View allAll Photos Tagged Mathematics
We have to live with the idea that we can rely on our intelligence and our senses (otherwise normal living wouldn't be possible). Our intelligence says that 2x3 is the same as 3x2. But if we see with our senses that 2x3 can be different from 3x2 (two different underlying structures) then we can get confused. Is there more than we can see or reason?
CSX W001 approaches Gettysburg on it's way from Baltimore to Hagerstown on a track inspection assignment.
Relatively prime. The 13th of 25 mathematic Lego mini mosaics (20 inches square). When completed the entire montage will stretch over 42 feet.
Randomness II: The 6th of 25 mathematic Lego mini mosaics (20 inches square). When completed the entire montage will stretch over 42 feet.
Law of cosines. The 5th of 25 mathematic Lego mini mosaics (20 inches square). When completed the entire montage will stretch over 42 feet. (THIS IS A 3-D AND DIFFICULT TO PHOTOGRAPH FROM ABOVE)
Ponendo x= π, con calcoli banali si ottiene, dalla formula di Eulero:
exp(ix)=cosx + i senx
la formula che compare nella immagine.
E' una formula che mi ha sempre affascinato, porchè contiene i 5 numeri con cui possiamo riassumere la matematica:
O e 1, con cui costruire l'aritmetica dei naturali;
π, il rapporto tra la lunghezza di una circonferenza e il suo diametro;
i, l'unità immaginaria, radice quadrata dell'unità negativa;
e, il numero di Nepero...
Sustituyendo x = π, con cálculos triviales, se obtiene, de la fórmula de Euler:
exp (ix) = cos x + i senx
la fórmula que aparece en la imagen.
Es una fórmula que siempre me encantò, ya que contiene 5 números con los que podemos resumir las matemáticas:
O y 1, con los que construir la aritmética de los numeros naturales
π, la relación entre la longitud de una circunferencia y su diámetro;
i, la unidad imaginaria, la raíz cuadrada de la unidad negativa;
e, el número de Napier
Substituting x = π, with trivial calculations we obtain, from Euler's formula:
exp (ix) = cosx + i senx
the formula which appears in the image.
It 'a formula that has always fascinated me, because it contains 5 numbers with which we can summarize the math:
O and 1, with which to build the arithmetic of natural numbers;
π, the ratio between the length of a circumference and its diameter;
i, the imaginary unit, square root of negative unity;
e, the number of Napier
I know this photo has been done a million times, but I wanted to give it a try anyways. I alone have seen several variations on DPC. I got the inspiration for this photo from photographer Konador on DPC. I believe he received his inspiration from an artist by the name of Ilona Wellmann.
I swear "Advanced Mathematics" is exactly what it is. Haven't the slightest idea why the shot turned out like that!
*cough*
And now, a message from the model:
Hello! Online registration is open. Only accepting 20 students max, first-come-first-serve basis.
You know the drill, lovelies.
(I know the class will be full of beautiful, bright minds.)
As usual, lessons will be via-login, livestream and/or podcast.
Examinations are still every end of the week, unfortunately. Post-qualifications at end of term for PhDs!
Please submit all application forms and 1st term payments to the Applied Mathematics Department. You may also send them through email and Paypal. No "personal meet-ups" or "requests for consultation." I've grown weary of repeating myself, ladies. A no is a no. :)
Filming the Lectures and tagging them as "eyeporn" or other similar terms is strictly prohibited.
And PLEASE, no hacking the Cardiff Uni website for extra slots. Be mindful, this is for Advanced Maths NOT Computer Science.
I am a Mathematician, not the head of the Cyber Crime Department. Hence, I've a lack of appreciation for such skills.
Also, remember:
Flowers/Chocolates/private voice calls on Skype or intimate items strewn across your keyboard during livestream recitation will not give you an A.
Only your knowledge will. :)
Thank you and see you online this semester.
- Prof. Chlopher P. Sierren
Galleria Continua San Gimignano
Human Mathematics, ipaekre - roib 1982-2015, installation, mixed media
This was towards the end of a pure mathematics lecture in the Centre for Mathematical Sciences, University of Cambridge
Too many variables?
I handheld the 3.0kg Nikon 500mm f/4.0P lens, to take this shot.
Imagine trying to find a target, with a 4.94° Diagonal Field of View, get the focus and exposure right, just in the very small amount of time the target has been acquired. Challenging, to say the least.
I originally had the lens mounted on a tripod, but just could not change it fast enough along a horizontal and vertical axis.
So, I do not know the angular inclination of the lens above the horizon.
If the aircraft was straight overhead, it might be simple mathematics, knowing the physical dimensions of the aircraft, comparing the number of pixels the wing or fuselage length occupy to the dimensions of the Sensor.
The Wings are nearly parallel to the Hypotenuse formed by the diagonal line on the sensor. and in reality that rectangular area would be perpendicular to the axis from the Circle of Confusion on the Sensor to the middle underside of the aircraft's fuselage. I depicted it in this way merely to give you some idea what I have been up to for the past couple days or so.
Because the aircraft is flying away at some angle and the wings is not precisely aligned with the Hypotenuse, actual pixel count seems more of a guess.
For example, knowing the wing from tip-to-tip is 36 feet (10.9728 meters) and the fuselage length is 28.25 feet (8.6106 meters) . . . I have measured between 987.5 pixels for the wing span . . . When comparing that to the length of the fuselage, the pixel count seems incorrect . . . Or, doing a pixel count of the fuselage, taking a ratio and applying that to the wing, the pixel count will be off. Frustrating.
There must be an accurate and predictable method, but I have not played with mathematics on this level in a very, very long time. Believe it or not, I designed a rocket, at university, but that was more than four decades into my past. I was smarter, then, or had many more active brain cells working for me.
What this demonstrates is that photography is more than just pretty pictures.
Possible solution:
Right Triangle ⊿, a Base, b Height, c Hypotenuse
tan = b/a
tan(2.47°) = 0.043136357952622
b = 3,634.45
a X tan(2.47°) = b
isolate a
a = 84254.9109962412 pixels
(Note: this will be from the Focal Plane to the belly of the aeroplane, between the main landing gear))
sin = b/c
sin(2.47°) = 0.043096280984403
isolate c
c X sin(2.47°) = b
c = 84333.2630329656
Taking a piece of paper, hold it parallel to the Span of the Wing, that is taken from a point in the middle of each wingtip and one gets 987.5 pixels. We know the Wing Span of this NACA 2412 type of wing on a Cessna U206G is 35 feet 10 inches (+/- 2 inches, depending on references) and is 987.5 pixels. The Wing Tip uses a NACA 0012 type of aerofoil and measures 3 feet, 8.5 inches.
The Wing Span is a known number.
So, I would want to know the number of pixels/foot of wing span.
If 987.5 pixels ÷ 35.83333 feet, then I would have 27.5581395348837 pixels/foot
Taking excerpts from the above:
a = 84254.9109962412 pixels ÷ 27.5581395348837 pixels/foot = 3,057.3512007072 feet from the Focal Plane to the belly of the Cessna, between the Main Landing Gear.
This is plausible. That still does not give me the height above the ground.
This may not be correct, though.
Why?
In the photo of the aeroplane, the Fuselage length appears longer than the Wing Span, but we know this is not true, as the Fuselage Length is 28.25 feet, as compared to the Wing Span of 35.83333 feet.
Measuring the Fuselage Length, I arrive at 865 pixels.
865 pixels ÷ 28.25 feet = 30.6194690265487 pixels/foot
35.833333 ÷ 28.25 = 1.26843657699115
How does this ratio compare with the pixels/foot count?
987.5 ÷ 865 = 1.14161849710983
Do you see the difference and the dilemma I have???
This is probably due to a parallax. How do I resolve that???
84254.9109962412 ÷ 30.6194690265487 = 2751.67772906799
3,057.3512007072 - 2751.67772906799 = 305.673471639207
Big difference! Which is correct? Is there a mathematical solution to know for certain?
How about this addition to the confusion? Measuring the Right Wingtip it is found to be 95 pixels and 3.70833333333333 feet or 25.6179775280899 pixels per foot.
Let's average the three pixels/foot counts. That would equal 27.9318620298408 pixels per foot
So, my best guess for the distance from the Focal Plane to the aircraft belly would be:
84254.9109962412 ÷ 27.9318620298408 = 3016.44447857533 feet
This is more plausible, though still does not give me the height above the ground of the aircraft.
3016.44447857533 feet becomes the new hypotenuse (c) to determine Height of the aircraft above the ground.
It would not be too far fetched to assume 3,000 feet above ground. And, the angle I held the big 500mm lens at could have been as much as 84° above the horizon. Maybe. All a guess, really.
I would guess 2,500 feet, considering the surrounding hilltops are approaching 1,500 feet or 457 meters.
This required two cups of strong morning coffee.
If anyone has a better solution, other than asking the pilot, I would like to learn from you.
A gentle reminder about copyright and intellectual property-
Ⓒ Cassidy Photography (All images in this Flickr portfolio)
The University of Manchester's Mathematics Building on Oxford Road in 1969. Designed by Scherrer & Hicks and built 1967-68.
Wooden footbridge across the River Cam, between two parts of Queens' College, Cambridge. A popular myth is that the bridge was designed and built by Sir Isaac Newton without the use of nuts or bolts.
Agfa 100 (50), D-76 (1+1), 13', 20°C. Expired from 2003, developed at 2015, always preserved in freezer.
Leica M6, Summicron 50mm (5th gen).
My Palette. The 8th of 25 mathematic Lego mini mosaics (20 inches square). When completed the entire montage will stretch over 42 feet. Not so mathematical. Organizational
Quadrature: The 9th of 25 mathematic Lego mini mosaics (20 inches square). When completed the entire montage will stretch over 42 feet.
I wedged myself between two walls on top of a sewer grate to find this little spot. In general the building (mathematics and computer building at University of Waterloo) is very square, but it has some lines to exploit.
This is currently my favourite shot around campus.
Entry for "Chuck Norris vs." contest, resulting in winning Category 1 ^_^
Thanks a lot to the judges! :^D
This picture came out a bit too dark.. I should have better light when working.
I think the paper is from a Barbie office table set?
Charles Solomon: Mathematics
Hamlyn all-colour paperbacks
Paul Hamlyn, The Hamlyn Publishing Group Ltd - London, 1969
by Samuel Musungayi.
Captured with a Konica Big Mini HG BM-300 and an expired roll of Agfachrome CTx 100 from May 1998.
CanoScan 8800F.
Explore Aug 1, 2012 #170
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“Happiness is when what you think, what you say, and what you do are in harmony.” by Mahatma Gandhi
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“Always aim at complete harmony of thought and word and deed. Always aim at purifying your thoughts and everything will be well.” by Mahatma Gandhi
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“In the end we shall have had enough of cynicism, skepticism and humbug, and we shall want to live more musically.” by Vincent van Gogh
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“Mathematics expresses values that reflect the cosmos, including orderliness, balance, harmony, logic, and abstract beauty.” by Deepak Chopra