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31 items are filed under this topic.
Subdividing land 2019-05-09
From Reuben:
This is the measurements of my plot, A-B 46.7M, B-C 193.1, C-D 198.5 & D-A 208.25 (Clockwise naming of sides) angle A at 90 degrees. My questions is how do i subdivide this plot from the bottom having lines running parallel to C-D, eg two 2acre plots. the the remaining part becomes my compound (Uper part at line A-B)
Answered by Harley Weston.
7 spheres on a hexagonal tray 2019-01-14
From herm:
what is the length of each side of a hexagonal tray, with the height of each side 0.75 inch, to hold seven spheres, each with a diameter of 3.00 inches? The spheres are placed such that each side of the hexagon is touched by one sphere at its midpoint (and the seventh sphere is place in the center of the "ring" of the other six spheres.
Answered by Harley Weston.
Two intersecting tubes 2018-08-15
From Tommy:
Hi, I am trying to determine a mathematical model for two metal tubes joining at various degrees for weld.
For instance, if I am trying to join the end of a tube to the side of another at a 90 degree angle, it will be a simple profile cut out of the joining tube.
Where it gets tricky is if you want to join the new tube at a given angle.
It would be very helpful if you could give insight as to how I can solve this problem or an equation I could work off of.
Thanks for the help!!

Answered by Edward Doolittle.
The volume of a berm 2018-06-04
From Mike:
Have a berm that is 700’ long, 29’ y’all and needs a 20’ top with 5-1 slope on one side and 3-1 slope in the other
Answered by Penny Nom.
Volume of liquid remaining in a tilted cylinder 2016-11-08
From Brian:
I am trying to determine the amount of a liquid remaining in a 55 gallon drum when it is tilted at 45 degrees and the liquid level is low enough so that the liquid does not completely cover the bottom of the drum.

Your help is greatly appreciated.

Answered by Harley Weston.
A deck that is half an ellipse 2016-02-28
From Steve:
On your website, I was reading a question and your response from a girl named Angela in which you provided a formula by which her father, a welder, could figure out points on an arc corresponding to equal 3' intervals on a 30' chord where the vertex was 1' off the chord. Is there an equivalent formula when working with an ellipse? I suspect this change will make the calculations significantly more complex. I am building a deck that is half an oval, and would like to be able to mark out the perimeter by measuring the distance from regular intervals on the primary access to a corresponding point on the perimeter. I will then connect the points on the perimeter and cut a reasonably smooth arc. The length of the primary access will be 22' and width of the deck at the vertex is 9'. I would like to be able to know the distance from the primary axis to a point on the perimeter at equal intervals of 6" along the primary axis. Can you help?
Answered by Penny Nom.
Filling a pool with dirt 2015-05-01
From Mike:
I have a hole which a 24 ft pool in it is 10" deep in the the centre and goes to 1" inch at the edge want to fill it in with dirt how many yards of dirt would I need to fill it in
Answered by Penny Nom.
A wireless fence 2015-04-18
From Dave:
I'm buying a wireless fence to keep my pet in my yard. It has a half acre range. In a straight line how far would that be?
Answered by Penny Nom.
Bricks around a fire pit 2015-03-05
From Jayson:
I have a round fire pit. It measures 25 inches in diameter. I have 12 inch long square bricks to go around it . My question is what degree do I cut the ends of these bricks to make them fit around this circle? The brick dimensions are 12"Lx6"Wx4"D.
Answered by Harley Weston.
Cutting a round cake so that it doesn't dry out 2014-08-26
From James:
I'm wondering if there's a simple way to calculate the area between two parallel chords of a circle equidistant from its diameter, or if I have the area, to find the distance between the two chords.
Here's my "problem". You may have heard of the way of cutting a round cake so that it doesn't dry out - make two parallel cuts (chords) the length of the cake, take the middle piece, then push the two pieces together.
So I know the area of a 12" cake, and I want say, exactly an eighth of the cake. How wide do I cut the centre piece? Now to get even more difficult, the next day I want another eighth from the centre. How wide do I cut the next pieces, and so on...? Thanks, James

Answered by Harley Weston.
An oval pool 2014-06-21
From steve:
I have a 16' x 28' oval pool that is buried 24" deep inground. The dig site is dug 2' wider all the way around the pool. I need to back fill this area with stone. I want to fill this area with 6 to 8" of stone. How many tons of stone will this take?

Thanks you

Answered by Penny Nom.
An octagonal pad 2014-04-25
From George:
Hi, I need to pour a cement pad in the shape of an octagon that allows for 12" of clearance around the tank I will be putting on it. The tank has a radius of 16'.
Answered by Penny Nom.
A sand trench around a pool 2014-04-13
From steve:
How sand is needed to back fill a trench around a 24ft dia. pool that is 26" deep by 2ft wide. Making the outside dia. 26ft.

Thank You

Answered by Penny Nom.
Cutting a hexagon from a disk 2014-04-05
From Paul:
I am a machinist and sometimes need to make a hex from round material.
If I know the distance of the flat sides opposite one another of my hex, how can I calculate the size of material I need to turn to give me the right diameter to finish the part with six sides?

Answered by Penny Nom.
The area of a 5 sided lot 2014-03-15
From Michael:
Question from michael:

This lot is in feet. 59x154x109x188x137 per the plot plan

Answered by Harley Weston.
Rolls of window film 2014-02-14
From Travis:
This question is probably close to the same question as "roll of paper"

We have Rolls of Window Film that we are trying to figure out an equation for a spreadsheet that we can use to "inventory" our window film.

We use a caliper tool to measure the thickness of the roll in millimeters.

the core thickness = 1.90mm
Full Roll thickness(including core) = 9.08mm to 9.12mm
Film thickness = 0.06

Full Roll of Film is supposed to average 1200" of film

What equation could we use to get the approximate inches left remaining on the roll if we measured the roll including the core with the Caliper tool in Millimeters?

Answered by Harley Weston.
conical lamp stand/staved wood 2013-12-07
From Henry:
need to make lamp stand that is wooden staved; need it to be 25 inches at bottom and 10 inches at top; need to know angles for staves to be cut; the lamp stand will be rounded on a lathe and will be 40 inches tall John Lucas built one and it is pictured on his web page. thank you for any help/direction; I checked out the answered for cone shaped objects on your page but didn't find what I could use. thanks again. Henry--woodturner, parent teacher student . . . . .
Answered by Harley Weston.
A square inscribed in a circle 2013-10-14
From Jenn:
Hello! I am about to buy a 7'9" round rug, but I want to have it cut down into a square. What's the largest square I can obtain from this? Thank you!
Answered by Penny Nom.
Dirt to fill a pool 2013-07-13
From Neil:
I had a 24 foot diameter pool. The perimeter of it was at ground level. The pool sloped 1 foot deeper to the middle. In other words "a 1 ft. dish. How many cubic yards of dirt do I need to fill this hole?
Answered by Penny Nom.
A gravel pile in the shape of a triangular pyramid 2013-04-04
From Casey:
Right now I am stuck and I feel embarrassed because I feel like the answer is so easy I should know it.

I am working on a project and need to find a volume of gravel it will take to occupy this triangular prism like area. I am not sure what formulas I should use whether it be that for the volume of a pyramid or something more complex? Basically it forms a right triangle at one side then from there all points slope to one singular point about 10412mm away.
I am attaching a picture drawn up in paint with the actual dimensions to clear up any confusion.

Thank you for any help. Casey

Answered by Penny Nom.
Question 2013-04-04
From Casey:
Right now I am stuck and I feel embarrassed because I feel like the answer is so easy I should know it.

I am working on a project and need to find a volume of gravel it will take to occupy this triangular prism like area. I am not sure what formulas I should use whether it be that for the volume of a pyramid or something more complex? Basically it forms a right triangle at one side then from there all points slope to one singular point about 10412mm away.
I am attaching a picture drawn up in paint with the actual dimensions to clear up any confusion.

Thank you for any help. Casey

Answered by Penny Nom.
A label to cover a plastic cup 2012-10-23
From Kevin:
I'm trying to make a label to cover the entire outer area or a plastic cup. I know there must be a way to figure out the dimensions needed, but I can't seem to figure it out. The circumference of the bottom of the cup is 21.4cm and the circumference at the top of the cup is 29.8cm. The cup is 14.5cm tall. What should the height of the arc from the plane connecting the two ends of the 21.4cm arc. I attached a diagram where x is the value I'm looking for. I'm guessing there is some simple relationship between the length of a line and the arc needed to turn that line into a perfect circle, but I don't know what it is. Can you figure this out and share it with me? Thanks.


Answered by Penny Nom.
A tank with an inner walled compartment 2012-10-12
From don:
I have a tank 20 feet diameter, 19' 8" tall with an inner walled compartment that has a 7' 6" radius arc with in the tank. I need to figure out the volume of the inner area and the volume of the larger area.
Answered by Harley Weston.
Making a wind sock 2012-08-28
From John:
I am trying to build a wind sock and need to be able to lay the shape out on cloth. I need the wind sock front opening (diameter) to be 3 1/2" and the rear opening diameter to be 1". The windsock needs to be 9 1/2" long. I tried using the example of the person trying to make a crayfish trap but got confused and could not figure out my numbers. Any help would be greatly appreciated.



Answered by Penny Nom.
Measuring the liquid in a horizontal tank 2012-02-28
From Philip:
I have a steel gas tank that is 3' dia X 5' length. The total volume is 1000 litres. But how much is left when I use a stick and measure 6" from the bottom or 12" or 24" ?? Is there a formula to use for this task?


Answered by Harley Weston.
Geometry Related Careers 2011-09-01
From Richard:
Good morning,

I am hoping to find a list of careers that would relate to geometric constructions and tessellations.

I have searched the internet some and some of the sites I have found are a bit questionable or dated. Are there sites that Math Central would recommend?

I could not find anything when I did a search on the site.

Thanks in advance,


Answered by Walter Whiteley.
Calibrating a conical tank 2011-02-05
From Bill:
Hi, I have a round tank with tapered sides where I know the diameter at the top and bottom. Is there a formula I can use to calculate the volume by measuring from the bottom up the side (at the angle of the side) to any given point? Thanks, Bill
Answered by Stephen La Rocque and Penny Nom.
liquid in a 3/4 inch pipe 2009-06-30
From junior:
We are in dilemma at my job. We need to figure out the formula for how much water can our pipe hold. It's inside diameter is 13/16" and is 50ft. long?
Answered by Robert Dawson.
Concrete around a pipe 2009-01-14
From Doug:
How much concrete will i need for a hole that is 20 feet deep 20 inches in diameter with a 8-inch pipe in it. I need to know how much concrete on the outside of the 8-inch pipe.
Answered by Penny Nom.
A barrel on its side 2008-11-13
From Dave:
Question from Dave:

How many gallons are left in a 36x60 in. barrel (laying on its side) and has 16 in. of gasoline left. I have attached a diagram.

Answered by Harley Weston.
A perpendicular intersection of two barrel vaults 2006-07-21
From Neal:
I'm wanting to build a series of architectural models of different roman and medieval buildings out of cardboard. Once I have perfected the models I want to print them out on card stock so that school kids (or anyone else) can make the buildings.
A feature of many of these models is the cross or groin vault (a perpendicular intersection of two barrel vaults).
A single barrel vault is easy to imagine as a plane (a rectangular piece of cardboard) that will be folded into a semi-circular arch.
The intersection of a second barrel vault and this one is presenting me with problems. The second plane needs to have an ellipse cut into it so that when it is folded into the arch, it will mate up with the curve of the first barrel vault.
Given that the two pieces of card have identical widths (and therefore identical arcs in cross section) is there a way to calculate the ellipse that needs to be cut so that it can be cut before the second arch is folded?

Answered by Edward Doolittle.



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