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[Mr Skrunts]

Doing some calcs to help out with a specialist wheel chair.

Working on a wheel chair base being 36" x 36" (square) need to work out the minimum door opening required for clearance and the radius required to go through the door and turn into a hall way at a right angle.

So basic questions are?

1.  Minmum width of door required for the wheel chair

2.  Minimum width of hall required during turning wheel chair into the hall.

3.  What is the diagonal length of the chair from corner to corner.

Very much appreciated for any help.

TIA.  

[Elite Pete]

Is it a public building and do you want to know the legalities?

[Mr Skrunts]

No, it's just for inside a home

There is talk of door widening to be done but I think the dtails given by the builder are getting out of hand as he wants to move the wall in the hall way etc.

So I tried to do some home work and got stuck.

As a matter of interestis there any legislation regards wheel chair useage in the home v building work.?

Sorry to be a pain.

TIA.  

[MikeDundee]

Have a look here; http://www.wheelchairaccess.fsnet.co.uk/page37.html

It explains DDA and who it applies to, nothing for private house though.

[Elite Pete]

As Mike says, the DDA generally applies to Service providers, employers and Landlords.

[Elite Pete]

The absolute minimum width of a door in a public building should be 750mm, the recommended width is 800mm plus

[Mr Skrunts]

Cheers, am actually trying to do the working out for a custom built wheel chair.

My Mathematics has completely dried up and the grey cells need recharging.

[HolyCount]

if its 36" x 36" the diagonal will also be 36" -- the base would form two equilateral triangles.

As a wheelchair can revolve around its centre the turning circle would also be 36" ----- remember to allow for elbow room though.

You local building control office might be able to help

[MikeDundee]

Cheers, am actually trying to do the working out for a custom built wheel chair.

My Mathematics has completely dried up and the grey cells need recharging.

I think you'll need a minimum of 4ft (48'')into the hallway, that allows you getting the wheelchair right through the door, allowing 1ft (12'')clearance, you may even get away with less. Obviously if your front entrance door is standard size, then you'll need to get a larger FED, and maybe a ramp aswell, dependant upon the threshold

[MikeDundee]

if its 36" x 36" the diagonal will also be 36" -- the base would form two equilateral triangles.

As a wheelchair can revolve around its centre the turning circle would also be 36" ----- remember to allow for elbow room though.

You local building control office might be able to help

Avoid them if you can, they will only want to take your money

[Bandit127]

Doing some calcs to help out with a specialist wheel chair.

Working on a wheel chair base being 36" x 36" (square) need to work out the minimum door opening required for clearance and the radius required to go through the door and turn into a hall way at a right angle.

So basic questions are?

1.  Minmum width of door required for the wheel chair

2.  Minimum width of hall required during turning wheel chair into the hall.

3.  What is the diagonal length of the chair from corner to corner.

Very much appreciated for any help.

TIA.  

Taking your qustions literally...

1. 36" plus an allowance for fingers on wheels. Say 2" each side = 40" or
1016 mm
2. Wheelchairs pivot from (effectively) the rear corners of the 36" square, so the radius is the same as the next question. This is assuming fingers are not required on the inside wheel...
3. Diagonal is the side times square root of 2, according to Pythagoras. So 36 x 1.4142 = 50.9" or 1293 mm.

[HolyCount]

Doing some calcs to help out with a specialist wheel chair.

Working on a wheel chair base being 36" x 36" (square) need to work out the minimum door opening required for clearance and the radius required to go through the door and turn into a hall way at a right angle.

So basic questions are?

1.  Minmum width of door required for the wheel chair

2.  Minimum width of hall required during turning wheel chair into the hall.

3.  What is the diagonal length of the chair from corner to corner.

Very much appreciated for any help.

TIA.  

Taking your qustions literally...

1. 36" plus an allowance for fingers on wheels. Say 2" each side = 40" or
1016 mm
2. Wheelchairs pivot from (effectively) the rear corners of the 36" square, so the radius is the same as the next question. This is assuming fingers are not required on the inside wheel...
3. Diagonal is the side times square root of 2, according to Pythagoras. So 36 x 1.1412 = 41.1" or 1043 mm.

[/highlight]


Disagree with 3 --- diagonal is the hypotenuse (?) of the triangle formed by the two sides. Therefore, according to Pythagoras it is the square root of 36" squared + 36" squared

[jereboam]

if its 36" x 36" the diagonal will also be 36" -- the base would form two equilateral triangles.

As a wheelchair can revolve around its centre the turning circle would also be 36" ----- remember to allow for elbow room though.

You local building control office might be able to help

Back to school! Completely wrong, I'm afraid, so the turning circle calculation is wrong as well.

Bandit127 has it absolutely right - no question.  

Anyway, 36*36 + 36*36 = 2596, and the square root of this is 50.9

 

[Bandit127]

Doing some calcs to help out with a specialist wheel chair.

Working on a wheel chair base being 36" x 36" (square) need to work out the minimum door opening required for clearance and the radius required to go through the door and turn into a hall way at a right angle.

So basic questions are?

1.  Minmum width of door required for the wheel chair

2.  Minimum width of hall required during turning wheel chair into the hall.

3.  What is the diagonal length of the chair from corner to corner.

Very much appreciated for any help.

TIA.  

Taking your qustions literally...

1. 36" plus an allowance for fingers on wheels. Say 2" each side = 40" or
1016 mm
2. Wheelchairs pivot from (effectively) the rear corners of the 36" square, so the radius is the same as the next question. This is assuming fingers are not required on the inside wheel...
3. Diagonal is the side times square root of 2, according to Pythagoras. So 36 x 1.1412 = 41.1" or 1043 mm.

[/highlight]


Disagree with 3 --- diagonal is the hypotenuse (?) of the triangle formed by the two sides. Therefore, according to Pythagoras it is the square root of 36" squared + 36" squared

You were right. I did a quick edit when I spotted the mistake, but you beat me to it. The sqaure root of 2 is 1.4142, not 1.1412.

Either way it's 50.91"

[Entwood]

Both answers are the same .. in the case of a square ...

36 * 1.4142 =  50.91

root((36*36)+(36*36)) = root (1296+1296) = root(2592) = 50.91

simply proved by

root(x² + x²) = root(2x²) = root(2) * root(x²) = root2 * x

if I could use a root sign it would be even easier to read