Tidal House, envisioned by architectural designer Margot Krasojević, draws inspiration from some of nature’s weirdest sea creatures — echinoderms like starfish and sea urchins whose symmetrical shapes have long fascinated biologists. Learn more

This took me a while to wrap my head around. Basically, the shapes are generated every time the faster one “passes” the slower one. In the first example, we had 13 and 8, so in one full cycle the faster one passes the slower one 5 times. Each time, it generates a cardioid. Cardioid are epicycloids, as seen here.

Since each pass will occur in an identical amount of time, we end up with rotational symmetry. Forgive my poor drawing skills: 

It’s easy to miss the fact that they are cardioids because we are instead just seeing emergent patterns from all the overlap. More cardioids = more pedals and more layers of pedals. The outermost pedals are easy: 5 cardioids = five overlaps = five outermost pedals. Each cardioid overlaps with its nearest neighbor. As we move inward, we see overlap of non-nearest neighbors (ie, the first is overlapping with the third). Note: because of the cyclic nature here, “third” in one direction can also be considered “fourth” in the other direction, so two colors will have two different overlaps. 

To summarize: 

Inner pedals = Red and Green

Next Layer = Red and Yellow

Outer Layer = Red and Blue

Hope that helps!

This took me a while to wrap my head around. Basically, the shapes are generated every time the faster one “passes” the slower one. In the first example, we had 13 and 8, so in one full cycle the faster one passes the slower one 5 times. Each time, it generates a cardioid. Cardioid are epicycloids, as seen here.

Since each pass will occur in an identical amount of time, we end up with rotational symmetry. Forgive my poor drawing skills: 

It’s easy to miss the fact that they are cardioids because we are instead just seeing emergent patterns from all the overlap. More cardioids = more pedals and more layers of pedals. The outermost pedals are easy: 5 cardioids = five overlaps = five outermost pedals. Each cardioid overlaps with its nearest neighbor. As we move inward, we see overlap of non-nearest neighbors (ie, the first is overlapping with the third). Note: because of the cyclic nature here, “third” in one direction can also be considered “fourth” in the other direction, so two colors will have two different overlaps. 

To summarize: 

Inner pedals = Red and Green

Next Layer = Red and Yellow

Outer Layer = Red and Blue

Hope that helps!

dinosauradvocate:

proofmathisbeautiful posted this today: http://ift.tt/1rM7rXQ

I felt the need to do the math out. m is the weight of the corner weights, M is the table’s weight, and L is the side length of the table. h is how far the table is from the ceiling, and theta is the shown angle. Theta seemed useless to me, so my final answer included the other variables. Now, if you want to do this with a specific table and a specific height, you can solve for the masses to put on the corners… I’m tempted to!

dinosauradvocate:

proofmathisbeautiful posted this today: http://ift.tt/1rM7rXQ

I felt the need to do the math out. m is the weight of the corner weights, M is the table’s weight, and L is the side length of the table. h is how far the table is from the ceiling, and theta is the shown angle. Theta seemed useless to me, so my final answer included the other variables. Now, if you want to do this with a specific table and a specific height, you can solve for the masses to put on the corners… I’m tempted to!

curiosamathematica:

It is possible to define a consistent addition of points on certain kinds of curves (elliptic curves). This arithmetic plays an important role in modern mathematics. For instance, Wiles’ proof of Fermat’s last theorem is a consequence of the modularity theorem (once known as the Taniyama-Shimura-Weil conjecture), which gives a strong connection between elliptic curves and modular forms. Elliptic curves over finite fields also have cryptographic applications, or can be used for integer factorization.

curiosamathematica:

It is possible to define a consistent addition of points on certain kinds of curves (elliptic curves). This arithmetic plays an important role in modern mathematics. For instance, Wiles’ proof of Fermat’s last theorem is a consequence of the modularity theorem (once known as the Taniyama-Shimura-Weil conjecture), which gives a strong connection between elliptic curves and modular forms. Elliptic curves over finite fields also have cryptographic applications, or can be used for integer factorization.

I’m happy being myself, which I’ve never been before. I always hid in other people, or tried to find myself through the characters, or live out their lives, but I didn’t have those things in mine.

I’m happy being myself, which I’ve never been before. I always hid in other people, or tried to find myself through the characters, or live out their lives, but I didn’t have those things in mine.

beahbeah:

this website SAVED MY BRAIN when i was a stressed out college student who couldn’t stop flipping out long enough to prioritize. quite a few of you are still suffering through college so i hope this helps you too!! c:

beahbeah:

this website SAVED MY BRAIN when i was a stressed out college student who couldn’t stop flipping out long enough to prioritize. quite a few of you are still suffering through college so i hope this helps you too!! c:

Hello! So the first equation you’d be looking at would be the first derivative of y=ax^2+bx+5, since the first derivative of something is the same as its tangent/slope. You already know the answer to this equation is 4, and you know that this is 4 when x = 5. So: 

The second equation (the one below the red text) happens when x = 5, and y = 0. This is a point of the function, because it’s where it has a tangent (therefore (5,0) has to be part of the function). So now you have a in terms of b from your first equation, which you then substitute into your new equation (as seen below the red equation). The last part would be to solve the algebra:

You can prove that your solutions for a and b are correct by finding the value of the tangent at the point (5,0). Since it indeed results in 4, then your solutions are correct.

Hello! So the first equation you’d be looking at would be the first derivative of y=ax^2+bx+5, since the first derivative of something is the same as its tangent/slope. You already know the answer to this equation is 4, and you know that this is 4 when x = 5. So: 

The second equation (the one below the red text) happens when x = 5, and y = 0. This is a point of the function, because it’s where it has a tangent (therefore (5,0) has to be part of the function). So now you have a in terms of b from your first equation, which you then substitute into your new equation (as seen below the red equation). The last part would be to solve the algebra:

You can prove that your solutions for a and b are correct by finding the value of the tangent at the point (5,0). Since it indeed results in 4, then your solutions are correct.

Photo

Photo

lightprocesses:

Hyperbolic paraboloid

lightprocesses:

Hyperbolic paraboloid

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Photo