# Another Geometry Question **URL:** \ **Category:** Factual Questions\ **Created:** [April 25, 2012, 2:27pm UTC](https://boards.straightdope.com/t/another-geometry-question/619906 "2012-04-25T14:27:10Z")\ **Posts on this page:** 10\ **Page:** 1
**Author:** ![dolphinboy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/dolphinboy/32/330_2.png) [@dolphinboy](https://boards.straightdope.com/u/dolphinboy)\ **Post date:** [April 25, 2012, 2:27pm UTC](https://boards.straightdope.com/t/another-geometry-question/619906/1 "2012-04-25T14:27:10Z")
No, this is not homework… I graduated from college 30 years ago and I struggled with math back then. Here’s the question: I am building a PVC pipe structure over my raised garden beds that will be used to hold a clear plastic tarp over the young plants. (Even though it’s spring here in the Northern Rockies, we still get the occasional hailstorm and sub-freezing temperatures.) My beds are 8 feet by 11 feet in size, so what I want to do is use 1/2" Schedule 40 PVC tubing that is glued together and bent into an arch over the beds. I will then place the plastic over the PVC arches and life will be good. 🙂 You can scroll down to the middle of this [webpage](http://blog.foodierachel.com/2012/03/garden-diy-a-hot-house-for-late-winter-seeds/)to get a rough idea of what I am planning to build. The PVC piping needs to span an 8 foot width, and by adjusting the length of the PVC piping that will change the height of the arch over the bed. I can do this by trial and error, but let’s say I want the height of the arch to be exactly 4 feet over the raised bed. What is the formula to determine how long the PVC pipe has to be assuming the distance between the two ends is 8 feet? And is there a calculator somewhere that I can just plug in the numbers to get a good approximation? I can’t seem to find one anywhere. :dubious: ---
**Author:** ![SmellMyWort](https://avatars.discourse-cdn.com/v4/letter/s/ea5d25/32.png) [@SmellMyWort](https://boards.straightdope.com/u/SmellMyWort)\ **Post date:** [April 25, 2012, 2:41pm UTC](https://boards.straightdope.com/t/another-geometry-question/619906/2 "2012-04-25T14:41:33Z")
If you want the arch to resemble a half circle, just calculate the circumfrence with diameter 8 and then divide it in half. Or in the website you linked to the blogger’s garden is 4 feet wide and she used 10-foot pvc, so you could just double that. ---
**Author:** ![blasto9000](https://avatars.discourse-cdn.com/v4/letter/b/9fc348/32.png) [@blasto9000](https://boards.straightdope.com/u/blasto9000)\ **Post date:** [April 25, 2012, 3:07pm UTC](https://boards.straightdope.com/t/another-geometry-question/619906/3 "2012-04-25T15:07:40Z")
The formula for circumference is 2 \* pi \* radius. You want a 4 foot radius, so the circumference of a complete circle would be 2 \* 3.14 \* 4 = 25.12 feet. Multiply by 0.5 since you only want a half circle so the length of pipe needed would be 12.56 feet, or roughly 12 feet 7 inches. ---
**Author:** ![Buck\_Godot](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/buck_godot/32/6573_2.png) [@Buck\_Godot](https://boards.straightdope.com/u/Buck_Godot)\ **Post date:** [April 25, 2012, 3:08pm UTC](https://boards.straightdope.com/t/another-geometry-question/619906/4 "2012-04-25T15:08:53Z")
I did some quick google searching and came up with a lot of links to complicated models, suggesting that this is more of an engineering problem than a strict geometry problem. You’re probably best off just doing it by trial and error. ETA: SmellMyWort, Is there any reason we should expect the resulting curve to resemble a circular arc rather than say a parabola or more complicated curve? ---
**Author:** ![CookingWithGas](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/cookingwithgas/32/485_2.png) [@CookingWithGas](https://boards.straightdope.com/u/CookingWithGas)\ **Post date:** [April 25, 2012, 3:49pm UTC](https://boards.straightdope.com/t/another-geometry-question/619906/5 "2012-04-25T15:49:34Z")
I am thinking that a parabola is a better approximation, although it might also be a [catenary arch](http://en.wikipedia.org/wiki/Catenary). But I think you would have to integrate over the curve to determine its length; I’m also 30 years rusty on this kind of thing. ---
**Author:** ![johnpost](https://avatars.discourse-cdn.com/v4/letter/j/f17d59/32.png) [@johnpost](https://boards.straightdope.com/u/johnpost)\ **Post date:** [April 25, 2012, 3:54pm UTC](https://boards.straightdope.com/t/another-geometry-question/619906/6 "2012-04-25T15:54:18Z")
PVC arc will sag. if actual 4 feet height is important then trial and error might be used. on a raised platform (porch, patio, tables spread 8 feet apart) anchor one end of the longer than needed tube, place the other tube end below 8 feet mark, raise until middle height is 4 feet. there may be some engineering data somewhere but this is easy. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [April 25, 2012, 5:13pm UTC](https://boards.straightdope.com/t/another-geometry-question/619906/7 "2012-04-25T17:13:06Z")
It’s neither a parabola nor a catenary. Consider the extreme case where you bring the endpoints entirely together, forcing the pipe into a teardrop-like shape: Neither a parabola nor a catenary can include that extreme. ---
**Author:** ![SmellMyWort](https://avatars.discourse-cdn.com/v4/letter/s/ea5d25/32.png) [@SmellMyWort](https://boards.straightdope.com/u/SmellMyWort)\ **Post date:** [April 25, 2012, 5:24pm UTC](https://boards.straightdope.com/t/another-geometry-question/619906/8 "2012-04-25T17:24:45Z")
> [@Buck\_Godot](#): > > I did some quick google searching and came up with a lot of links to complicated models, suggesting that this is more of an engineering problem than a strict geometry problem. You’re probably best off just doing it by trial and error. > > ETA: SmellMyWort, Is there any reason we should expect the resulting curve to resemble a circular arc rather than say a parabola or more complicated curve? No, just an easy starting point. ---
**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\ **Post date:** [April 25, 2012, 6:00pm UTC](https://boards.straightdope.com/t/another-geometry-question/619906/9 "2012-04-25T18:00:02Z")
> [@dolphinboy](#): > > … let’s say I want the height of the arch to be exactly 4 feet over the raised bed. If you really need an exact height the problem is difficult - it requires that you figure out how to predict the exact shape PVC will bend to, adjusted for such effects as the rigidity of the ends of the PVC, the weight and strength of the plastic covering, the way plastic changes with temperature, etc. If precision in height doesn’t much matter (which is typically the case for backyard garden projects), the half-circle estimate should, as SMW suggests, work just fine. ---
**Author:** ![dolphinboy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/dolphinboy/32/330_2.png) [@dolphinboy](https://boards.straightdope.com/u/dolphinboy)\ **Post date:** [April 25, 2012, 6:45pm UTC](https://boards.straightdope.com/t/another-geometry-question/619906/10 "2012-04-25T18:45:51Z")
> [@Xema](#): > > If you really need an exact height the problem is difficult - it requires that you figure out how to predict the exact shape PVC will bend to, adjusted for such effects as the rigidity of the ends of the PVC, the weight and strength of the plastic covering, the way plastic changes with temperature, etc. > > If precision in height doesn’t much matter (which is typically the case for backyard garden projects), the half-circle estimate should, as SMW suggests, work just fine. Thanks! That works for me.