curved wheelchair ramp
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c=2∏r
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Are you going to try to put in a "bank" for curve to eliminate rubbing on the railing?
Dick
Dick
what i have is... an ADA wheelchair ramp that has a rise of 42" and an outer radius of 20 ' and an inner radius of 15' it will be a 1:12 slope. it is to be built from wood framing members to be finished with oak vaneer's ballasters floor walls etc. I am to build only the substructure frame for this. What i want to know is how to calculate the radius of the curved slope so i can make top plates for joist to rest on and to be able to stud in a perfect curve along the underside of the ramp. these plates will mach or be plumb up from the given floor plan flat radius.
whew! i hope i nailed the explanation . thanks
whew! i hope i nailed the explanation . thanks
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Not sure I understand what you are asking for. You want the radius but you say the radius is 15' and 20'. The run would be the percentage of the circle, is that what you are asking?
yes i have a given radius of the on the flat topo one demension view.. blueprints floorplan call out the inner and outer radius and has a given rise of ramp. what i am looking for is how to acheive the radius of the curve while on a slope. i'm trying to figure out how to explain it properly but picture a curved staircase with no treads and no risers just a flat plane all the way down, ok it has a given rise and a given inner and outer radius. what would be the formula for figureing the radius or is it an elipse of the plane or slice of the cylinder that is created from the curved slope. thanks
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Yes it would be an ellipse but that would be overcomplicating the problem.
First think of it as a straight run.
You have a given rise, 42" and a given pitch, 4.76 degrees. That would make the length of the ramp 506" 1/8 and the run 504" 3/8, Now curve that ramp at a 15' radius.
Determine a centerpoint and draw a 15' arc. On the radial line draw the point where the ramp ends. You now what to know where point is that 506" will hit the arc in plan view.
circumference=2∏r
c=2∏x15'
c=1131"
Take the run of the ramp and divide by circumference to get the portion of the circle 504 3/8 divided by 1131= 44.6
44.6x360=160.54 degrees
Now from the previously determined centerpoint swing 160.54 degrees and on that radial line will be the point where the ramp begins.
This starting point can be altered slightly by determining if the 15’ is the inside of the stringer or the outside and if the ramp pitch is measured at what would be considered the walkline.
First think of it as a straight run.
You have a given rise, 42" and a given pitch, 4.76 degrees. That would make the length of the ramp 506" 1/8 and the run 504" 3/8, Now curve that ramp at a 15' radius.
Determine a centerpoint and draw a 15' arc. On the radial line draw the point where the ramp ends. You now what to know where point is that 506" will hit the arc in plan view.
circumference=2∏r
c=2∏x15'
c=1131"
Take the run of the ramp and divide by circumference to get the portion of the circle 504 3/8 divided by 1131= 44.6
44.6x360=160.54 degrees
Now from the previously determined centerpoint swing 160.54 degrees and on that radial line will be the point where the ramp begins.
This starting point can be altered slightly by determining if the 15’ is the inside of the stringer or the outside and if the ramp pitch is measured at what would be considered the walkline.
I'm no stairbuilder but I have made wheelchair straight ramps volunteering for Habitat for Humanity. The 504"run is exactly 42'. I would consider the radius to be 17.5', the middle of the ramp. The beginning and end of the ramp must be shown on the plan view of the plans. I would have a support every one foot to make the math easy. I would build the first support one foot from the top of the ramp at a height that the finished ramp would be 41". I would build the next support one foot away and one inch lower until I got to the bottom. Also I would chamfer the top of the joists so the ramp deck would lay flat.
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The run will be more than 42’ with the 5’ (min) landing that’s required somewhere in the middle. You can’t have more than a 30” rise without a landing.
kw excellent point about the 30' + rise for ADA ramp.my last post i did not have exact measures but the ris is at exactly 30".
Keith would it be an ellipse or a Helix or as i heard a "Helicoid" shape or for mathematical equasion figuring. because i can figure and build an ellipse structural members once given a rise and span. also i am still working my way through your explanation of how to figure the result shape and calculation of arc/curve ? when given an inner and outer radius and rise.
thanks: rjn
Keith would it be an ellipse or a Helix or as i heard a "Helicoid" shape or for mathematical equasion figuring. because i can figure and build an ellipse structural members once given a rise and span. also i am still working my way through your explanation of how to figure the result shape and calculation of arc/curve ? when given an inner and outer radius and rise.
thanks: rjn
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I think you’re over-complicating this ramp. Instead of building two curved walls with a rise and setting joist on top of them, I’d eliminate the joist altogether and just build a series of pony walls to act as joist. Once you know the run of each side, figuring out the rise for any given point of the run would be easy.
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I've never built a wheelchair ramp so I don't know the codes but what I can suggest is not to start building until after you have drawn a "stair Plan". All your decisions are made in plan view. You need to know your total rise, desired pitch, either the inside or outside radius and ramp width. From there it is a matter of finding the circumference of those radius then the portion of the circle you need.
For example:
30" rise
15' inside radius
desired pitch 4.76 (where is this measured- inside radius, walk line or center or ramp)
Lets pick the inside radius the point where the pitch is measured, that will make to others lower.
Lets say a 48" ramp width which will make the outside radius 19'
30" rise and 4.76 pitch= 360 1/4" run in plan view
total circumference of a circle with a 15' radius (inside stringer) (2Pixr) is 1131"
360.25/1131=.318523
.318523x360°=114.7°
114.7° is the run of the ramp without landings, etc.
outside stringer circumference 19' 2π19'= 1432 9/16"
divide 1432 9/16" by 360°=4" per degree, multiply by 114.7°= 456 7/16"
456 7/16" run, 30" rise= 3.76° pitch of outside stringer.
The middle of the ramp pitch would be?
attached is a typical single point radius stair plan.
For example:
30" rise
15' inside radius
desired pitch 4.76 (where is this measured- inside radius, walk line or center or ramp)
Lets pick the inside radius the point where the pitch is measured, that will make to others lower.
Lets say a 48" ramp width which will make the outside radius 19'
30" rise and 4.76 pitch= 360 1/4" run in plan view
total circumference of a circle with a 15' radius (inside stringer) (2Pixr) is 1131"
360.25/1131=.318523
.318523x360°=114.7°
114.7° is the run of the ramp without landings, etc.
outside stringer circumference 19' 2π19'= 1432 9/16"
divide 1432 9/16" by 360°=4" per degree, multiply by 114.7°= 456 7/16"
456 7/16" run, 30" rise= 3.76° pitch of outside stringer.
The middle of the ramp pitch would be?
attached is a typical single point radius stair plan.
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Ok Gentlemen; I finally was able to get a copy of the detail page from blueprints on this curved ramp i am to build. Again what i want to do is be able to calculate the length of and the curve of in a flattened state the the Helix stringer lines of the outer and inner sloped curved edges of the ramp. thanks. rjn
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rjn64,
as you have posted on a couple of sites and received a considerable amount of good information you may already have everything you need to build this ramp. That being said, from the information you posted it would appear that you have a fairly small rise and a fairly large radius. From what I can see your only concern is to maintain level along the radial line, I.E. twist. With such a small rise i would think that 3/4" plywood would easily bend to that amount of twist. It would seem that the only thing left is to start construction.
as you have posted on a couple of sites and received a considerable amount of good information you may already have everything you need to build this ramp. That being said, from the information you posted it would appear that you have a fairly small rise and a fairly large radius. From what I can see your only concern is to maintain level along the radial line, I.E. twist. With such a small rise i would think that 3/4" plywood would easily bend to that amount of twist. It would seem that the only thing left is to start construction.
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