How would I integrate different values into 4 tables?

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Michael Sabol
Michael Sabol 2020-4-20
回答： BALAJI KARTHEEK 2020-4-21
I have this code
clc
clear
close
a = 40*(pi/180);
v = 1200;
g = 32.2;
k=[0,2e-6,10e-6,20e-6];
k=0;
dt=.5;
t = 0:dt:55;
v = zeros(length(t),4);
vx = zeros(length(t),4);
vy= zeros(length(t),4);
x = zeros(length(t),4);
y = zeros(length(t),4);
v(1,1) = 1200;
vx(1,1) = v(1,1)*cos(a);
vy(1,1) = v(1,1)*sin(a);
x(1,1)=0;
y(1,1)=0;
for i=2:length (t)
x(i,1) = x(i-1,1)+vx(i-1,1)*dt-.5*k*v(i-1,1).^2*cos(a)*dt.^2;
y(i,1) = y(i-1,1)+vy(i-1,1)*dt-.5*k*v(i-1,1).^2*sin(a)*dt^2-0.5*g*dt.^2;
vx(i,1) = vx(i-1,1)-k*v(i-1,1).^2*cos(a)*dt;
vy(i,1) = vy(i-1,1)-k*v(i-1,1).^2*sin(a)*dt-g*dt;
a=atan(vy(i,1)./vx(i,1));
v(i,1)=sqrt(vx(i,1).^2+vy(i,1).^2);
end
vx(1:round(5/dt):end,:)
vy(1:round(5/dt):end,:)
x(1:round(5/dt):end,:)
y(1:round(5/dt):end,:)
That results in
ans =
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
ans =
771.35 0 0 0
610.35 0 0 0
449.35 0 0 0
288.35 0 0 0
127.35 0 0 0
-33.65 0 0 0
-194.65 0 0 0
-355.65 0 0 0
-516.65 0 0 0
-677.65 0 0 0
-838.65 0 0 0
-999.65 0 0 0
ans =
0 0 0 0
4596.27 0 0 0
9192.53 0 0 0
13788.80 0 0 0
18385.07 0 0 0
22981.33 0 0 0
27577.60 0 0 0
32173.87 0 0 0
36770.13 0 0 0
41366.40 0 0 0
45962.67 0 0 0
50558.93 0 0 0
ans =
0 0 0 0
3454.23 0 0 0
6103.45 0 0 0
7947.68 0 0 0
8986.90 0 0 0
9221.13 0 0 0
8650.35 0 0 0
7274.58 0 0 0
5093.81 0 0 0
2108.03 0 0 0
-1682.74 0 0 0
-6278.52 0 0 0
My issues is whenever I change the K value to [2e-6,10e-6, or 20e-6] it results in the first column changing to the respective k values. However, if I change the x(i,1) positions to say x(i,2) for [2e-6] (and all subsequent notations) then I get every k = 2e-6 value in the 2nd column and the first column becomes zero. Question is how do I get the answers to add on to the table made from previous calculations?
  2 个评论
BALAJI KARTHEEK
BALAJI KARTHEEK 2020-4-20
Just give sample image of the table (what ur expecting in the result), so it will easy to understand and modify the code written by u..
Michael Sabol
Michael Sabol 2020-4-20
The answer to the above code is listed. I would like to have the three other columns filled with the coressponding k values (2e-6, 10e-6, and 20e-6)

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采纳的回答

BALAJI KARTHEEK
BALAJI KARTHEEK 2020-4-21
I modified the code for you and i posted the ans also, i hope this is what u wanted.....
clc
clear
close
format long
k1=[0,2e-6,10e-6,20e-6];
dt=.5;
t = 0:dt:55;
g = 32.2;
v = zeros(length(t),4);
vx = zeros(length(t),4);
vy= zeros(length(t),4);
x = zeros(length(t),4);
y = zeros(length(t),4);
for j=1:length(k1)
a = 40*(pi/180);
k=k1(j);
v(1,j) = 1200;
vx(1,j) = v(1,j)*cos(a);
vy(1,j) = v(1,j)*sin(a);
x(1,j)=0;
y(1,j)=0;
for i=2:length (t)
x(i,j) = x(i-1,j)+vx(i-1,j)*dt-.5*k*v(i-1,j).^2*cos(a)*dt.^2;
y(i,j) = y(i-1,j)+vy(i-1,j)*dt-.5*k*v(i-1,j).^2*sin(a)*dt^2-0.5*g*dt.^2;
vx(i,j) = vx(i-1,j)-k*v(i-1,j).^2*cos(a)*dt;
vy(i,j) = vy(i-1,j)-k*v(i-1,j).^2*sin(a)*dt-g*dt;
a=atan(vy(i,j)./vx(i,j));
v(i,j)=sqrt(vx(i,j).^2+vy(i,j).^2);
end
end
vx(1:round(5/dt):end,:)
vy(1:round(5/dt):end,:)
x(1:round(5/dt):end,:)
y(1:round(5/dt):end,:)
ans =
1.0e+02 *
9.192533317427737 9.192533317427737 9.192533317427737 9.192533317427737
9.192533317427737 9.087437198565105 8.688194539448844 8.232369634405721
9.192533317427737 8.992683198820997 8.271606474129895 7.515600132134569
9.192533317427737 8.906142118918728 7.919402869533692 6.956903130394356
9.192533317427737 8.825438093191325 7.611669801834986 6.497663684523137
9.192533317427737 8.747994917296394 7.330900766229695 6.094617348287314
9.192533317427737 8.671191392627970 7.062225373628980 5.716607364427984
9.192533317427737 8.592588049635696 6.794295817589458 5.344401073260919
9.192533317427737 8.510130827164266 6.519858929429604 4.969536950219919
9.192533317427737 8.422255391440707 6.235515863016696 4.591412572439150
9.192533317427737 8.327886941553478 5.940899030969548 4.213955536697458
9.192533317427737 8.226377828143628 5.637702715980655 3.842942121374995
ans =
1.0e+02 *
7.713451316238471 7.713451316238471 7.713451316238471 7.713451316238471
6.103451316238469 6.023419388685378 5.719372277363858 5.372185731794696
4.493451316238467 4.358060751416640 3.869349418126093 3.356497018654969
2.883451316238464 2.713013480784294 2.125068241342515 1.550149352309664
1.273451316238464 1.084948020280437 0.460462096125176 -0.114597066820229
-0.336548683761536 -0.528223533354256 -1.139760914385037 -1.672277710014929
-1.946548683761535 -2.127194956892446 -2.681175666264873 -3.132854106675353
-3.556548683761536 -3.711278137353608 -4.161511360325129 -4.490534191556887
-5.166548683761538 -5.278613302973931 -5.573538912210908 -5.733287819377662
-6.776548683761540 -6.826502519553036 -6.908138272916148 -6.850276415575496
-8.386548683761543 -8.351753997529265 -8.156695586961243 -7.835785114702563
-9.996548683761544 -9.850966102443524 -9.312567227724861 -8.690247541356209
ans =
1.0e+04 *
0 0 0 0
0.459626665871387 0.456952509660115 0.446601570599481 0.434388168668790
0.919253331742773 0.908917067719421 0.870286135308686 0.827286469285742
1.378879997614160 1.356358536870645 1.274838854896066 1.188582146706933
1.838506663485547 1.799629168051811 1.662970072505337 1.524634799788792
2.298133329356934 2.238956995531161 2.036456085149122 1.839283305588884
2.757759995228320 2.674439272489372 2.396260757987131 2.134509243829672
3.217386661099707 3.106045768985768 2.742688806528300 2.411035206651566
3.677013326971094 3.533633248951142 3.075579187877789 2.668900501599922
4.136639992842481 3.956967824055119 3.394507100925129 2.907931858633309
4.596266658713867 4.375749794739506 3.698957566875462 3.128051985769669
5.055893324585254 4.789636722968088 3.988452648518830 3.329435350798285
ans =
1.0e+04 *
0 0 0 0
0.345422565811923 0.343309821893089 0.335128870726984 0.325469355578336
0.610345131623847 0.602754750055932 0.574341837338965 0.542628698537214
0.794767697435771 0.779455549117251 0.723828388470866 0.664600119087606
0.898690263247694 0.874339858815437 0.788177334077686 0.699997622986476
0.922112829059617 0.888199047586014 0.770947619004973 0.654920194605361
0.865035394871541 0.821754585219541 0.675181896918930 0.534387622251712
0.727457960683465 0.675728278727461 0.503850396003435 0.343355037091190
0.509380526495388 0.450906840766734 0.260173936315252 0.087260126553497
0.210803092307311 0.148192401693298 -0.052206230259897 -0.227864175692256
-0.168274341880766 -0.231364338893490 -0.429197553309034 -0.595561742932445
-0.627851776068843 -0.686546835784460 -0.866321819027821 -1.009244632798958

更多回答（1 个）

darova
darova 2020-4-20
Try this solution
  2 个评论
Michael Sabol
Michael Sabol 2020-4-20
Hello thanks for the answer, one question. is the k=K1(j); inside the for loop you gave or my original one?
like so?
for j=1:length (k)
for i=2:length (t)
k=K1(j);
x(i,j) = x(i-1,j)+vx(i-1,j)*dt-.5*k*v(i-1,j).^2*cos(a)*dt.^2;
y(i,j) = y(i-1,j)+vy(i-1,j)*dt-.5*k*v(i-1,j).^2*sin(a)*dt^2-0.5*g*dt.^2;
vx(i,j) = vx(i-1,j)-k*v(i-1,j).^2*cos(a)*dt;
vy(i,j) = vy(i-1,j)-k*v(i-1,j).^2*sin(a)*dt-g*dt;
end
end
darova
darova 2020-4-20
It doesn't matter. More reasonable be inside first
for j=1:length (k)
k=K1(j);
for i=2:length (t)
% code

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