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75 485 N = 109 lbf 100 Figure 744 shows the performance of bearings when the diametrical clearance, 2ho = 0001 in (25 m) The airflow is proportional to the cube of the bearing clearance so that a reduction in clearance provides a significant reduction in the demand for the compressor power as well as an increase in the radial stiffness The value of the airflow is read on the curve (023 cfm) at the point of intersection of the length (3 in) and diameter (2 in) For a supply pressure gauge of less than 100 lbf/in2, a correction factor is required (indicated at the bottom of Figure 744) This final value of the air flow would be less than the airflow that is available for our design At a supply pressure of 75 lbf/in2, the airflow needed is 023 065 = 015 scfm 00042 m3/min This value is within the capability of the air supply system that is available In addition, the clearance that is chosen is also five times greater than the available manufacturing error In fact, it is possible for us to design a bearing that has a clearance, which is better than the chosen value However, it is always a good idea to design something that is in the middle range and not at the extreme ends Next, it is necessary to verify that the clearance chosen will gives an adequate radial stiffness The stiffness of the bearing is inversely proportional to the clearance where the reduction in radial clearance will increase the radial stiffness This reduction in radial clearance is limited by the cost and the difficulty in manufacturing From Figure 75, it can be seen that the stiffness of an aerostatic journal bearing is constant up to an eccentricity ratio of 05 The aerostatic stiffness, K, can then be defined as.

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resulting in the following output voltage: vout = vout, dif + vout, com with vout, dif = RF and vout, com = RF RF = R R + RF R Avcom RF + R + R RF Avcom R R + RF R Ava dif RF + R + R RF Avb dif R (1513) (1512)

(1514)

Cytokinesis and mitosis result in two daughter cells, each with genetic material identical to that of the parent cell. This exact distribution of the genetic material, in the form of chromosomes, to the daughter cells, ensures the stability of cells and the inheritance of traits from one cell generation to the next. Cells have evolved complex

The common-mode rejection ratio (CMRR; see Section 126) is given in units of dB by CMRRdB = Adif Adif = 20 log10 Acom vout, com /vcom Adif

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Gamete formation presents an entirely new engineering problem to be solved. To form gametes in animals (and, for the most part, to form spores in plants), a diploid organism with two copies of each chromosome must form daughter cells that have only one copy of each chromosome. In other words, the genetic material must be reduced by half so that when gametes recombine to form zygotes, the original number of chromosomes is restored, not doubled. If we were to try to engineer this task, we would rst need to be able to recognize homologous chromosomes. We could then push one member of each pair into one daughter cell and the other into the other daughter cell. If we were unable to recognize homologues, we would not be able to ensure that each daughter cell received one and only one member of each pair.The cell solves this problem by pairing up homologous chromosomes during an extended prophase. The spindle apparatus then separates members of the homologous chromosome pairs, just as it separates sister chromatids during mitosis. But there is one complication. As in mitosis, cells entering meiosis have already replicated their chromosomes. Therefore, two nuclear divisions without an intervening chromosome replication are necessary to produce haploid gametes or

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where Adif is the differential gain (which is usually assumed equal to the nominal design value) Since the common-mode gain, vout, com /vcom , should ideally be zero, the theoretical CMRR for the instrumentation ampli er with perfectly matched resistors is in nite In fact, even a small mismatch in the resistors used would dramatically reduce the CMRR, as the Check Your Understanding exercises at the end of this subsection illustrate Even with resistors having 1 percent tolerance, the maximum CMRR that could be attained for typical values of resistors and an overall gain of 1,000 would be only 60 dB In many practical applications, a requirement for a CMRR of 100 or 120 dB is not uncommon, and these would demand resistors of 001 percent tolerance (see Check Your Understanding Exercise 153) It should be evident, then, that the discrete design of the IA, employing three op-amps and discrete resistors, will not be adequate for the more demanding instrumentation applications

145

Compute the common-mode gain and common-mode rejection ratio (CMRR) for the ampli er of Figure 1516

cell, 2n 4. Maternal chromosomes are red; paternal chromosomes are blue. (b) Fluorescent microscope image of a cultured cell in anaphase. Microtubles are red; chromosomes (DNA) are stained yellow. ([b] John M. Murray, Department of

Find: vout, com /vcom , CMRRdB Analysis: The common-mode gain is equal to the ratio of the common-mode output

Anatomy, University of Pennsylvania. Cover of BioTechniques, volume 7, number 3, March 1989. Reproduced with permission.)

The CMRR (in units of dB) can be computed from equation 1515 where Adif = A and therefore, CMRR = Adif Acom = 20 log10

Direction of chromosomal movement Figure 3.15 The kinetochore acts as a microtubule motor,

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