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A shooting house is 20 ft off the ground. If you build a staircase that sets at 45 degrees, how long should the staircase be? What is the equation used to calculate this? Am I right in thinking that the steps should be built at 45 degrees so they will be level when the staircase is lifted and connected to the stand?

20^2 +20^2 = c^2 so right about 28.284271247461900976033774484194'

a(squared) + b(squared) = c(Squared)
where C is your ladder.
Pathagorean theorum

This is the best example I could find for your situation
http://www.google.com/imgres?imgurl=htt ... A&dur=7683

rjohnson wrote:20^2 +20^2 = c^2 so right about 28.284271247461900976033774484194'

ahh, an educated man.......

What they said. You're dealing with an isosceles right triangle.

That's gonna be a long set of steps.

If you cut the distance down to 10' out from the bottom of the stand it will still be 22.360679774997896964091736687313'

Yes 28.3' may be too long. If the distance from the bottom of the stand is cut to 10' and the ladder is 22.36' then what would the angle be?

bigoak wrote:Yes 28.3' may be too long. If the distance from the bottom of the stand is cut to 10' and the ladder is 22.36' then what would the angle be?

I would think that if you dropped the bottom of the stand to 10', then to have a 45 degree angle, your ladder would be 14.14' long.

Your triangle would have two sides that would be 10' (10 from the ground to the bottom of the stand and 10 from the bottom of the stand to the where the ladder meets the ground) and the hypotenuse would be 14.14.

Or am I missing something?

We have a shooting house that's about 20' off the ground. We have a ladder to it. I would like to build wrap around stairs to get to it, probably will at some point.

deltadukman wrote:a(squared) + b(squared) = c(Squared)
where C is your ladder.
Pathagorean theorum

This is the best example I could find for your situation
http://www.google.com/imgres?imgurl=htt ... A&dur=7683

Vickers said Pathagorean theorum

novacaine wrote:

deltadukman wrote:a(squared) + b(squared) = c(Squared)
where C is your ladder.
Pathagorean theorum

This is the best example I could find for your situation
http://www.google.com/imgres?imgurl=htt ... A&dur=7683

Vickers said Pathagorean theorum

hahahahaha

I think your ladder would be at about 63 degrees. cos-1(10/22.36)

Maybe I said that wrong. If the stand is still 20' off the ground and the distance out from the bottom of the stand to the bottom of the ladder is 10' then the ladder would have to be 22.36'. Need to know what angle to weld the steps when the ladder is laying flat on the ground so they will be level when the ladder is erected and attached to the stand. We are gonna have to build this thing ahead of time and transport it to the stand.

bigoak wrote:Maybe I said that wrong. If the stand is still 20' off the ground and the distance out from the bottom of the stand to the bottom of the ladder is 10' then the ladder would have to be 22.36'. Need to know what angle to weld the steps when the ladder is laying flat on the ground so they will be level when the ladder is erected and attached to the stand. We are gonna have to build this thing ahead of time and transport it to the stand.

You didn't say it wrong, I misread what rjohnson suggested. I believe Double_D has your answer.

I remember having to do the same thing many years ago. The Rise and Run is extremely important - so check to make sure your steps are "Comfortable" as it relates to the dimension step to step. As important, make sure your steps (treads) are adequate to handle large, muddy, hunting boots (2 x 12's would be my recommendation).

Just my .02 cents worth of....

just build a damn ladder straight up, by screwing steps between the back 2 posts!!! How we do all of ours and no problems yet!!