Help with a practical problem for my job

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?
I've already spent many hours on this problem, and I am sure if we add the hours other posters spent on it will be a pretty large number. Do people call you names if you refuse to do free work for them? That would be pretty rude in my opinion.
 
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.
I assumed that [imath]45^\circ[/imath] is the pitch of the hip, not the base angle of the hip's horizontal projection. But in your drawing in #26 you show the hip's pitch as [imath]25^\circ[/imath], which I assumed to be the pitch of the gable. Looks like I misunderstood the problem from the beginning :(
 
I assumed that [imath]45^\circ[/imath] is the pitch of the hip, not the base angle of the hip's horizontal projection. But in your drawing in #26 you show the hip's pitch as [imath]25^\circ[/imath], which I assumed to be the pitch of the gable. Did I misunderstand the problem from the beginning?
Yes. See my post #29.

I initially tried searching for a site that would have formulas for angle A or side AC, thinking it might be a common need, but found nothing -- just lots of sites on how to calculate rafter lengths and angles, and so on. But this particular detail was visible in all of them, though I don't think any used the term "hip angle" or identified what angle is 45 degrees.
 
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?
It seems that I've misunderstood the problem: I thought that the pitch of the hip is always 45 degree, but it seems more likely that the projection of the hip always has [imath]45^\circ[/imath] at the base. The corner might be mine to occupy...
 
Yes. See my post #29.

I initially tried searching for a site that would have formulas for angle A or side AC, thinking it might be a common need, but found nothing -- just lots of sites on how to calculate rafter lengths and angles, and so on. But this particular detail was visible in all of them, though I don't think any used the term "hip angle" or identified what angle is 45 degrees.
That would explain why your result is close to @rooferpaul 's empirical observations.
 
It seems that I've misunderstood the problem: I thought that the pitch of the hip is always 45 degree, but it seems more likely that the projection of the hip always has [imath]45^\circ[/imath] at the base. The corner might be mine to occupy...
No corner for you unless it is 45 degrees. You tried to help but got lost in the original description of the problem, which completely stumped me.
 
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.
The formula is sensitive to small errors in measurement of angles. 23 degrees gives 1125 as a result.
 
No corner for you unless it is 45 degrees. You tried to help but got lost in the original description of the problem, which completely stumped me.
This is exactly why I direct people (with practical problem) to face-to-face to meetings - misunderstanding "jargon" which changes from company to company. After 46 responses (thread/name-calling started Tuesday) - we have not given the op (or is it OP) a spoon-fed formula to regurgitate.
 
This is exactly why I direct people (with practical problem) to face-to-face to meetings - misunderstanding "jargon" which changes from company to company. After 46 responses (thread/name-calling started Tuesday) - we have not given the op (or is it OP) a spoon-fed formula to regurgitate.
Good point. In case of someone studying math we can ask for the original statement of the problem, which in most case is pretty clearly stated. Communicating practical problems has often been a frustrating exercise in my experience ("why did not you say so from the beginning?") even in face-to-face meetings.
 
Good point. In case of someone studying math we can ask for the original statement of the problem, which in most case is pretty clearly stated. Communicating practical problems has often been a frustrating exercise in my experience ("why did not you say so from the beginning?") even in face-to-face meetings.
Which, of course, is why patience is needed ... on both sides. Even when it can't be face-to-face, it's best to have a one-on-one conversation about such a problem, in order to efficiently negotiate the details of the problem and iron out communication issues. (Lags due to moderation don't help, either.)

Ultimately, this is a quick problem to solve, and I'm still amazed that I couldn't find the formula ready-made on one of the many sites I looked at. The hard parts are human interaction and translation to and from problem-specific language. (Which is true in word problems generally! Math is easy; people and English are hard.)
 
... 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].
I agree with this. I thought it was probably worth repeating, in a large font, because it gets lost in the conversation...

[math]762\sqrt{1+\sec^2(P)}[/math]
...and obviously anyone who decides to use this should check the result with a scrap piece of material/ paper etc before using any expensive building supplies.

And once again, here's the link to a brilliant online calculator...
To atone for my sins I've corrected my online calculator using @Dr.Peterson formula : https://sites.google.com/view/roof-rake-calculator/home. It confirms that one needs [imath]23^\circ[/imath] pitch to get 1125mm rake.

--

This is exactly why I direct people (with practical problem) to face-to-face to meetings - misunderstanding "jargon" which changes from company to company. After 46 responses (thread/name-calling started Tuesday) - we have not given the op (or is it OP) a spoon-fed formula to regurgitate.
My wife is a business analyst, and her work is basically full time "translation" between the people who have a business need and the people who can fulfill it (banking business speak <-> maths/ computer speak). But, we did get there in the end with this problem, I think :unsure::confused::D

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?
There's no reason to insult people who choose to only help students. These volunteers give up their free time to help people. If you were nice to them they might be more likely to help the next non-student. Anyway, I hope the expression/ calculator above are helpful to you. Please come back if you think we've made some kind of error.
 
...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
NOTE: Then the calculated "rake", or cut-length, will also be from the centre of the first ridge to the centre of the last. The actual cut-length will be longer (across the whole width of the sheet)
 
Next time when you see a roofer's truck in your neighborhood with "FREE CONSULTATION" written in bold and colorful writing - try to get a FREE advice regarding application of flashing at the chimney-roof joint.
 
Next time when you see a roofer's truck in your neighborhood with "FREE CONSULTATION" written in bold and colorful writing - try to get an advice regarding application of flashing at the chimney-roof joint.
I've used various "diy help forums" to get helpful free advice. Although with DIY it's hard to get a professional looking finish without having the experience that full-time builders have. Certainly plastering is harder than it looks, as I have found out :ROFLMAO:
 
I've used various "diy help forums" to get helpful free advice. Although with DIY it's hard to get a professional looking finish without having the experience that full-time builders have. Certainly plastering is harder than it looks, as I have found out :ROFLMAO:
I did too - but those came from another homeowner - not from professionals touting free consultation (specially not roofers).
 
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