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Charles Loder
Charles Loder

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CSS: 3 Sided Card

The other day I was tasked with creating a "3 sided card" that flips. It proved more challenging than I thought, though the solution isn't that complex.

What we're building

Calling it a "3 sided card" is misleading, but when I first started trying to build this, that's all I could think to call it.

In reality, it is pentahedron — in particular, a triangular prism. Imagine a cube where you remove one side, and then fold the adjoining two together.

The final product can be found at the end.

The gist

The structure consists of three divs making up the sides, wrapped in a div for the prism.



<div class="prism">
  <div class="side one">1</div>
  <div class="side two">2</div>
  <div class="side three">3</div>
</div>


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We don't need to worry about the top and bottom.

The css is little more complicated. First, some non-critical styling to make it look nice.



.prism {
  margin: auto;
  width: 100px;
  aspect-ratio: 1 / 1;
}

.side {
  border: 1px solid black;
  display: flex;
  flex-direction: column;
  align-items: center;
  justify-content: center;
}

.one {
  background-color: #63cdda;
}

.two {
  background-color: #ea8685;
}

.three {
  background-color: #f5cd79;
}


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The good stuff

Making this relies heavily on the rotate3d function, which I'll talk about in the next section.

For now, let's define the css for the prism and it's sides.

css



.prism {
  transition: transform 0s linear;
  transform-style: preserve-3d;
}


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On the prism div, we define a transition for when the transform property changes. This will define how fast the card changes.

It's currently set at 0s for accessibility.

Below it we can define a media query for if the user has no preference for reduced motion.



@media (prefers-reduced-motion: no-preference) {
  .prism {
    transition: transform 0.5s linear;
  }
}


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This will give the card a spin.

We will apply a transform in the next section.

Now the sides.



.side {
  position: absolute;
  height: 100%;
  width: 100%;
}


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This ensures each side takes up the whole face of the prism.

Lastly, we add a transform property to each side.



.side.one {
  transform: rotateY(0deg) translateZ(29px);
}

.side.two {
  transform: rotateY(120deg) translateZ(29px);
}

.side.three {
  transform: rotateY(240deg) translateZ(29px);
}


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Each side is rotated on it's Y-axis 120deg (i.e. 360/3).

The translateZ property was a bit of a guessing game. It's related to the width of the prism, but I had to figure out the right value by trial-and-error.

Html and JavaScript

Now, I haven't talk about rotate3d property yet. Before we add it, we're going to add a data attribute to the prism.



<div class="prism" data-degrees="0">
  <div class="side one">1</div>
  <div class="side two">2</div>
  <div class="side three">3</div>
</div>


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Now we can manipulate the data attribute with JS and apply a rotate3d value based off of the data attribute.



function toggleDegrees(e) {
  const currentDegrees = +e.currentTarget.dataset.degrees || 0;
  const newDegrees = currentDegrees + 120;
  e.currentTarget.style.transform = `rotate3d(0, -1, 0, ${newDegrees}deg)`;
  e.currentTarget.dataset.degrees = newDegrees;
}

document.querySelector(".prism").addEventListener("click", toggleDegrees);


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By constantly increasing the degree, instead of setting it back to 0, the card will keep spinning in one direction.

Live

Conclusion

I hope someone else finds this useful. I wanted to get thoughts typed out before I forget how to do this.

Any suggestions are welcomed!

Top comments (8)

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anonymous123 profile image
Anonymous • Edited

thanks for the tutorial! correct me if i am wrong: the formula for 'translatez(x)' could be the same formula as for the radius of the incircle of an equilateral triangle [ r = a / (2 * sqrt(3) ]. In your example that would be: 100 / (2 * sqrt(3)) = 28.868 ~ 29.

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charlesloder profile image
Charles Loder

Honestly, I have no idea! But that math seems write to me

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teseu profile image
Tulio

It's really a fine idea. Congrats!

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charlesloder profile image
Charles Loder

Thanks! I appreciate it

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tracygjg profile image
Tracy Gilmore

Hi Charles,
I kindly suggest you might want to show off your example in codepen.
Regards, Tracy

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charlesloder profile image
Charles Loder

Thanks! Is the fiddle not visible?

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tracygjg profile image
Tracy Gilmore

It is and it looks really good but it would not be out of place amongst other fine demonstrations at codepen.

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charlesloder profile image
Charles Loder

It's been updated — thanks for the suggestion