Wikipedia:Reference desk/Archives/Mathematics/2021 July 22

From Wikipedia, the free encyclopedia
Mathematics desk
< July 21 << Jun | July | Aug >> Current desk >
Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.


July 22[edit]

Center of area in Venn diagram[edit]

I have a venn diagram. Each of the circles is proportional to the size of the population the circle represents. I call them circles 1, 2, and 3. So, I have the center points: x1, y1, x2, y2, x3, and y3. I have three radii: r1, r2, and r3. I want to calculate the center of the area for the parts where only two of the circles overlap. As an example, see Venn diagram. Using the center points and radii, what is the center of the trianglish shape with Π in the middle of it? What about the one with C in the middle of it? I've worked out how to calculate the intersection points of the circles. Each of those shapes has three points, but if I treat it as a triangle, I don't get to the center because those shapes can be very curved if you make one circle large and another small. What I am doing right now (and I haven't completely figured it out) is trying to calculate the center point of the three arcs and then find the center between them. But, I'm not sure that will work when the slice is curved. I will likely get a point outside the slice. 97.82.165.112 (talk) 17:51, 22 July 2021 (UTC)[reply]

There is no standard notion of the centre of an arbitrarily shaped region. For various notions, see Centre (geometry). They agree if the region has central symmetry, which the regions you are considering have not. If you can clarify why you are seeking these centres, it may help to determine an appropriate notion for this specific case. The centroid and the Chebyshev centre are both reasonable candidates. The average of the midpoints of the three bounding arcs, which is relatively easy to compute, should always be internal if the three circle interiors have a non-empty intersection.  --Lambiam 20:32, 22 July 2021 (UTC)[reply]
Why? This is a Venn diagram. I want to put a label in each region. If every circle was the exact same size, it is easy to hard-code the center of each region. But, I'm trying to do it with circles that are sized based on the population of that group and overlap the circles relative to how much overlap there actually is. So, I'm not getting regions with pretty convex sides. I often get regions with concave sides. If I take the three points of the curved triangle, I get a point outside the region. My goal is to get a point inside the region so I can place the label approximately where it belongs. 97.82.165.112 (talk) 11:13, 23 July 2021 (UTC)[reply]
If all you want is a point inside the region, then taking the centroid of the midpoints of the three bounding arcs should work for you. Note though that there is no a priori guarantee that the region is large enough to accommodate any label with readable text. Probably, the Chebyshev centre is then visually better in extreme cases (e.g. when the central region is very small), but more work to determine. (See Problem of Apollonius, for which the algebraic road is probably the easiest.) If the label is oblong, it may also pay off to scale so that the label becomes equally wide and long (and the Venn circles ellipses) before determining the Chebyshev centre and then rescale – but then the calculations become complicated. --Lambiam 12:15, 23 July 2021 (UTC)[reply]
Thanks. Using the three corners of each slice, I calculated a line to the midpoint of the opposite arc. Then, I calculated the average intersection point (being curved, the three lines form a little triangle, not a point like a triangle with straight sides). That point is almost always inside the curved triangle. It fails if the triangle is very skinny and very curved, but the "center" point is close enough. I've now turned the three circles into bezier "circles" so I can make them more egg shaped. So, I can make a Venn diagram where the area inside each region is an accurate representation of the size of the population inside that region. The hardest problem is ensuring it is exactly accurate so when a region has a population of zero, the lines touch perfectly and I don't get overlap or a little gap. On another note: calculating the size of the area contained inside two overlapping egg shapes is very complicated. I found that it is best to rotate the overlap so both intersection points have a Y axis value of zero. Then, I calculate the area under the positive curve, which is one egg border, and the negative curve, which is the other egg border. That jumps from geometry and algebra, which I was using, to calculus, but it works. 97.82.165.112 (talk) 17:10, 27 July 2021 (UTC)[reply]

Conversion factor for torque[edit]

Hi all, I'm trying to work out the torque in Kilopondmetres (abbreviated in my sources as mkg), of various old Maybach engines with power given in metric horsepower (PS). I realise that max. torque is often reached at lower revs than max. power.
If Torque (lb·ft) = 5252 x Power (hp) / Speed (rpm), what would be the equivalent formula using mkg and PS? An approximation calculation taking 5252 lb·ft = 726.12 mkg seems to work, but 1 hp = 1.04 PS, and that's where my maths reaches its upper bound. I've stared blankly at Torque#Derivation and Horsepower#Definitions but there's a mental block in the way. If anyone could also explain in very basic terms how the conversion factor is arrived at in the way that 5252 ≈ 33,000/2π, I would be most grateful. Cheers, >MinorProphet (talk) 21:22, 22 July 2021 (UTC)[reply]

This is not really a maths question. Does this video answer your question? If not, I suggest you ask the question at the Science section of the Reference desk.  --Lambiam 00:09, 23 July 2021 (UTC)[reply]
1 kpm = 7.233 ft-lbf (commonly calls ft-lb) per GNU Units. Basically 1 kpm = 1*kg*g*1 meter = 2.2026 lbf*1 meter = 2.2026 lbf * 3.2808 feet = 2.2026 * 3.2808 ft*lbf = 7.2262 ft*lbf which is about the same as the GNU Units number except for some likely rounding errors. 2601:648:8202:350:0:0:0:2B99 (talk) 00:48, 23 July 2021 (UTC)[reply]
Thanks for your replies, but I'm trying to avoid conversions between different systems, and to use a specific constant, thus:
  • Torque (lb.ft) = 5,252 * power (hp) / speed (rpm)
  • Torque (N.m) = 9.5488 * power (kW) / speed (rpm)
  • Torque (kpm) = X * power (PS) / speed (rpm) - what is X? Thanks, MinorProphet (talk) 01:34, 23 July 2021 (UTC)[reply]
As Lambiam says, this should go to the Science desk. I'll repost my question there, thanks for your replies. Cheers, MinorProphet (talk) 01:57, 23 July 2021 (UTC)[reply]