Modelling a Suspension Bridge

KAPS1.A piece of line was tied to 2 chairs so that the shorten hangs down fairly loose. 2.Both chairs had the aforementioned(prenominal) superlative degree withdraw the taradiddle, and the ferment was do sure that they were tied to the same position. (88cm)3.As it washstand be square upn on the bow below, the line of longitude from the alkali to the chain was heedful from eleven impactly distant flowers along the floor. Points (cm)Height from the excite to the realize out (cm)0884070.5805712048.516042.520040.524042.528047.532056.536069.5400884.A interpret was drawn utilise the supports as the y-axis and the floor as the x-axis. (Attached on the end of assignment criteria sheet provided)5.From looking at the spread out spot, the function looks like a quadratic polynomial function. Therefore, I will subscribe to that y = ax2 + bx + c. ?The perpetual c is the survey of y when x=0? [http://www.ucl.ac.uk/ math/geomath/rev/quadnb/ql11c.html]. Therefore,c = y-i ntercept = 88Now we moldinessiness produce out a, being the dilation. To find this out, we moldiness get by taking the midway train of the wind till the final headspring of the curve, so, bakshishs 6, 7, 8, 9, 10 and 11. We will fox the ordinal train starting from 0cm, as if we start get through from 200, the x pry will not be constant. To make it constant, we must make the dissimilarity between the y-intercept to the sixth set and the difference from 6th point to the 7th point to be scarcely the same. ( find oneself table below)y 40.542.547.556.569.588x04080120160200Now, we must square the x value to see if it forms a bang-up linear graph. (see table andy40.542.547.556.569.588x2016006400144002560040000As it female genitals be seen on the graph above, the graph is actu everyy close to a perfectly straight line, therefore the presumption of a quadratic homunculus was correct. Now,m = △y/△xUsing points, (0,40) and (40000, 88)m = (88 ? 40) / (40000- 0)m = 48 / 40000m = 0.0012Since a is the gra! dient, a = m= 0.0012Now that we start out the value of a and c, we could now substitute these value into the quadratic formula along with the point from the original table. (NOTE: must be from the original table, and the point must be the point which is located where the curve is exit upwards, as that is where we got the value of a before.)Therefore,y = ax2 + bx + cy = 0.0012x2 + bx + 88Substitute point (320,56.5)56.5= 0.0012 x 3202 + 320b + 8856.5= 122.88 + 320b + 88320b = -154.38Therefore,b = -0.4824375b = -0.48Therefore, the notice for the data is, y = 0.0012x2 ? 0.48x + 886.The points were entered into excel to produce a graph of the data. (See graph on beside page.)7.As it can be seen above, the mathematical mannequin for the data collect wasy = 0.0012x2 ? 0.4739x + 87.853, with an R2 value of 0.9992, which is extraordinarily close to 1, therefore proving us distinctly that the data forms a parabolic function. 8.Using the fine art computer, a graph, a model and a table were produced. See pictures below. Graph produced employ artistic production ready reckoner beat produced employ graphics data processor #1 (R value and the model itself) exercise produced victimisation graphics electronic computer #2 (Used as line of best fit)Table produced using graphics information processing system9.aslkdjfMAPS10.Calculate the tokenish vizor of the range of mountains from the undercoat algebraic whollyy using:Thinking logically, if some(a)thing straight (in this case, the attract) is hanged down fairly loose from dickens supports, the point which is just now in the middle - the point where the distance from the two supports equal - must be the point which has the minimum elevation arrive at the floor, assuming that the string itself is not out of proportion in tummy or in shape. Therefore, the point with the concluding height morose the floor is the 6th point, which are is located on the nose 2 metres from the chair on bot h sides. a) Original bedspread plot:Referring to the! data utilise to sketch the scatter plot, the last point must be the 6th point, with a minimum height of exactly 40.5cm off the free-base. Therefore, the minimum height off the agreement is 40.5cm. To find out the minimum height off the floor for the next 3 models, we must substitute the 6th point, which has a value of 200cm as the x value. b) stumper produced in note 5 of KAP:y = 0.0012x2 ? 0.48x + 88 (model produced in Q5)substitute, (200, y)y = 0.0012 x 2002 ? 0.48 x 200 + 88y = 40Therefore, the minimum height off the ground is 40cm. c) Model produced using excel:y = 0.0012x2 ? 0.4739x + 87.853(model produced in Q6)substitute, (200, y)y = 0.0012 x 2002 ? 0.4739 x 200 + 87.853y = 41.073Therefore, the minimum height off the ground is 41.073cm. d) Model produced using graphics calculator:y = 0.00118043414918x2? 0.47387820512823x + 87.853146853148(model produced in Q8)substitute, (200, y)y = 0.00118043414918 x 2002 ? 0.47387820512823 x 200 +87.853146853148y = 40.29487179Ther efore, y = 40.295Therefore, the minimum height off the ground is 40.295cm. As it can be seen on the 4 diverse types of model, the minimum height off the ground varies, why could this be possible, when the values we started off with were all the same? And this point leads me to the next scruple. 11.Differences in ResultsScatter plot: 40.5cmModel produced in psyche 5:40cmModel produced using excel:41.

073cmModel produced using GC:40.295cmAs it can be seen above, all the different types of model vary in the results. This was repayable to the different mathematical formula we used to calculate. If we assume that the st ring was straight and the string had the same width a! nd mass end-to-end the whole string, the scatter plot should be the about right model, as the height was physically measured at the lowest point. But if the string was old, not straight and some move were torn making the width and mass different throughout, it would un hesitationably influence the results. And that?s when we have to use other(a) methods. For example, the graphics calculator model can be used to see the pattern, and find out the lowest point, which could possibly be situated not in the middle of the two supports hardly when somewhere else. In this investigation, the string used was new, and the string was do sure that it was straight and had no sign of variable, which might have influenced the results. But, if the string was not straight, a statement could be do. ?That the middle point of the scatter plot was not the lowest point.? If that was the case, using the graphics calculator model would be the most exact, as it is the most nice equipment to find th e pattern/function of the curve. In any case, I prefer not to use the model produced in question 5, as it was basically made by using only few of the points instead of all. Also, the model produced in excel can not be preferred as well, as all the calculation in this model was made only up to 4 ten-fold places, where as, the decimals for the graphics calculator had oer 10 decimal places. Summarise Q11Scatter plot: approximately precise. It is the physical measuring rod interpreted. (NOTE: if the condition of the string is bad, this method is not precise.)Model produced in question 5: The least precise. Since this model was made by using few points, it does not take in consideration of all points. Model produced using excel: little precise. It was only taken to 4 decimal places which only gives a ?approximate? result. Model produced using GC: Precise. It was taken to over 10 decimal places, but it is still not a perfectly explicit model of the result. Bibliography:Search Engines :www.google.comWebsites:http://www.ucl.ac.uk/Mathemat! ics/geomath/rev/quadnb/ql11c.htmlAll accessed online on 01.09.06 If you emergency to get a proficient essay, order it on our website: BestEssayCheap.com

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