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Math Quiz Thread
- Thread starter steve_bnk
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Another real world practical problem.
There are three boxes with three lengths of rods. The lengths of the rods have normal distributions as follows.
Rod1 mean length 1 foot with a standard deviation of 0.08 feet.
Rod2 mean length 2 feet with a standard deviation of 0.16 feet.
Rod3 mean length 3 feet with a standard deviation of 0.32 feet.
You randomly pick rods from each box and connect them in line using sleeves.
There are three boxes with three lengths of rods. The lengths of the rods have normal distributions as follows.
Rod1 mean length 1 foot with a standard deviation of 0.08 feet.
Rod2 mean length 2 feet with a standard deviation of 0.16 feet.
Rod3 mean length 3 feet with a standard deviation of 0.32 feet.
You randomly pick rods from each box and connect them in line using sleeves.
- What is the mean length of the assembled rods?
- What is the standard deviation of the assembled rods?
- What percentage of assembled rods will be less than 6.37 feet?
beero1000
Veteran Member
For arbitrary pin shapes and tolerances, the minimal hole shape is given by the
Minkowski sum of the two shapes i.e. the set of all points whose radius vectors (from a specified origin) are the sums of radius vectors of some pair of points from the two given shapes. This is the same mathematics behind finding the cut shape in computerized machining - the Minkowski sum of the tool head along with the tool trajectory gives the material removed. You can also detect collision in computer design or modeling, or even robot motion (if you're in configuration space) - if the Minkowski difference contains the origin, the two shapes collide...
For arbitrary pin shapes and tolerances, the minimal hole shape is given by the
My guess is you found that after anumber of searches. Same response as to Lauren. Reduce tit o anactual numerical solution.
beero1000
Veteran Member
For arbitrary pin shapes and tolerances, the minimal hole shape is given by the
My guess is you found that after anumber of searches. Same response as to Lauren. Reduce tit o anactual numerical solution.
Actually, no. To be honest, I've been able to solve all of your problems immediately and just prefer to give others a chance on such simple questions.
Minkowski sums are a well-known, hundred-year-old piece of very useful mathematics that would be immediately familiar to any working mathematician, especially a geometer like myself. As for numerical solutions, there are multiple implementations in various software packages to efficiently compute sums and differences of arbitrary geometric shapes in any dimension.
lpetrich
Contributor
Presumably one from each box. The sleeves make them touch each other.
My solution:
You did not solve the problem, you linked to an a relatively obscure piece of math that may have applied. Years back knowing the solution was readily available I worked through the solution. I found doing so in general where practical and time permitting improves understanding and problem solving skills, which is the point of the thread. Knowing aboutsomething is not the same as knowing something and being able to use it.
What do you do when you have no solution available?
I gave you a chance to recover. Thesolution I gave is widely used. I have never seen the topic youlinked to. However if you actually understood what you posted the solution I gave is actually such a sum which you apparently did not see. The theoretical position of each pin is the center of acoordinate system with the tolerance box diagonal and radii of the circles at each corner vectors.
NC machines do not calculate tolerances, tolerances are predetermined. My 3D CAD tool Alibre Design will check for interference between parts in an assembly, but does not tell what tolerances should be.
Back in the 80s I did some work on an NC diamond turning machine. The company made aluminum reflective optics. You will be hard pressed to convince me you really know about this stuff.
Any experienced engineer would see ageneral solution for arbitrary shapes as sums of vectors, but that is not what I asked. If you just want to post links you can start a thread for that
On your Zeno's Paradox based problemyou were unable or unwilling to show how you derived what you claimedto be an exact number of steps.
On your problem involving points o ahalf sine curve you claimed the series of points was actually a Riemann Sum. The Riemann Sum, or integral, of a half sine. is 2 and that makes your solution bogus. You were unable or unwilling to justify your leap. And ignored my question. I just dropped it, no matter to me.
You used a calculator to try to prove 3 * .333... = 1 proving infinities could be used in arithmetic calculations. I showed you how what you were seeing was due to finite arithmetic and rounding, no response. You also claimed that for some reason RPN provided the results you wanted.
My point of the above being my conclusion is you do not know much math at all. You can respond to my posts as you please, I will not be responding unless you begin to demonstrate substance.
Last edited:
beero1000
Veteran Member
I'm devastated.
While I'm sure if you dredge through enough of my posts you'll find errors that I've made, I am confident that none of your examples are instances of that. In fact, I don't think the last one was even me. In any case, some solutions are really that obvious to me, and I avoid writing out every last detail in many posts, but if someone asks me to expand on a statement, I usually will. It is not my fault if you cannot follow, nor do I care if that makes you conclude that I don't know much math. In particular, I do not need your certification of my knowledge - my PhD and math professorship do that just fine.
My posting history will show that if someone is genuinely interested in learning something, I am the first one to take the time to try and help them. In fact, that was what I was trying to do with the above post. Instead of an ad hoc solution based on an instance with the simplest possible input shapes, I posted a general method of solution for arbitrary shapes in any dimension. I hoped that some people would follow the wiki link and learn something interesting. I did not write out the specific solution to your trivial problem because there was no need to, the specific solution is painfully apparent.
I usually try not to waste my time with trivialities, nor with arguing with obstinate people on the internet. I have found that your posts often fit both categories, so as a general rule I have tried to avoid engaging with you. I will continue to do so.
This might interest you.
You can verify numerically. Scilab script, you can convert to your math tool.
It is not an idle problem. In high volume manufacturing it can be cheaper to have looser tolerances andthrow away parts that do not fit.
It leads into Monte Carlo Simulation.When the system gets large and the variables can have different distributions it is easier to create randomized vectors for each variable and then sequentially combine each vector entry per the process. The system could be electrical, mechanical, or a human process like queuing.
clear;
n = 1000;
rod1_mean = 1;
rod2_mean = 2;
rod3_mean = 3;
rod1_sd = .08;
rod2_sd = .16;
rod3_sd = .32;
sd_rss = sqrt(rod1_sd^2 + rod2_sd^2 +rod3_sd^2);
// create randomized vectors of normal distributions.
rod1 =grand(1,n,"nor",rod1_mean,rod1_sd);
rod2 =grand(1,n,"nor",rod2_mean,rod2_sd);
rod3 =grand(1,n,"nor",rod3_mean,rod3_sd);
// sequentially add each vector creating a combined distribution
for i = 1:n
rod_sum(i) = rod1(i) + rod2(i) +rod3(i);
end
m = mean(rod_sum); // mean of total lengths
sd_sum = stdev(rod_sum); // standard deviation of combined lengths
figure(1);
histplot(50,rod1);
figure(2);
histplot(50,rod2); //histogram
figure(3);
histplot(50,rod3);
figure(4);
histplot(50,rod_sum);
m
sd_sum
-->m
m =
6.0076993
-->sd_sum
sd_sum =
0.3678730
y(t) = [1 + .4 * cos(2*pi*1000*t) ] *100 * sin(2**pi * 500000*t)
where 1000 and 500000 are frequenciesin Hertz.
What are the resultant frequencies and their magnitudes?
The number of frequencies is > 2 and there are two approaches.
where 1000 and 500000 are frequenciesin Hertz.
What are the resultant frequencies and their magnitudes?
The number of frequencies is > 2 and there are two approaches.
lpetrich
Contributor
= (1/2)*(sin(x+y) + sin(x-y))
Google it. I do not keep the continuous time FT on the top of my head. Beyond simple text bookproblems it is not very useful.
The FT is commonly applied numerically using the Discrete Fourier Transform and the Fast Fourier Transform.You can find plenty of discussion online.
My favorite.
Brigham, E. Oran,
The Fast Fourier Transform , Prentice-
Hall, 19
One of most important mathematicaldevelopments in the rise of modern technology
Cooley, J.W. and Tukey, J.W.,
“An Algorithm for the Machine
Calculation of Complex Fourier
Series”, Mathematics of Computation,
Vo. 19, No. 90, p. 297,
April 1965.
For shits and giggles Fourier analysisof the modulation equation. You can find code examples of the FFTonline and substitute for fft() if you want to experiment/
// Amplitude Modulation
clear;
clf;
M = .5;
A = 2;
fc =50000; //carrier frequency
fm = 2000; // modulation frequency
fs = 500000; // sample rate in Hertz
ts = 1/fs; // sample time interval inseconds
record_length = 2^16; // 64k transform
time_index = 0;
for i = 1:record_length;
t(i) = time_index;
time_index = time_index + ts;
end;
for i = 1:record_length;
s(i) = [1 + M * cos(2*%pi*fm*t(i)) ]* A * sin(2*%pi * fc*t(i));
end;
y = fft(s);
f =fs*(record_length/2))/record_length; //associated frequencyvector
n = size(f,'*')
clf()
figure(1);
plot(f,abs(y(1:n)))
figure(2);
plot(s(1:1000));
beero1000
Veteran Member
The Fourier Transform via evaluating the integral yields a trigonometric Fourier Series not a time domain function. The inverse transform, or La Place transform gets you back to the timedomain.
What you seem to have done is convert to phasors, rearrange and go back to trigonometric form. I do not see where you actually performed the transform to frequency domain. When you take the transform the variable is jw, there is no t or time domain element.
Given a rectangular wave f(t) what are the frequency components and phases via the integral?
t 0 → 1 seconds f(t) = 0
t 1 → 2 seconds f(t) = 1
t 2 → 3 seconds f(t) = 0
t 4 → 5 seconds f(t) = 1 …...
lpetrich
Contributor
I've decided to revisit the SAS Postulate again, something I'd mentioned in an earlier post. The problem is to construct a triangle in an arbitrary space or manifold and then find out how much it departs from a flat triangle. I've tried doing that generally, but it's a big headache. So I've considered how to construct a 2D subspace that contains a triangle. I now have a solution.
Take triangle ABC. Construct a geodesic through B and C, with s2 as the parameter that controls point locations on it. Construct a geodesic through A and some point D on BC, with s1 as the parameter that controls point locations on it. A point X on geodesic AD thus has its position controlled by two parameters, s1 and s2. All points X thus form a subspace of the original space. Since each pair of vertices is connected by a geodesic, that means that the tangent directions at each vertex is also in the subspace, along with the geodesics.
Thus, one can solve this general-space problem by solving a related 2D-space problem. From the Gauss-Bonnet theorem,
(sum of polygon's angles) = 180d*((number of polygon sides) - 2) + (curvature integrated over the polygon's area)
That integral becomes, in general,
\( \int R_{ijkl}e^{ij}e^{kl}dA \)
where the e is the surface's "tangent bivector", composed from two tangent vectors:
\( e^{ij} = e_1^i e_2^j - e_2^i e_1^j ,\ (e_1.e_1) = (e_2.e_2) = 1, \ (e_1.e_2) = 0 \)
and R is the Riemann curvature tensor.
As I'd posted earlier, the SAS postulate means that Rijkleikekl is constant, independent of location and the orientation of e. That gives us covariantly-constant curvature: Rijkl = K*(gikgjl-gilgjk) where K is a constant.
Take triangle ABC. Construct a geodesic through B and C, with s2 as the parameter that controls point locations on it. Construct a geodesic through A and some point D on BC, with s1 as the parameter that controls point locations on it. A point X on geodesic AD thus has its position controlled by two parameters, s1 and s2. All points X thus form a subspace of the original space. Since each pair of vertices is connected by a geodesic, that means that the tangent directions at each vertex is also in the subspace, along with the geodesics.
Thus, one can solve this general-space problem by solving a related 2D-space problem. From the Gauss-Bonnet theorem,
(sum of polygon's angles) = 180d*((number of polygon sides) - 2) + (curvature integrated over the polygon's area)
That integral becomes, in general,
\( \int R_{ijkl}e^{ij}e^{kl}dA \)
where the e is the surface's "tangent bivector", composed from two tangent vectors:
\( e^{ij} = e_1^i e_2^j - e_2^i e_1^j ,\ (e_1.e_1) = (e_2.e_2) = 1, \ (e_1.e_2) = 0 \)
and R is the Riemann curvature tensor.
As I'd posted earlier, the SAS postulate means that Rijkleikekl is constant, independent of location and the orientation of e. That gives us covariantly-constant curvature: Rijkl = K*(gikgjl-gilgjk) where K is a constant.
You mean a planar Euclidian triangle or any triangle and by geodesic you mean a the shortest distance between two points?
lpetrich
Contributor
A triangle with geodesics as its sides.