The Naked Scientists Forum

swalker

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« on: 08/09/2006 18:17:31 »
I am a firefighter with a practical knowledge of how ladders react at different angles; however I wonder if someone can give me some theory?

When the ladder is steeply pitched there is little load on the head and it moves easily. With a flat pitch the deflection increases, it is harder to move the head and if the head is resting on the rollers I worry about the heel kicking out.

How does the load on a ladder transfer to the head and the heel at different angles?

If, for example, there was a 100kg load in the middle of the ladder I presume that 100% of this would be on the heel with the ladder at 90 degrees (to horizontal) and 50% on the head and heel with the ladder at 0 degrees. What % would be on the head/heel at angles between these extremes?

If the rollers (at the head of the ladder) are on a vertical wall what determines the horizontal force causing the heel to slip out?

Thanks Steve

neilep

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« Reply #1 on: 08/09/2006 20:18:11 »
Steve,

First I want to WELCOME you to the site, Secondly, I personally can't help you but I am sure an answer will come, and THIRDLY...YOU guys are the greatest... !!!

You do an amazing job......YAYYYYYYYYYYYYYY !!

Men are the same as women, just inside out !

syhprum

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« Reply #2 on: 08/09/2006 20:29:16 »
My arithmatic is not very good at this time of night but two things strike me, If we look at the vertical component of the force the heel of the ladder exherts on the ground it will always be very nearly 100 Kg at all normal working angles but the horizontal component will become very large as the angle becomes small, the horizontal component of the the force that the top of the ladder exherts no the wall must always be the same as the bottom else the ladder would not be stationary.
now you have fitted wheels on the top the vertical component at the top is very small unless the angle is very shallow in which case it approaches half of that which the bottom applies to the ground provided the ladder was prevented from moving.
I think a graph would be the best way of representing these findings, at shallow angles the forces involved would be much greater than 100Kg

syhprum

Soul Surfer

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« Reply #3 on: 08/09/2006 23:48:49 »

To calculate the horizontal thrust at the top it is half the weight of the ladder plus the weight of the man divided by the distance he is up the ladder expressed as a proportion of the total ladder length.

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syhprum

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« Reply #4 on: 09/09/2006 08:12:33 »
Too late I was just about to go into sines, tangents,etc

syhprum

syhprum

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« Reply #5 on: 09/09/2006 08:15:50 »
Wrong! the horizontal thrust at the top is the same as that at the bottom See Sir I Newton (laws of motion)

syhprum

swalker

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« Reply #6 on: 09/09/2006 17:26:22 »
Thanks for your welcome neilep and your responses syhprum and Soul Surfer.

Can you give me some worked examples:

70kg firefighter 9m along 10m ladder at 45 degrees.

Are there formulas I could apply with the four variables(perhaps something I could use on Excel)?
Giving:

Vertical force at the heel.
Horizontal force pushing heel out.

syhprum

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« Reply #7 on: 09/09/2006 19:09:59 »
Excuse me if I am wrong but I will assume you are not familiar with trignometric functions hence I will define what a tangent is (tan).
Imagine a right angled triangle with a base on the ground, a vetical and a diagonal the tan is the ratio in length of the base and the vertical in relation to the angle the diagonal makes to the vertical (tables are published of these values and many pocket calculators will compute them)To give some examples tan 5°=0.0874, tan 45°=1, tan 85°=11.43.
Now you have made the rather simplistic assumption that the ladder is of zero weight so I will allow 20 Kg for it and 80 Kg for the climber.
At 45° the vertical force at the base would be 10Kg due to the ladder plus (1/10)*80Kg due to the climber a total of 18Kg while at the top the vertical force (that would have to be resisted by friction between the ladder and the wall ) would be 10Kg due to the ladder plus (9/10)*80Kg due to the climber a total of 82Kg.
The horizontal force exherted on both the ground and the wall would be tan 45°*100Kg = 100Kg.
Now let us see what the result would be if you were foolish enough to set up an experiment to test this scenario.
at the top the 100Kg thrust might be sufficient to resist the 82Kg down force if the wall was rough but at the bottom you would only 18Kg downforce to generate enough friction to resist 100Kg of thrust which would be hopelesly inadequate the climber would hit the ground at about 10 meters per second.

syhprum

swalker

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« Reply #8 on: 09/09/2006 21:55:44 »
syhprum thanks for that. You were right, I didnt know what a tan was -but I do now. I think that I follow your thread.

The downward force on the head and heel depend on the weight of the ladder and the weight of the load (firefighter). The proportion of this weight on the heel or head depends on where that weight acts along the length of the ladder. If it acts in the centre of the ladder then the downward force on the heel and head is equal. However if it acts 3/4 of the way up the ladder then 3/4 of the force will act on the head and 1/4 on the heel. The angle of the ladder has no effect on this downward force.

However the horizontal force is a product of the total weight of ladder and load * tan angle of ladder. This force acts against the wall at the head and tries to overcome the friction of the heel against the ground at the heel.

Our standard 135 ladder(14.1m max extension) actually weighs 102kg and since it is a triple extension ladder the weight is not distributed equally but is bottom heavy. Ideally the heel will be out 1/3 of the working height. We always have someone to foot the ladder when it is being used. This entails them putting their weight on the heel.
« Last Edit: 09/09/2006 22:02:22 by swalker »

syhprum

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« Reply #9 on: 09/09/2006 23:32:11 »
I think that a heavy ladder at the angle you suggest would be quite safe without any additional weight on the bottom but of course if a person is available to do this so much the better, my experience with ladders has been for antenna instalation which is normaly a one man job.
Of course if you normaly think of the angle in terms of the ratio of height compared to distance from the wall you already have the tan!, in your case 0.66

syhprum

swalker

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« Reply #10 on: 10/09/2006 09:43:30 »
Is this right? I have assumed the weight of ladder and load as 180kg. The angle is degrees from horizontal. The force is the horizontal force in kg.

Angle    Force

90.0    0.0
80.0    31.7
70.0    65.5
60.0    103.9
50.0    151.0
40.0    214.5
30.0    311.8
20.0    494.5
10.0    1020.8
1.0    10312.2

When I tried 0 degrees the force was 2938419409971430000.0 kg which seems a bit excessive. I thought that if the ladder was horizontal the force would be 0. However I can see that it is assuming that the rollers are in contact with a wall. In reality when we bridge (low angle) a ladder we always rest the strings (sides) of the ladder on top of the wall (or window cill ect).

« Last Edit: 10/09/2006 09:45:06 by swalker »

syhprum

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« Reply #11 on: 10/09/2006 19:16:01 »
I get exactly the same results with my calculator except for a zero angle it displays an infinity character (an "8" rotated by 90°).
When the top rest on window sill if there are wheels so that the weight of the ladder is directed in a truly vertical direction the horuzontal component would be zero but in the more realistic case where the top of the ladder rests at an angle another computation would be required to take than angle into account. ( I am wracking my brain how to do it ).

syhprum

syhprum

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« Reply #12 on: 11/09/2006 08:04:53 »
This is rather a guess based on observation I would estimate that for a ladder resting on the ground and leaning against a window sill the vertical force would be as previously computed in my post 44 but the horizontal force would follow a sine/cosine law i.e the sine of the base angle times the cosine.
Sample at 45° sin=0.707 cos 45°=0.707 product=0.5 sin 10°=0.173,cos=0.985 product=0.171
sample at 80° sin 80°=0.985, cos=0.173 product=0.171, as experience shows that the force peaks at 45° and falls to zero at 90° and 0°.

Errata sines are the ratio of the base to the diagonal while cosines are the ratio of the vertical to the diagonal ( available on your pocket calculator ).

syhprum

eric l

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« Reply #13 on: 11/09/2006 14:09:34 »
quote:
When I tried 0 degrees the force was 2938419409971430000.0 kg which seems a bit excessive. I thought that if the ladder was horizontal the force would be 0.

You are right, and the reason why this does not show in your calculation is that you have not taken into account that the weight on the ladder is divided in a force perpendicular to the ladder (and resulting in flexion of the ladder) and a force along the beams.  It is this force along the beams that can be divided into a downward force and a horizontal force trying to push the foot of the ladder away.

For a weight W in the middle of the ladder, we have :

• at an angle of 30°  :  W*0.25 downward thrust, W*0.43 forward thrust and W*0.86 for flexion of the ladder.

•  at an angle of 45° :  W*0.5 downward thrust, W*0.5 forward thrust and W*0.70 for flexion
•  at an angle of 60° :  W*0.75 downward thrust, W*0.43 forward thrust and W*0.50 for flexion.

at 90° we have only downward thrust.

I am afraid I must have left out something (the position of the weight on the ladder probably) but it would take me some time consulting my old books.

syhprum

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« Reply #14 on: 11/09/2006 14:39:08 »
We are using tangent's as the angle tends to zero tangents become infinite, the fact that the computer did not give an infinity symbol when it should have done is a quirk of the software used in it, every computer I have tried has always recorded infinity when I have divided 1 by 0 which is what you are doing when you try to compute the tangent of 0.0°.
For the purposes of calculation we are using a ladder of infinite rigidity as there is no need to complicate the calculation by introducing flexing which would have very little effect with practical ladders

syhprum

syhprum

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« Reply #15 on: 11/09/2006 14:59:03 »
I have tested out several trigometric function calculators availabe on the web and indeed several fail to record infinity when they should.
This one is O.K http://www.analyzemath.com/Calculators/Trigonometry_Cal.html

syhprum

swalker

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« Reply #16 on: 11/09/2006 20:35:01 »
Ok so if the head is resting on top of the sill / wall etc and the load is 180 kg then the horizontal force (kg) on the head and heel is:
Angle   -    Force

90.0   -     0.0
80.0   -    30.8
70.0   -    57.9
60.0   -    77.9
50.0   -    88.6
45.0   -    90.0
40.0   -    88.6
30.0   -    77.9
20.0   -    57.9
10.0   -    30.8
0.0   -    0.0

Quite a difference when resting the head on top of the wall as opposed to leaning the head against the wall.
« Last Edit: 11/09/2006 20:36:48 by swalker »

syhprum

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« Reply #17 on: 11/09/2006 20:47:13 »
I am pretty sure that these calculated values tie up with what you have found in practice.
Are you going to write a pamphlet or article in a house magazine for the instruction of of your fellow firefighters ?

syhprum

syhprum

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« Reply #18 on: 12/09/2006 16:46:03 »
May I suggest that your ladders incorporate a small extension of about 15 CM set at an angle of 135° to the upper section, this would sit on the window sill and re-direct the downforce to a vertical direction and therefor reduce the horizontal force to zero when the ladder is at 45°

syhprum

swalker

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« Reply #19 on: 12/09/2006 18:00:50 »

As for what to do with my newfound knowledge - I will have to think about that. Certainly I will try to pass it on but the method needs consideration.

My interest in this matter was rekindled by a thread (Operational / Bridging drills) at:

I am not sure that I have a clear picture of your suggestion to redirect the downforce. Is it some type of hook?
« Last Edit: 12/09/2006 18:03:18 by swalker »

syhprum

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« Reply #20 on: 12/09/2006 19:02:33 »
No not a hook but just a bending of the last 15 CM so that when the ladder is at an angle of 45° to the ground this short section lays flat on the sill, I think that this would introduce less instability than having wheels at the top.

syhprum

swalker

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« Reply #21 on: 12/09/2006 19:52:58 »
Ok, I see.

The wheels are there to make it easier when moving the head. For example; we can leave the head in contact with the wall and change the height and angle by shifting the heel in or out, Only the "135" ladder has wheels, the small ones don't. I imagine that the maths theory doesn't change much if it has wheels or not.
If there was a projection it could catch on something. If the top extension is pulled up it disengages the pawls that hold it up ( this can be a bad thing).

Ladders are definitely far more complicated than they appear at first sight.

Out of interest (and before my time): we used to use a hook ladder that was hung from the window sill - that looks a bit scary. We also used to use a ladder called an "escape" that had big wheels at the heel.

syhprum

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« Reply #22 on: 12/09/2006 20:07:49 »
As you say wheels are probably the best bet, how do you erect a 105 KG ladder? too much for one normal man, do two do it?

syhprum

syhprum

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« Reply #23 on: 12/09/2006 20:11:12 »
P.S I had a trip up an 'escape' when I was a young lad installing a TV in the canteen of my local fire station, no big deal for me as I was used to installing antennas.

syhprum

swalker

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« Reply #24 on: 13/09/2006 06:31:08 »
The 135 needs 4 people to pitch it. Quite a team effort - especially with a bit of wind on an uneven site. There are "props" that are hinged levers (about 5 m long) attached to the top of the bottom extension. These are used to help steady the ladder.
« Last Edit: 13/09/2006 06:36:21 by swalker »