Help with a practical problem for my job

[rooferpaul]

The main purpose of this site is to offer help & guidance to students with their Maths problems. We would normally only offer new members advice on how to proceed with their work once they have submitted their own attempt(s) to solve a problem (so we can assess what help they need to reach a solution).
In my (personal) opinion, if you are part of a commercial organisation (eg: a roofing company) then you should employ someone who is capable of carrying out any calculations you need for your operations or pay an external (consulting?) agent for such services rather than looking for free help on the internet.
Some other forum member may well take a different view and be willing to offer comment on the problem you describe though I do suspect that your description leaves much to be desired. For example: did you mean that the width of your sheets is 762 mm or .762 m (not .762 mm) and a "guess" at the rake being 1020 (mm?) for a 25° pitched roof is a much use as a chocolate teapot! Only accurate information is of any use to us! Furthermore, I would expect your Rake length to depend not only on the pitch of the roof but also on the height of the roof (apex) above the gable and the width of the gable itself.
But those are just my opinions. ?
Good Luck.

I'm apologise for my ignorance. I just googled math help and the first site that came up was called free math help so I clicked on it.
I'm merely a labourer that wanted to find a theoretical answer to a problem that I encounter daily but haven't the knowledge to structure a formula for it. Usually we use an existing sheet as a template but I find it a little cumbersome to move around and it also causes a lot of scratches on the new sheet.
You are correct with the width, I should have reread my post but they are 762mm. Technically they are wider than that but we measure from the centre of the first ridge to the centre of the last but that's irrelevant to the problem as the remaining width is lapped under the next sheet.
The height of the roof doesn't matter as the only factor really is the pitch of the hip of the roof. There is no gable. Perhaps think of the hypotenuse of a triangle which has 762mm sections that run perpendicular to its adjacent side. As the angle increases, so does the length of the angled part of those sections so its all relative.
The hip is always 45 degrees but the pitch changes length of the rake if that makes sense

 

[blamocur]

I tried running your code...

python /tmp/a.py 25 762
Angle : 56.60
Rake : 912.8

...but this rake seems impossible. Imagine that the pitch angle of the roof is 0° (it's a flat roof). The diagonal cut where two purpendicular metal sheets meet will be at 45° to both sheets. The length of this diagonal (label "b" in post#11) will be [imath]\sqrt{762^2+762^2} \approx 1077.63[/imath] by Pythagoras. As the pitch increases, this minimum length should only get bigger?

If one of us gets sent to the corner for a blunder - then hopefully the roof there will be watertight

We can verify the result:

Assume the roof is 2 units (mm?) wide. At [imath]25^\circ[/imath] gable pitch the height of the roof is about 0.466.

The height of the [imath]45^\circ[/imath] hip triangle is the about [imath].0.423 \times \sqrt{2} \approx 0.659[/imath].

The side of the hip triangle is approximately [imath]\sqrt{1^2 + 0.659^2} \approx 1.198[/imath].

If we multiply this value by 762 we get about 912.78.

I am not going to the corner yet

 

[Cubist]

We can verify the result:

Assume the roof is 2 units (mm?) wide. At [imath]25^\circ[/imath] gable pitch the height of the roof is about 0.466.

The height of the [imath]45^\circ[/imath] hip triangle is the about [imath].0.423 \times \sqrt{2} \approx 0.659[/imath].

The side of the hip triangle is approximately [imath]\sqrt{1^2 + 0.659^2} \approx 1.198[/imath].

If we multiply this value by 762 we get about 912.78.

I am not going to the corner yet

Please just consider the flat roof scenario for a minute. The plan view would look like this (if we stick to the same layout that would be used for a pitched roof)...



...the width of each metal strip is 762mm. The length of each of the highlighted red diagonals would be approximately 1077.63mm (see calculation in post#18). This length would only become longer if the roof had depth and pitch?

Maybe I'm not visualising the problem correctly. Maybe I'm headed for the corner. Hmmm, perhaps we can get a sofa for the corner?

 

[blamocur]

This length would only become longer if the roof had depth and pitch?

I disagree. Imagine a very flat roof (I prefer [imath]0.1^\circ[/imath] instead of 0) and a [imath]45^\circ[/imath] hip pitch -- how would you expect the length to change?

And if you want to know how bored I can occasionally get (or you are bored with Python scripts) here is an interactive web page which does the same : https://sites.google.com/view/blamocur-roof-rake/home

 

[Cubist]

And if you want to know how bored I can occasionally get (or you are bored with Python scripts) here is an interactive web page which does the same : https://sites.google.com/view/blamocur-roof-rake/home

Wow, I LOVE the webpage calculator! I didn't know that kind of thing could be done with google sites! So cool.

I disagree. Imagine a very flat roof (I prefer [imath]0.1^\circ[/imath] instead of 0) and a [imath]45^\circ[/imath] hip pitch -- how would you expect the length to change?

As the pitch angle increases, then the plan view of post#21 will still appear identical (although some points will now have depth). All lines that have their endpoints at different depths will have their "real-life 3D" length increase. This length could never be smaller than that measured on a flat, orthographic, projection.

EDIT: tomorrow I'll try to make a scale model with cardboard. IF your calculator is wrong then I'm sure it can be easily tweaked.

 

[Dr.Peterson]

We can verify the result:

Assume the roof is 2 units (mm?) wide. At [imath]25^\circ[/imath] gable pitch the height of the roof is about 0.466.

The height of the [imath]45^\circ[/imath] hip triangle is the about [imath].0.423 \times \sqrt{2} \approx 0.659[/imath].

The side of the hip triangle is approximately [imath]\sqrt{1^2 + 0.659^2} \approx 1.198[/imath].

If we multiply this value by 762 we get about 912.78.

I am not going to the corner yet

Please explain these quantities in terms of the picture in #11. I don't know what you mean by "45 degree hip triangle", or where you got 0.423. Your numbers disagree with my own formula or GeoGebra construction:

 

[blamocur]

Please explain these quantities in terms of the picture in #11. I don't know what you mean by "45 degree hip triangle", or where you got 0.423. Your numbers disagree with my own formula or GeoGebra construction:

View attachment 32995

I believe your drawing shows [imath]25^\circ[/imath] hip pitch, whereas I am using [imath]45^\circ[/imath] for the hip and [imath]25^\circ[/imath] for the gable.

 

[blamocur]

As the pitch angle increases, then the plan view of post#21 will still appear identical...

Disagree again. I've attached a drawing of horizontal projections for [imath]1^\circ[/imath] gable pitch and varying hip pitches -- let me know it makes sense to you:

 

[Dr.Peterson]

I believe your drawing shows [imath]25^\circ[/imath] hip pitch, whereas I am using [imath]45^\circ[/imath] for the hip and [imath]25^\circ[/imath] for the gable.

It's assumed that the pitches are the same, resulting in the 45 degree angle in the top view, as mentioned in the OP (as I understand it), and as in all of Cubist's drawings.

See here, for example: "Most hip roofs are equal pitch. This means the angle of slope on the roof end or ends is the same as the angle of slope on the sides. Unequal-pitch hip roofs do exist, but they are quite rare."

 

[rooferpaul]

I second that opinion ! Go to the nearest University or College and consult for commercial problem - FaceToFace !!

I thought that this would be a simple problem for somebody to work out and help me out. I'm not going to take a day off work to go to a university and approach another prick like you and highlander that tell me to go and pay somebody else. How hard could this possibly be to work out. The time it took you to reply to me would be the same time it would take to find the formula and tell me what it it? What's wrong with you people?

 

[rooferpaul]

It's assumed that the pitches are the same, resulting in the 45 degree angle in the top view, as mentioned in the OP (as I understand it), and as in all of Cubist's drawings.

See here, for example: "Most hip roofs are equal pitch. This means the angle of slope on the roof end or ends is the same as the angle of slope on the sides. Unequal-pitch hip roofs do exist, but they are quite rare."

The angle of the hip is always 45 degrees.. the pitch of the roof itself changes almost every day (pitch=steepness)

 

[rooferpaul]

While I agree with @The Highlander I got curious and came up with a relatively simple formula (but which I hope someone will on this forum will double-check) :

[math]\alpha_R = \arctan\left(\frac{\sin \alpha_H}{\tan \alpha_G}\right)[/math]
where [imath]\alpha_R[/imath] is the rake angle (half of the angle of the top of the hip)
[imath]\alpha_H[/imath] is the angle of the hip with the horizontal plane (usually 45 degrees)
[imath]\alpha_G[/imath] is the angle of the gable with the horizontal plane.

If you implement this keep in mind that many programming languages use radians, not degrees, for computing trigonometric functions.
I list a simple Python code below, but for practical purposes I'd write this in Javascript and make a local web page on a computer since every web browser understands Javascript.

import numpy as np
import sys

## Usage : rake-angle.py <gable angle> <hip angle>,
## where <hip angle> defaults to 45%

aH = 45;
aG = float (sys.argv[1]);
if len (sys.argv) > 2:
aH = float (sys.argv[2]);
aHrad = (np.pi/180) * aH;
aGrad = (np.pi/180) * aG;

aR = (180/np.pi) * np.arctan (np.sin (aHrad) / np.tan (aGrad));

print (aR);

I don't know what any of that means sorry

 

[rooferpaul]

I must say that I do not agree at all, either in substance or in tone.

I have a bar buddy. We got to know each other talking about classical music. After the army, he had became a house painter. He was one when we first met. A few years later, he went out on his own with a few guys from the crew he normally worked with. It is a small business, no more than five people including him. He spends eight hours a day at job sites painting, and, many evenings, drives around to make estimates for new contracts. The idea that he has the time, money, or knowledge to find, vet, and hire a professional mathematician, who in many cases is paid largely by public taxes, is absurd.

I do not share the contempt that so many academics seem to have for people who do manual labor (probably because I never was an academic and am personally inept at almost all manual tasks.) If I had any clue what this problem was about, I’d help the guy out. I do not decline to help those who work with their hands. And if anyone wants to know why much of blue-collar America despises the academic world, they need only observe the classism in this thread. Help a roofer? How declasse.

Unfortunately, I was totally unable to understand the question posed so I can be no help in this case.

Thanks jeff for your sympathy. I find it difficult to understand why people on this site are willing to help somebody with a complicated theoretical problem but have no time for a labourer that's trying to make his life easier. This can't be a difficult problem. I have found that a 25 degreed roof pitch equals an 1125mm rake across a 762mm roof sheet if that helps anyone

 

[rooferpaul]

We can verify the result:

Assume the roof is 2 units (mm?) wide. At [imath]25^\circ[/imath] gable pitch the height of the roof is about 0.466.

The height of the [imath]45^\circ[/imath] hip triangle is the about [imath].0.423 \times \sqrt{2} \approx 0.659[/imath].

The side of the hip triangle is approximately [imath]\sqrt{1^2 + 0.659^2} \approx 1.198[/imath].

If we multiply this value by 762 we get about 912.78.

I am not going to the corner yet

I know for certain that 25 degrees equals an 1125 mm rake

 

[rooferpaul]

I disagree. Imagine a very flat roof (I prefer [imath]0.1^\circ[/imath] instead of 0) and a [imath]45^\circ[/imath] hip pitch -- how would you expect the length to change?

And if you want to know how bored I can occasionally get (or you are bored with Python scripts) here is an interactive web page which does the same : https://sites.google.com/view/blamocur-roof-rake/home

Looking from above wont help. Picture a pyramid or built from stone blocks that are 762mm wide. If the length of the angle that they were cut at didn't change, you wouldn't be able to change the steepness off the pyramid or church steeple or anything that has a steep pitch.

 

[rooferpaul]

Please just consider the flat roof scenario for a minute. The plan view would look like this (if we stick to the same layout that would be used for a pitched roof)...

View attachment 32994

...the width of each metal strip is 762mm. The length of each of the highlighted red diagonals would be approximately 1077.63mm (see calculation in post#18). This length would only become longer if the roof had depth and pitch?

Maybe I'm not visualising the problem correctly. Maybe I'm headed for the corner. Hmmm, perhaps we can get a sofa for the corner?

this is exactly right. can you please post a formula for me or at least help me work it out. Yes, the pitch increase is relative to the red diagonals. The depth and the pitch are synonymous. The further the apex moves from the gutter the less the pitch and the same applies reciprocally. You maybe the only person that understands this on this page. I know that 25 degrees is an 1125 rake or red diagonal but I cant work it out into a formula as I'm a simple labourer. As you can see ,I accidentally have turned my keyboard into italics and I don't know how i did it or how to fix it.
I am just a **** kicking worker Please help me

 

[blamocur]

I know for certain that 25 degrees equals an 1125 mm rake

While there is always a chance that I've made a mistake somewhere in this particular case I've checked the formulas more than once. I'll post my derivations later for others to check. Meanwhile, do you have any other data for the [imath]25^\circ[/imath] pitch so I can make sure that I understand the problem correctly?

 

[Dr.Peterson]

I know for certain that 25 degrees equals an 1125 mm rake

This is one of the things we've needed, to confirm our understanding. This agrees with my own formula, and my drawing.

The main trouble (apart from the comments about not wanting to help you) is that your phrase "The angle of the hip is always 45 degrees" does not communicate well to us ignorant mathematicians who don't know all the terms used in your field! And, in return, I think you misunderstand my statement, which I think is correct:

It's assumed that the pitches are the same, resulting in the 45 degree angle in the top view, as mentioned in the OP (as I understand it), and as in all of Cubist's drawings.

See here, for example: "Most hip roofs are equal pitch. This means the angle of slope on the roof end or ends is the same as the angle of slope on the sides. Unequal-pitch hip roofs do exist, but they are quite rare."

I was not saying that the pitch is the same on every roof, but that the two pitches on one roof are the same, as the quotation from a knowledgeable site says. But please tell us as clearly as you can what specific angle is 45 degrees, if I'm wrong. (This is why I asked you to clarify with a sketch.)

So here's the derivation of my formula (for the sake of those who can check it:

​

The rake is proportional to AC, so if the width is 762 mm and the pitch angle is [imath]P[/imath], then the rake is [imath]762\sqrt{1+\sec^2(P)}[/imath].

For P = 25 degrees, this is [imath]762\sqrt{1+\frac{1}{\cos^2(25^\circ)}}=1.489\cdot762=1134.7[/imath].

This is at least very close to your 1125. The difference may be due to the inaccuracy of measurements.

You may need more help in carrying out this calculation, depending on whether you are using a calculator, spreadsheet, or whatever.

 

[blamocur]

I thought that this would be a simple problem for somebody to work out and help me out. I'm not going to take a day off work to go to a university and approach another prick like you and highlander that tell me to go and pay somebody else. How hard could this possibly be to work out. The time it took you to reply to me would be the same time it would take to find the formula and tell me what it it? What's wrong with you people?

This is one of the things we've needed, to confirm our understanding. This agrees with my own formula, and my drawing.

The main trouble (apart from the comments about not wanting to help you) is that your phrase "The angle of the hip is always 45 degrees" does not communicate well to us ignorant mathematicians who don't know all the terms used in your field! And, in return, I think you misunderstand my statement, which I think is correct:

I was not saying that the pitch is the same on every roof, but that the two pitches on one roof are the same, as the quotation from a knowledgeable site says. But please tell us as clearly as you can what specific angle is 45 degrees, if I'm wrong. (This is why I asked you to clarify with a sketch.)

So here's the derivation of my formula (for the sake of those who can check it:

View attachment 33003​

The rake is proportional to AC, so if the width is 762 mm and the pitch angle is [imath]P[/imath], then the rake is [imath]762\sqrt{1+\sec^2(P)}[/imath].

For P = 25 degrees, this is [imath]762\sqrt{1+\frac{1}{\cos^2(25^\circ)}}=1.489\cdot762=1134.7[/imath].

This is at least very close to your 1125. The difference may be due to the inaccuracy of measurements.

You may need more help in carrying out this calculation, depending on whether you are using a calculator, spreadsheet, or whatever.

What does [imath]a[/imath] stand for? The length of AD or the length of ED?

 

[Dr.Peterson]

What does [imath]a[/imath] stand for? The length of AD or the length of ED?

I showed AB as 2a, so AD is a. ED is equal to this because of the equal pitches/45 degree angle in the horizontal plane (BDE is a right isosceles triangle). I should have stated that explicitly.