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The number $555,555$ can decompose, as the product of two factors of three digits, in how many ways?

The number $555,555$ can decompose, as the product of two factors of three digits, in how many ways?

I've seen the answer to the question, and there is only one way: Since $555, 555 = 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 37$, the only way to combine the factors to achieve expressing it as a product of two three-digit numbers is $(3 \cdot 7 \cdot 37) (5 \cdot 11 \cdot 13)$. Regardless of this, I struggle to understand how the answer was formulated. Can someone show me the procedure?

Sorry if the question is poorly phrased, it is a rough translation of the original problem in Spanish.

• list all the divisors in order. $\sqrt{555555} \approx 745.36.$ One of your divisors must be larger than that but still smaller than 1000. The other divisor in your pair will be between 555.56 and 745.36. I guess you can find them by hand, look at all the products of two primes, then three primes. You do not need to do four primes because you already did two – Will Jagy Apr 16 '18 at 2:34
• Yeah, two primes is too small, the biggest is $13 \cdot 37 = 481.$ So, three primes each – Will Jagy Apr 16 '18 at 2:38

Nothing wrong with the approach Will gives in the comments. Here's another way. Obviously $555,555=555\times1001$, but $1001=7\times11\times13$ is a little too large. The way to make it a little smaller is to swap the factor 7 with the factor 5 in 555, which gives you your solution, $(3\times7\times37)(5\times11\times13)$.

The factors $7,11,13$ can't be used together, since $(7)(11)(13)=1001$.

So one of the groups, group $1$ say, must have exactly two of the factors $7,11,13$.

Hence the factors $3$ and $5$ can't both be in group $1$, else the product of the factors in group $1$ would be at least $(3)(5)(7)(11) > (7)(11)(13)$.

Similarly, the factor $37$ can't be in group $1$, else the product of the factors in group $1$ would be at least $(37)(7)(11) > (7)(11)(13)$.

Label the other group as group $2$.

Thus, group $2$ contains

• $37$
• Exactly one of $7,11,13$.
• At least one of $3,5$.

But since $(37)(27) = 999$, the factors in group $2$ other than $37$ must have a product which is at most $27$.

It follows that neither of the factors $11$ or $13$ is in group $2$, since $(11)(3) > 27$, and $(13)(3) > 27$.

So the factor $7$ must be in group $2$, and the factors $11,13$ must be in group $1$.

Since the factor $7$ is in group $2$, the factor $5$ can't be in group $2$, since $(7)(5) > 27$.

Hence, the factor $5$ must be in group $1$, and the factor $3$ must be group $2$.

Thus, group $2$ has the factors $37,7,3$, and group $1$ has the factors $11,13,5$.

It is intelligent brute force. The largest a three digit number can be is $999$ so you need to find a factor of $555,555$ that is between $556$ and $999$. The other will also be in that range so you are done. Next note that $3 \cdot 5 \cdot 7=105$ which is too small by itself and too large multiplied by any of the other factors, so two of $3,5,7$ have to be in one factor and one in the other. $11\cdot 13 \cdot 37 \gt 999$ so again two of those need to be in one factor and one in the other. We are down to $18$ combinations to try, three singletons from $3,5,7$ times all the one or two combinations from $11,13,37$. I missed Will Jagy's point that you need three factors in each set, so that decreases the number to try to $9$.

• I also thought that trying to solve for the same question just that for 2 factors of 2 digits and then I would split one of the digits into 2. Would this help me solve this type of question faster? Also, the question doesn't require me to state the numbers, but it does require me to state the amount of ways I can decompose it. – Brian Blumberg Apr 16 '18 at 2:44
• I don't understand the first sentence. What does it mean to split one of the digits into $2$? Having gotten down to $9$ possibilities it won't take long to try them all. You could note that $3\cdot 11\cdot 37$ is too large, so $37$ must be by itself and $11,13$ must be in the other set. Now just try each of $3,5,7$ times $11\cdot 13=143$ to see which are in range and find that only $5$ works. – Ross Millikan Apr 16 '18 at 2:53
• Yeah, never mind my question. I forgot to delete it after I realized that. Thanks. – Brian Blumberg Apr 16 '18 at 3:01

The prime factor are $3 · 5 · 7 · 11 · 13 · 37$ so there are $2^6=64$ factors and $32$ complement pairs. Just list them all but don't bother with those that are less than $555555/999=556$

Toss out, $1,3,5,7,11,13,37,3*5,3*7,3*11,3*13,3*37,3*5*7,3*5*11,3*5*13, 3*5*37,3*7*11,3*7*13$ (that's 18 that are too small).

$3*7*37=777$ and its compliment is $5*11*13=715$. (That's 1 that is acceptible)

We can continuing tossing out $3*11*13$ and $3*11*37$and $3*13*37$ are too high so we toss them. (That's 21 that are unaccptible)

$3*5*7*11$ is too high so there wonvt be any more factors that are multiples of $3$ in range. Hence no other complements which aren't multiples of $3$ will be in range either.

Of the 9 we haven'haven't considered: $3*5*7*13,3*5*7*37,3*5*11*13,3*5*11*37,3*5*13*37,3*7*11*13,3*7*11*37,3*7*13*37,3*11*13*37$ are all too big.

So that was exhaustive.

We can write $$555,555=5\times111,111$$ and notice that $111=37\times3$, so we have $$555,555=5\times37\times3003$$ and since $1001=7\times11\times13$, the prime factorization is $$555,555=3\times5\times7\times11\times13\times37.$$

If we multiply each of the three combinations $11\times13$, $11\times37$ and $13\times37$ by $3$, we see that only $3\times(13\times37)$ exceeds $1000$ which is a four-digit number.

Hence $37$ must pair with two other one-digit numbers, and $11\times13$ must pair with either $3$ or $5$ since $7\times11\times13>1000$.

If $11\times13$ pairs with $3$, then the other product must be $$37\times(5\times7)>1000$$ which is not a three-digit number.

Therefore, the only possible combination for $555,555=P_1\times P_2$ is $$P_1=5\times11\times13,\quad P_2=3\times7\times37.$$

I think the procedure might be based on Fermat method.

We can calculate that the number $555555$ can be expressed as a difference of two squares:

$746^2 - 31^2=556516-961$

Number $31$ is the distance from $746$ both ways which gives $715$ and $777$ accordingly. From here the smallest prime factor of:

$777$ is number $3$ which equals to $3\cdot259$ and for

$715$ is number $5$ which equals to $5\cdot143$

Further factorisation results in final solution where $143=11\cdot13$ and $259=7\cdot37$ therefore:

$(5\cdot11\cdot13)(3\cdot7\cdot37)$ can satisfy equation.

If $555{,}555=ab$ with $a,b\lt1000$, then we must also have $a,b\gt555$ (e.g., if $b\lt1000$, then $a=555{,}555/b\gt555{,}555/1000\gt555$). Now since $555{,}555=3\cdot5\cdot7\cdot11\cdot13\cdot37$, we may assume, without loss of generality, that $a=37k$. From $a\gt555=15\cdot37$, we see that $k\gt15$, and from $a\lt1000$, we see that $k\le\lfloor1000/37\rfloor=27$. The only product $k$ of the primes $3$, $5$, $7$, $11$, and $13$ that falls in the interval $15\lt k\le27$ is $k=3\cdot7=21$. Thus $a=37\cdot21=777$, $b=5\cdot11\cdot13=715$ is the only factorization of $555{,}555$ into two three-digit numbers.


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# 1 km is equal to how many hours

View Answer. What is a Meter? if each step he takes is equal to 2.25 ft, how many steps does he take? A speed of exactly 1.609344 kilometers per hour equals one mile per hour, and this follows from the definition of a mile as 5280 feet, of a feet as 0.3044 meters, and of a kilometer as 1,000 meters and performing the necessary mathematical operations.These are internationally agreed upon unit definitions established by treaty in 1959. How Far Is 1 Km. Answer : We know hectare is unit of area . We can say that one hundred twenty kilometers is approximately seventy-four point five six five miles: 120 km ≅ 74.565 mi Somehow they convert 1 mile = 1.609344 km = 80 chains = 5280 ft. Then they take the distance covered by a single step assuming a stride length of 2.5 feet. How Many Hours In 6 Days. How Far Is A Km . Meter (metre) is a metric system base length unit. On the other hand, 1 meter equals 0.000621371 mile. 1 Meter (m) is equal to 0.001 kilometer (km). Most experts agree modern engines do not handle idling very well, and … The velocity in miles per hour is equal to the distance in miles divided by time in hours. How to convert kilowatt hours per kilometer to kilometers per kilowatt hour [kWh/km to km/kWh]:. A kilometer (abbreviation km), a unit of length, is a common measure of distance equal to 1000 meters and is equivalent to 0.621371192 mile or 3280.8398950131 feet. Here are some more examples of calculating how many hours are between two specified points in time. If 1 mile is 1.609 km, how many terameters is in 1 parsec? Another way is saying that 4 kilometers is equal to 1 ÷ 0.402336 miles. Definition of mile. You can view more details on each measurement unit: km or miles The SI base unit for length is the metre. 4 years ago. Minute to Hour Conversion Example Task: Convert 240 minutes to hours (show work) Formula: minutes ÷ 60 = hours Calculations: 240 minutes ÷ 60 = 4 hours Result: 240 minutes is equal to 4 hours Share 1. If you want 1/4 of just the light hours, 12/4 = 3 hours. Select am or pm. Share with your friends. 60 mm = 1 hour 24 hours = 1 day 7 days = 1 week 52 weeks = 1year 10 years = decade 7 71 _____ k Z_ - 6. How Far Is Kilometer. 5000km is about 3125 miles to be precise 3106.85 km How many square miles are in 2 million acres? MB has the prefix Mega. meters to kilometers formula. On a poorly maintained one it is. A mile is a unit of length in a number of systems of measurement, including in the US Customary Units and British Imperial Units. Approximate result. Miles are generally split into yards, with each yard equalling 0.000568182 mile. Example. We can say that four kilometers is approximately two point four eight five miles: 4 km ≅ 2.485 mi NaN = … In this case 1 mile is equal to 0.402336 × 4 kilometers. Km To Miles Formula . How many hours are in a fortnight (2 weeks)? Converting 2.1 mi to km is easy. 1 kilometer is equal to 0.62137119 miles: 1 km = (1/1.609344) mi = 0.62137119 mi. 2000000 acres x 1 square miles/640 acres = 3125 square miles 8 years ago. How many Megabytes in a Kilobyte. For example, to convert 100 m to km, multiply 100 by 0.001, that makes 0.1 km is 100m . Difference Between KB and MB. Megan. All In One Unit Converter ⇨ Please, choose a physical quantity, two units, then type a value in any of the boxes above. Since a day is 24 hours, 24/4 = 6 hours if you want 1/4 of a full day. Question: QUESTION 1 How Many Seconds Are In 22.5 Hours? Miles per hour can be expressed using the formula: v mph = d mi t hr. We assume you are converting between mile and kilometre. Note that rounding errors may occur, so always check the results. 0 0. 2.1 miles equal 3.3796224 kilometers (2.1mi = 3.3796224km). How many miles in 1 km? Simply use our calculator above, or apply the formula to change the length 2.1 mi to km. Simply use our calculator above, or apply the formula to change the length 1 mi to km. 1 hour of engine idle time is equivalent to 25 miles/40 km according to Ford. 1 Meter = 0.001 Kilometer . Another way is saying that 120 kilometers is equal to 1 ÷ 0.0134112 miles. How To Convert Km To Miles. 1 Kilobyte is equal to 0.0009765625 megabytes (binary). mile = kilometer * 0.62137119223 Each book has approximately 79 chapters. 1 metre is equal to 0.00062137119223733 miles, or 0.001 km. How much are 1 miles in kilometers? At that kind of hours, I would be much more interested in how well it is has been looked after than concerned about the hours. Converting 1 mi to km is easy. Definition of mile. Click "Click to Calculate" button. 100000cm is the answer for 1km -1 ; View Full Answer 1 m=100 cm 1 km=1000 m So,100*1000=100000 . Source(s): https://shrink.im/badsx. Simply use our calculator above, or apply the formula to change the length 1 km to mi. Convert 20 kilometers to miles: d (mi) = 20km / 1.609344 = 12.4274mi. Next, let's look at an example showing the work and calculations that are involved in converting from minutes to hours (min to hr). The answer is 1.609344. A yard is 3 feet and was originally considered the average length of a man's stride. 1 week = 7 days; 1 day 24 hours — I / ‘ay 5. QUESTION 3 What Is The Density (in G/mL) Of An Unknown Solid That Has A Mass Of 125.1 Kg And A Volume Of 511 ML? 1 metre is equal to 0.001 km, or 0.00062137119223733 miles. The minutes entered must be a positive number between 1 and 59 or zero (0). matheis. For example, to calculate how many miles is 100 kilometers, multiply 100 by 0.62137119223, that makes 62.137119223 miles is 100 km. The velocity in kilometers per hour is equal to the distance in kilometers divided by time in hours. Note that rounding errors may occur, so always check the results. How many km in 1 miles? Kilometers to Miles Conversion. 1 Km Is Equal To . 1.6km. For practical purposes we can round our final result to an approximate numerical value. (4) 4 hrs. 200h on a properly maintained engine is not a whole lot. If you leave them at 7AM and pick them up at 3PM then the question is how many hours is 7AM to 3PM and the answer is eight hours (15-7 = 8). QUESTION 2 6,892 Centimeters Is Equal To How Many Kilometers? But you're probably looking for the first one. 1 Kilometer Equals Miles . a kilometer equals 1312 steps because 1 times 1312 (the conversion factor) = 1312 . There are 0.0166666667 hours in a minute. Definition of kilometer. And the answer is 0.6213711922 mi in 1 km. How Long Is 1 Km. Megabyte is greater than Kilobyte. 1 km is equal to how many cm? kilometer = meter * 0.001. kilometer = meter / 1000. 8 Answers. Anonymous. How many Kilometers per hour equal one Mile per hour? Find the time taken by the truck to travel 310 kilometres. A kilometer (abbreviation km), a unit of length, is a common measure of distance equal to 1000 meters and is equivalent to 0.621371192 mile or 3280.8398950131 feet. How Many Km In A Mile . Approximate result. For example, 1 kilometer per hour can be written as 1 km/h or 1 kph. The answer is 0.62137119223733. 1 kilometers equal 0.6213711922 miles (1km = 0.6213711922mi). 1 KB = 2-10 MB in base 2. For practical purposes we can round our final result to an approximate numerical value. 1 miles equal 1.609344 kilometers (1mi = 1.609344km). (1 terameter = 1 trillion meters = 1 billion kilometers) View Answer. Explanation : Distance covered by truck in $$\frac{3}{2}$$hours = Speed × Time =90*$$\frac{3}{2}$$=135km. Share 1. Kilometers per hour can be expressed using the formula: v km/s = d km t hr. How Many Miles In A Kilometer. How many decades are equal to 1.7 x 1025 minutes? Kilometers per hour can be abbreviated as km/h, and are also sometimes abbreviated as kph. Kilometers to miles conversion table . For example, 1 mile per hour can be written as 1 mph, 1 mi/h, or 1 MPH. Source(s)::) 0 0. 1 mile=approx. On average, there are 3 pages in every chapter of a James Patterson novel. How many kilometers per kilowatt hour in a kilowatt hour per kilometer: If eF kWh/km = 1 then eF km/kWh = 1 km/kWh. geroge walks 1.5 km to scho. Kilobyte unit symbol is KB, Megabyte unit symbol is MB. How many steps in 1 km? Kilometers To Miles Chart How to Calculate Minutes to Hours 1 minute = 6 × 10 1 seconds 1 hour = 3.6 × 10 3 seconds 1 minute = (6 / 3.6) × 10 1 × 10-3 hours 1 minute = (1.6666666666667) × 10 1-3 hours 1 minute = (1.6666666666667) × 10-2 hours 1 minute = 1.6666666666667 × 0.01 hours 1 minute = 0.0166666667 hours How Many Hours in a Minute? Find other conversions here: Definition of Kilometer. A truck travels at 90 km/hr for the first 1$$\frac{1}{2}$$hours.After that ittravels at 70 km/hr. The hours entered must be a positive number between 1 and 12 or zero (0). KB has the prefix Kilo. Hello, I would like to solve this formula, I not so good at math so I need this help I am driving a car at 80 km / h, If in 60 km I spend 50 minutes, how many hours would I spend for 15.000 km? 1 km is equal to how many hectare? The number of hours and minutes between the two selected times will appear. As 1 kilometer is equal to 1000 meters, 1 mile equals 169.03 meters. Ans . 1/4 of a day is equal to how many hours? We can say that a car that has 0-80,000 km is in good/ok condition, any thing above 80,000 to 150,000 is too many km's. To convert kilometers to miles, multiply the kilometer value by 0.62137119223 or divide by 1.609344. You can view more details on each measurement unit: miles or km The SI base unit for length is the metre. Answer Save. kilometers to miles formula. Converting 1 km to mi is easy. Definition of kilometer. Enter hours and minutes. Relevance. Lv 4. (1) 2.5 hrs (2) 3 hrs (3) 3.5 hrs (4) 4 hrs. You can use our hours calculator above to compute the time between any two arbitrary hours and minutes. Share with your friends. 1 Kilometer (km) is equal to 0.62137119223 mile. 29 000 Kilometers Equals How Many Miles. km/kWh = 1 ÷ kWh/km. 1 KB = 10-3 MB in base 10 (SI). How To Easily Convert Kilometers To Miles. - 18591415 Likewise the question how many kilometer in 1 mile has the answer of 1.609344 km in 1 mi. The distance d in miles (mi) is equal to the distance d in kilometers (km) divided by 1.609344: d (mi) = d (km) / 1.609344 . 1 Kilobyte is equal to 0.001 megabytes (decimal). Square kilometre (International spelling as used by the International Bureau of Weights and Measures) or square kilometer (American spelling), symbol km 2, is a multiple of the square metre, the SI unit of area or surface area.. 1 km 2 is equal to: . How many miles in 1.61 kilometers. To convert meters to km, multiply the meter value by 0.001 or divide by 1000. 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## Monday, October 30, 2017

### Extinction and the Dimming of KIC 8462852

Extinction and the Dimming of KIC 8462852

Authors:

Meng et al

Abstract:

To test alternative hypotheses for the behavior of KIC 8462852, we obtained measurements of the star over a wide wavelength range from the UV to the mid-infrared from 2015 October through 2016 December, using Swift, Spitzer and AstroLAB IRIS. The star faded in a manner similar to the long-term fading seen in Kepler data about 1400 days previously. The dimming rate for the entire period reported is 22.1 ± 9.7 mmag yr−1 in the Swift wavebands, with amounts of 21.0 ± 4.5 mmag in the ground-based B measurements, 14.0 ± 4.5 mmag in V, and 13.0 ± 4.5 in R, and a rate of 5.0 ± 1.2 mmag yr−1 averaged over the two warm Spitzer bands. Although the dimming is small, it is seen at gsim3σ by three different observatories operating from the UV to the IR. The presence of long-term secular dimming means that previous spectral energy distribution models of the star based on photometric measurements taken years apart may not be accurate. We find that stellar models with ${T}_{\mathrm{eff}}=7000\mbox{--}7100$ K and ${A}_{V}\sim 0.73$ best fit the Swift data from UV to optical. These models also show no excess in the near-simultaneous Spitzer photometry at 3.6 and 4.5 μm, although a longer wavelength excess from a substantial debris disk is still possible (e.g., as around Fomalhaut). The wavelength dependence of the fading favors a relatively neutral color (i.e., ${R}_{V}\gtrsim 5$, but not flat across all the bands) compared with the extinction law for the general interstellar medium (${R}_{V}=3.1$), suggesting that the dimming arises from circumstellar material.


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Chemistry lewis structure

1. Dec 4, 2008

devanlevin

if i am told to draw a lewis diagram for
PCl$$^{+}_{4}$$
does this mean a melocule of 1P atom and 4 Cl atoms, the whole molecule with a charge of +1 or is it 1 P atom and 4 Cl$$^{+}$$ ions, giving the molecule a charge of +4

what does the lewis structure look like, is it P in the middle with Cl on each of its 4 sides, all in square brackets with a + sign after the brackets.

does anyone know of a site that has a tool where i can plug in the molecule and get the structure?

2. Dec 4, 2008

That's it.

Sounds OK.

3. Dec 4, 2008

devanlevin

what are the lewis structure for these ions?

F$$^{-}$$
is it [F]$$^{-}$$

Fe$$^{3+}$$
is it [Fe]$$^{3+}$$ with 5 dots around the Fe, (5 valence electrons) Fe$$^{3+}$$ : [Ar]4s$$^{2}$$3d$$^{3}$$

Co$$^{3+}$$
is it [Co]$$^{3+}$$ with 6 dots around Co

generally how are the transition metals displayed in lewis structure since they arent in any of the 7 colums of valance electrons

4. Dec 4, 2008

Joskoplas

PCl4+:

Write P in the center. Draw a single bond to each of four Cl atoms.

P normally has five electrons. With only four bonds, it has a formal charge of +1. Good thing that fifth electron took a hike, huh?

Each of the four chlorines is stuck at the end of a single bond, with three lone pairs. The formal charge is zero for each.

Therefore, the entire molecule has a charge of +1.

F-:

Draw F in the center. Surround it with 4 lone pairs. The formal charge is now -1. Done.

Iron, Cobalt:

Applying the rules for lewis dot structures to transition metals really isn't the best idea. Even atoms in the non-metals get sort of snarky as you go further down. Nitrogen has to stop at four bonds, but phosphorous can have five (using not just its s and p orbitals, but its brand new d-orbitals as well). Carbon is stable in methane with CH4, but Silicon in silane (SiH4) burns immediately on contact with air -- the extra d orbitals again. Sort of like extra parking spaces for reaction intermediates.

And the transition metals? Worst of all. They mostly ignore the octet rule (though some do kind of like the 18 electron rule, which you may hear about later). So trying to make them follow Lewis structure rules is kind of like tying a knot in water -- it's too easy to break the rules for the rules to make much sense.

5. Dec 4, 2008

chemisttree

Why are you using brackets in a Lewis dot structure? Don't!

In your examples, first you determine the oxidation state of the element(s). If the oxidation state is positive, you subtract electrons (or dots, 'x's', o's, etc...) from the neutral element. For example, NCl3 would have an oxidation number of +3 for nitrogen and -1 for each chlorine. The lewis dot structures for those would be an 'N' with two dots (5-3=2) and three 'Cl' with eight 'o's'(7+1=8). Together it would be a central 'N' surrounded by with two 'dots' and 6 'o's'; three sides would have a chlorine with 8 'o's', two of those being shared with the nitrogen.

Transition metals follow the 18 electron rule, not the 8 electron rule.

Last edited: Dec 4, 2008


text:
Section D

# System for Monitoring Avian Cardiac Output and Breathing Patterns Using Transmission-Type Microwaves

Isao Nakajima1,*, Ichiro Kuwahira2, Jun-ichi Hata3
1Seisa University, Yokohama, Japan, jh1rnz@aol.com
2Tokai University School of Medicine, Tokyo Hospital, Tokyo, Japan, kuwahira@tok.u-tokai.ac.jp
3Central Institute for Experimental Animals, Kawasaki, Japan, hata@ciea.or.jp
*Corresponding Author: Isao Nakajima, +81-90-8850-8380, jh1rnz@aol.com

© Copyright 2022 Korea Multimedia Society. This is an Open-Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License (http://creativecommons.org/licenses/by-nc/4.0/) which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.

Received: Sep 14, 2022; Revised: Oct 29, 2022; Accepted: Nov 08, 2022

Published Online: Dec 31, 2022

## Abstract

We report the development of a non-contact monitoring device for avian cardiac output and breathing patterns based on the anterior thoracic air sac pressure that uses transmission-type microwaves (2,400−2,500 MHz, continuous wave). Since the phase waveform represents the dielectric constant change, the phase reflects −j/ωc and the dielectric constant change is related to blood flow. The magnitude waveform is reflected from the electronic resistance of tissues due to the expansion of the anterior thoracic air sac, which mainly consists of the thoracic wall. To confirm these waveforms, pigeons and chickens were used for testing. To validate the output waveforms of the developed transmission-type microwave device, data from esophageal catheters and pressure sensors in the anterior thoracic air sac, abdominal air sac, and intraoral cavity were obtained. The waveform for the esophageal catheter, where electrocardiogram electrodes and an angular velocity sensor were installed, correlates with cardiac output. A heart sound microphone was used to confirm the closing sound of the arterial and mitral valves. The experimental results confirm that a linear waveform synchronized with the cardiac blood flow and the anterior thoracic air sac pressure of birds was obtained using transmission-type microwaves. The proposed device, which can monitor cardiac output and respiratory patterns, may enable the early screening of cytokine storms caused by avian influenza viruses. Existing devices use Doppler radar in the 10 to 77 GHz band; these high frequencies are reflected by the chest wall and do not reach deep into body, making it impossible to monitor the blood flow inside the body.

Keywords: Dielectric Constant; Non-Contact; Blood Pressure; Avian Influenza; Screening

## I. BACKGROUND

1.1. Vital Signs

When someone who has caught the flu seeks medical attention, the examining doctor will measure their blood pressure with a sphygmomanometer and listen to their breathing to conduct a pulmonary examination. In mammals and birds infected with the avian influenza virus, the immune system is activated and chemical messengers called cytokines are released from the mast cells of the tissue to fight the virus, increasing the permeability of the vascular wall and allowing amino acids to escape from the blood vessels. A mechanism that facilitates leakage activated. However, when the number of cytokines increases, blood pressure decreases, creating a cytokine storm. In this condition, the lungs become severely edematous and breathing becomes shallow and rapid. For this reason, in humans, the examining physician or nurse monitors blood pressure and observes breathing patterns.

1.2. Related Studies

Several devices for humans based on radar Doppler using continuous waves reflected from the skin of the chest have been developed. Radar Doppler is not limited to medical care; it is also used in health and sleep management. Through vibrations, it can sense whether a person is awake and moving or sleeping. In humans, since the vibration of the heart is transmitted to the chest wall, the heartbeat can be obtained from vibrations of the skin. High-frequency microwaves (10−77 GHz) are reflected at the surface of the body, so the propagation of radio waves is only in the air. The amount of power attenuation is relatively small for this Doppler system, so the microwave transmission power can be very low. However, if the inputs and reflected waves (nonlinear) are simply mixed with each other, the detected signal will be nonlinear waves. To compensate for this shortcoming, a method of simultaneously interfering waves with a phase shift of λ/4 (λ: wavelength) can be used; however, the improvement is limited. Since it is not possible to obtain waveforms of heart vibrations that hidden under the breathing waves, the frequency peaks (first peak is breathing rate and second peak is heart rate) are obtained by performing the fast Fourier transform (FFT).

Since FFT is an integration of the time axis, the resolution of the time axis is reduced. Therefore, in the RR interval, which requires the accuracy of the time axis, the waveform derived from the heart motion must be extracted, IFFT performed, the original waveform drawn, and the fluctuation of the RR interval must be measured. Since the evaluation of the RR interval is the standard deviation of the interval, accuracy cannot be obtained unless the heart-derived signal is large enough. Regarding radar Doppler, there are no reports on obtaining a blood pressure waveform even if the frequency is lowered to irradiate the inside of the thoracic wall.

1.3. Present Research

Measurement of esophageal impedance for chickens (monitoring of blood pressure and breathing patterns)

We previously developed a device for chickens that can monitor the cardiac output as the phase and breathing changes as the magnitude by adding a 1-MHz medium wave through electrocardiogram electrodes of the esophageal catheter [1-3].

In addition to electrocardiogram electrodes, this avian esophageal catheter incorporates an angular velocity sensor that correlates with stroke volume and a heart sound microphone that can confirm the closing sounds of the aortic and mitral valves.

Based on these physiological data, we can determine the blood flow of the cardiac output and the movement of the thoracic wall associated with respiration. Fig. 1 shows an example of measurement using an esophageal catheter.

Fig. 1. Esophageal impedance by 1 MHz. Phase is shown in light blue and magnitude is shown in green.

Fig. 1 shows (from top to bottom) the invasive axillary artery pressure, magnitude (impedance), angular velocity, electrocardiogram, phase (impedance), and heart sound. The phase waveform shows the outflow from the left ventricle and blood flow into the atria. The closing sounds of the mitral valve and the aortic valve can be confirmed. Since the closing sounds of the aortic valve match the notch timing of the waveform (shape) of the phase, the phase can be determined as the blood flow originating from the heart. Regarding the peak of the axillary artery pressure (mea- sured by invasively cutting down the axillary artery), a propagation delay was observed comparison from the peak of the phase waveform. The difference (L) between heart and axillary artery is the blood flow velocity V=L/Δt. Our previous paper demonstrates that there is a correlation between the magnitude of the amplitude of the angular velocity and the magnitude of the peak of blood flow [3].

In this paper, a carrier wave of 2.4 GHz is transmitted from outside the body, and microwaves that have undergone frequency modulation and amplitude modulation effected by the heart and chest wall, and the phase (frequency) and magnitude (amplitude) are separated and demodulated by the receiving device. Even if the voice is modulated with a medium wave carrier wave on 1 MHz, the voice can be demodulated in the same way even if the voice is modulated with a microwave on 2.4 GHz. Therefore, it is estimated that even if microwaves are used, the impedance waveform will resemble that of 1 MHz of the esophageal catheter. The difference is the site of measurement, that is, the spatial difference. Due to the difference between the measurement-site at a position close to the heart or ribcage and the measurement-site at a distance outside the body, the latter is generally considered to be physically blunted in the waveform. The solution of spatial loss, that could be solved by antenna directivity and penetration of tissue. The longer the wavelength, the better the penetration, and the shorter the wavelength, the sharper the directivity, so there is a spatial gain. These two factors are relationship of tread-off. A practical product for poultry farmers to understand the health status of their birds must be contactless. Because chickens are handled in cages. With this in mind, this paper describes the method and results of how to find the trend-off conflict.

1.4. Multimedia Data Analysis is Essential

The circulation and respiratory monitoring using with microwave are methods that do not contact the individual, and are spatially separated from the heart, lungs, and air sacs. So what does the microwave waveform mean? Is the temporal timing consistent with other multimedia data? These points must be watched. Occasionally, it may be observed with delays or skeletal muscle spasms as a noise. In order to evaluate, multimedia data measured in multiple dimensions, are essential. The further the distance from the heart, the more supplementary correction data is required. Therefore, the evaluation of microwave waveforms shall be recommended with data analysis by multimedia.

1.5. Noise Also Has Meaning in Birds without a Diaphragm

Birds do not have a diaphragm, so skeletal muscles in the abdomen and chest control balloons called air sacs to direct airflow into the lungs. Avian lungs are hard, and in addition to ventilation, which draws in outside air through the trachea, airflow is also exchanged between air sacs. Therefore, airflow flows into the lungs multiple times in one ventilation. Skeletal muscles are always in a state of tension, which suggests that the sympathetic nerves, which control the excitation state of skeletal muscles, maintain a dominant state in the autonomic nerves. The bar-tailed godwits tacked by GPS, a migratory bird that flies for 14 days without rest and sleep, flying from Alaska to the off-shore of New Zealand [28]. It is extremely difficult for humans, mammals, to run a marathon 24 hours a day, but birds have a physiological structure that allows them to maintain sympathetic excitation at all times. Therefore, in respiratory monitoring, skeletal muscle spasm may appear in the air sac pressure, but this is not noise but important information controlling the flow of air between air sacs. Therefore, air sac pressure monitors must be amplified even at relatively high frequencies without noise rejection.

## II. PRINCIPLE

2.1. Free-Space Propagation Loss

Here, we explain the principle of monitoring the blood flow associated with the heartbeat and the breathing pattern associated with the movement of the thorax. The relationship between the free-space propagation loss and frequency can be expressed as

$\text{Loss}\left(\text{dB}\right)=20×\mathrm{log}\left(4×\pi ×\frac{R}{\lambda }\right).$
$\text{λ=}\frac{c}{f}$

c: speed of light (m/s)

f: frequency (Hz)

R: distance (m)

λ: wavelength (m)

From this equation, the attenuation is 29.77 dB at 2,450 MHz and a distance of 30 cm.

2.2. Half-Power Depth

The proposed device passes microwaves through the bird’s tissue and measures changes in impedance. The tissue attenuates the microwave power. Under the assumption that the depth at which power is attenuated by half is D, the relationship between frequency f, dielectric constant εr, and the dielectric dissipation factor (tan δ) is as follows.

$\text{D}\text{\hspace{0.17em}}\text{=}\frac{3.31×{10}^{7}}{f×\mathrm{tan}\text{\hspace{0.17em}}\delta ×\sqrt{{\epsilon }_{r}}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le \delta \le \pi .$

The dielectric loss tangent is a numerical value that expresses one aspect of the ratio at which part of the energy becomes heat when an AC electric field is applied to a dielectric. If an AC electric field is applied to an ideal dielectric and only the capacitance component (capacitor) is considered, there is a 90-degree phase difference between the electric field and the current due to polarization/charge reversal.

The power consumption of the AC current is expressed as the vector product of the electric field and the current, so if the phase difference between the two is 90 degrees, the power consumption will be zero. It becomes an ideal dielectric. However, in actual materials, complete insulation is not possible. The resistance of the dielectric itself is finite, so a leakage current is generated. In addition, dielectric polarization and partial discharge create a parallel resistance component. The phase difference is thus shifted from 90 degrees. Let δ denote the shift from 90 degrees. Its tangent (tan δ), called the dielectric tangent, corresponds to the component where the electric field and current are in phase. Only this in-phase component contributes to power consumption or loss. The dielectric loss tangent is proportional to the dielectric loss [4-13]. In addition, the complex dielectric constant due to the parallel plate capacitor due to the difference in materials is described in the literature [14], the difference in the frequency characteristics of pure water and blood is described in [15], and the effects of weak microwave exposure are described in [16].

The dielectric constant of water is about 80. In living organisms, glucose in body fluids decreases the dielectric constant. Under the assumption that a body fluid has an average glucose content of 5%, the dielectric constant is 76.5 at around 2,450 MHz, as shown in Fig. 2.

Fig. 2. Glucose concentration and dielectric constant.

Furthermore, the non-dielectric constant of the chest is lowered by the air in the lungs, connective tissue fibers, fatty tissue, etc., so the average non-dielectric constant can be considered to be 25.5 or lower. For the estimation of the attenuation of microwaves, a factor of 25.5 may be chosen to maximize attenuation.

From Fig. 3, the power half-power depth is 26.2 mm with a non-dielectric constant of 25.5, so it is 4 times this (2 to the power of 4=16) 26.2×4=104.8 mm Approximately 105 mm has 1/16 of the power. The power is attenuated by −12 dB corresponds to half-power loss (1/16=−12 dB). It is assumed here that 105 mm is approximately the width of the anterior chest of a large chicken.

Fig. 3. Half-power depth versus frequency for dielectric constants of 76.5 and 25.5.
2.3. Scattering Loss

For microwaves, a higher frequency leads to more energy being reflected by the surface of a substance. Less energy is reflected when the wave is incident at right angles to the surface. Whether the reflective surface of an uneven material can be regarded as flat is determined by the Rayleigh roughness criterion, expressed as

$\text{g}=4\text{πσ}/\text{λsin}\text{\hspace{0.17em}}\Theta ,$

where σ is the standard deviation of the undulation amount in the first Fresnel zone and θ is the incident angle with respect to the normal direction of the surface.

If g < 1, the reflective surface is flat; if g>1, it is undulating. In chickens and pigeons, the probe is placed perpendicular to the chest wall.

Since the wavelength for 2.5 GHz waves is 12 cm and a bird’s skin is smooth, g < 1. In experiments with chickens and pigeons, the probe was intentionally set perpendicular to the chest wall, with a wavelength of λ 12 cm at a freq- uency of 2.5 GHz. Since the denominator of the above formula is large enough, the skin of birds is smooth, and g<1 is self-evident. There is no problem numerically assuming that scattered waves are relatively small at 2.5 GHz. However, the beam half width of the 10 dBi antenna is 49 degrees, so the scatter increases geometrically (it is not a pointing loss). Based on the past propagation experiments at 2.5 GHz show that the beam direction of antenna is incident perpendicularly (almost 90 degrees) and reflected or scattered by the skin, resulting in a halving of the power (i.e., a 3 dB drop).

2.4. Equivalent Circuit

In the equivalent circuit of avian thorax, Rbody is the change in distance mainly due to the movement of the ribcage. This change is very large. Many radar Doppler systems track only real numbers and blood flow cannot be seen logically because they calculated only magnitude (real number). Even if the change in blood flow is reflected in the real number ΔR, a small ΔR cannot be detected unless the dynamic range of the radar Doppler system is increased and the amplification circuit is not linear. The amount of change in blood flow on the real axis (Rleakage in Fig. 4) is about 1/100 of the resistance value of the thorax.

Fig. 4. Equivalent circuit of avian thorax.

In addition, the notch caused by the aortic valve closure cannot be seen in the real axis direction change unless the resolution is high. Previous studies have thus applied the FFT power spectrum to the received waveform. The first peak is taken as the respiratory rate and the second peak is taken as the heart rate.

In birds, the heart is covered by a large sternum and thus the vibrations of the heart are only slightly transmitted to the skin. The valve closing sound is even difficult to detect using a stethoscope. Reflective microwave radar, which detects skin vibrations, thus cannot be used for birds.

Using the equivalent circuit shown in Fig. 4, it is possible to measure the change on the complex plane, where the imaginary part is the phase and the change in the real number is the magnitude, using transmission-type microwaves (continuous waves), as described later.

2.5 Circuit Design

Specifically, when the transmit and receive antenna gains are 10 dBi, the frequency is 2,450 MHz, and the distance is 30 cm, the transmit power is −40 dBW and the received power is −67.76 dBW. The theoretical margin is 40.75 dB. Such a margin is difficult to obtain with radar Doppler.

Chickens are constantly moving and thus there is a pointing loss of the antenna toward the heart. In addition, the skeletal muscles of chickens constantly vibrate to control skeletal muscles which keep tension of air sacs [2], and those vibration causes fluctuations in the baseline of noise level. The actual measurement margin is about 30 dB, which is different from the theoretical value, but this 30 dB margin is necessary to draw the phase waveform including the notch of the aortic valve. Table 1 shows the line design of the system considering the free-space propagation loss, half-power depth, reflection/scattering, and pointing loss.

Table 1. Circuit design (theoretical values).
Target birds 5 pigeons (weight 450-470 g) 8 chickens (weight 1,000-2,500 g)
Frequency/ output 2,400-2,500 MHz-40 dW 2,400-2,500 MHz-40 dW
Distance be-tween probes 20 cm 30 cm
AD conversion 6 bits, quantization 500 /sec. or ,2,000 /sec 6 bits, quantization 500 /sec. or 2,000/ sec.
Measurement state Inhalation anesthesia or no anesthesia Inhalation anesthesia or no anesthesia
Signal detection Amplitude synchronous detection (obtain two orthogonal signals with PSN) Amplitude synchronous detection (obtain two orthogonal signals with PSN)

## III. METHODS

3.1. Verification Experiment

The conceptual diagram and block diagram of the implemented system are shown in Fig. 5. The reflection coefficient (S11) of the antenna built into the probe is shown in Fig. 6. The implemented device is shown in Fig. 7. The experimental conditions are summarized in Table 2.

Fig. 5. Conceptual and block diagrams of implemented system (f1 and f2 are 2,400-2,500 MHz ISM band continuous waves).
Fig. 6. The reflection S11 parameter of antenna in probe. The antenna is adjusted so that the dip comes to the frequency of 2.4 GHz.
Fig. 7. Implemented transmission-type microwave device for pigeons (chest antenna: horizontal polarization, abdomen antenna: vertical polarization).
Table 2. MRI scans used in experiments.
 Frequency (GHz) 2.45 Tx antenna gain (dB) 10 Tx power (dBW) -40 EIRP (dBW) -30 Free space pass loss (dB) 29.77 Reflection loss (dB) 3 Body attenuation (dB) 12 Rx Antenna gain (dB) 10 Antenna pointing loss 3 Receiving power (dBW) -67.76 Noise temp of Receiver (K) 1200 Noise power density (dBW/Hz) -197.81 G/T(dB/K) -20.79 Bandwidth (4 kHz) -36.02 C/No (dB Hz) 91.15 Required C/No (dB Hz) 50.4 Margin(dB)* 40.75
3.2. Insertion of Auxiliary Sensor

It is necessary to measure the heart and respiration with another sensor to corroborate the microwave measurements. In pigeons, an air pressure sensor and an electrocardiogram electrode were surgically sewn subcutaneously in the sagittal direction in the midline of the back under isoflurane inhalation anesthesia in advance. The implemented transmission-type microwave device for pigeons (chest antenna: horizontal polarization, abdomen antenna: vertical polarization) is shown in Fig. 7.

For chickens, an esophageal catheter can be inserted through the crop to the side of the heart. While maintaining inhalation anesthesia and spontaneous breathing, the operator guided the esophageal catheter to pass through the crop and into the lower esophagus. Isoflurane was then discontinued, sufficient air was aspirated, and the chicken was gently aroused and kept upright for the transmission-type microwave examination and data acquisition. The esophageal catheter was removed when it was noticed by the chicken. Data were then collected continuously. In addition, pressure sensors with silicon tubes were surgically inserted into the air sacs.

3.3. Experimental Results Pigeon

Respiration-induced changes in received power and heartbeat-synchronized waveforms were obtained from all target birds. Fig. 8 shows changes in the received magnitude waveform (top) and changes in the abdominal air sac pressure obtained by the pressure sensor, both of which are synchronous.

Fig. 8. Magnitude waveform from abdominal probe (top) and pressure of abdominal air sac (bottom) obtained with pressure sensor inserted into abdominal air sac.

Similarly, the change in the received power obtained from the chest probe of the pigeon (Columba livia), the FFT of the signal, the inverse FFT of the frequency components below 17 Hz, and the differentiation of this waveform are shown in Fig. 9. The waveform and the electrocardiogram obtained at the same time are also shown.

Fig. 9. Phase waveform (black) obtained from chest probe, its FFT, inverse FFT, and differentiated signal (synchronized with electrocardiogram R wave).

A heartbeat is obtained from a chest probe. The measured signal was about 20 dB lower than the calculated value. The reason for this was clarified during the post-experiment autopsy. Homing pigeons used for racing have well-developed breast wall meat, which is not much different from chicken breast wall thickness, and their wing meat is thicker than that of chickens. Nevertheless, a margin of about 30 dB was secured.

3.4. Chicken

Eight chickens were viewed using a circularly polarized probe and compared with esophageal catheter data. In Fig. 10, photograph shows the measurements taken on chest side of chickens.

Fig. 10. Photograph of measurements taken on chest side of chickens (isoflurane anesthesia device shown on left).
3.5. Evaluation of Magnitude Waveform for Chicken

The obtained magnitude waveform (black) is shown in Fig. 11. This waveform is strongly correlated with the anterior thoracic air sac pressure (yellow). In birds, the breathing pattern is inspiration, short duration, expiration, then long duration (one cycle) and both curves (black and yellow) suggest that principle. Amplitude modulation of the anterior thoracic air sac, indicated by asterisks, can be observed as turbulent airflow from other air sacs (posterior air sacs) to the precordial air sacs via the lungs. Such vibrations are due to phase differences in skeletal muscle contractures, which are described in detail in the paper [2]. Birds, unlike mammals, are efficient with more than one airflow to the lungs per breath.

Fig. 11. Normalized waves of Magnitude of microwave and pressure of anterior thoracic air sac. During the asterisks, air flows with vibration are inserted from another air sacs via lungs.

The magnitude waveform has a strong correlation with the anterior thoracic air sac pressure. It is speculated that the skeletal muscles that generate the anterior thoracic air sac pressure strongly affect the movement of the chicken’s chest wall. In more detail, it can be inferred from this graph that the breathing control muscles are different between the star-marked time period and the time period of inspiration and expiration.

3.6. Evaluation of Phase Waveform for Chicken

Fig. 12 shows the phase waveform of the microwaves (purple). There is a notch in the middle of the mountain-shaped waveform, as indicated by an arrow. This notch is synchronized with the aortic valve closing sound from the heart sound (green).

Fig. 12. Phase waveform for chicken (purple arrows indicate notches that occur at aortic valve closure; this timing can be deduced from heart sound).

The upper part of the graph shows the angular velocity of the axial rotation of the esophagus. Previous studies [3] have shown that its amplitude correlates with blood pressure.

When the measurement point was determined at the timing of breathing and the correlation between the amplitude of the angular velocity and the amplitude of the microwave phase was obtained, a linear correlation was established at R2=0.95 (Fig. 13).

Fig. 13. Correlation between potential of angular velocity and potential of phase.

This indicates that the microwave phase reflects the blood flow in the heart and that the notch measures the closing of the aortic valve and the subsequent inflow into the left atrium.

## IV. CONSIDERATIONS

4.1. Comparison with Reflected Radar Doppler Method

Many experiments have been conducted on heartbeat/ respiratory monitors using microwaves. We have developed a monitor that has a strong correlation between the cardiac output flow and the pressure of the anterior thoracic air sac in birds using transmission-type microwaves.

We can obtain not only the cardiac output flow but also the interval between beats (RR interval). This interval reflects the condition of autonomic nerves (sympathetic and parasympathetic nerve control) and is an important parameter for confirming health conditions. For radar Doppler, it is difficult to accurately measure interval fluctuations because increasing the frequency resolution leads to frequency components at intervals of Δt that are at least 1,024, 2,048, or more times the sampling time.

Therefore, the interval of time Δt becomes at least an error factor for time domain. Furthermore, since the RR interval evaluates the standard deviation of the 100 intervals, the error in Δt is included in the result of standard deviation. Therefore, the RR interval obtained from the heart rate by FFT using reflected radar Doppler is unreliable.

Since it is not fair to discuss only our strengths unilaterally, we will also introduce the advantages of the Doppler radar system. The biggest advantage is that only one antenna is required, so the high-frequency structure is extremely small. The distance setting range is also wide, so even if the target moves, it does not matter where it is within the line gain range. However, the target must be stationary at the moment of measurement [14-18]. To increase spatial resolution, increase the frequency. In this case, the penetration rate into the tissue decreases, so the distance should be shortened and the transmission power should be increased [19-20]. Furthermore, a detection method that emphasizes phase has been proposed to suppress magnitude wave due to respiration [21-22]. The paper [27] captures the waveform correlated with the blood flow with the reflection type, blood flow of intracranial space [7], and the paper [17] also reports the paper using the transmission microwave. A comparison of radar Doppler and the proposed method is given in Table 3 [14-26].

Table 3. Comparison of reflective radar Doppler and proposed method.
Frequency 10 GHz or more L band, 2.4 GHz
RR interval Inaccurate Exact
Aortic valve notch Not visible Visible
Heartbeat Indirect, Vibration of the chest wall Direct monitor, obtain flow wave (blood pressure)
Application Human Birds, human
4.2. Future Outlook

Birds do not have a diaphragm and their lungs are hard and thus barely expand. They thus breathe by contracting balloon-like air sacs that expand in their chest and abdomen.

The avian great vessel is the right descending aorta. The cells of the left ventricle and the interventricular septum look morphologically abnormal. The muscle fibers are short and thick. The normal avian heart presents macro-histological abnormalities in macroscopic tissue sections. T these abnormalities look similar to hypertrophic myocarditis in humans.

Cardiac ejection fraction is close to 100%, so the branch of the His bund from the atrial node is running outside of left ventricular descends laterally rather than medially. The heart rate of chickens is 200−300 beats/min, Hummingbird has more than 1,000/min, which is much faster than that of humans. Birds cannot accurately transmit electrical stimulation to the left ventricle with the same conduction pathway as mammals. Therefore, the outside of the left ventricular wall is a stimulus conduction path (myocardium degenerated like a nerve fiber).

The autonomic nervous system is controlled by a battle between the sympathetic and parasympathetic nervous systems. The indices are the standard deviation of the RR interval, systolic blood pressure, and heart rate. Parasympathetic nerves in birds are weakly suppressed and many birds suffer myocardial damage and large blood vessel damage when excited. Air sacs are controlled by the movement of respiratory muscles (skeletal muscles), but the subcutaneous tissue of birds has extremely weak connections and adhesions with the fascia, making it difficult to visually detect changes associated with respiration. Therefore, it is not possible to screen infected birds based on their appearance alone.

Infection in chickens with a highly pathogenic avian influenza virus propagates in type II alveolar epithelium and causes a cytokine storm. This leads to tachycardia, low blood pressure, shallow breathing, increased breathing rate, and slower skeletal muscle contraction. A diagnosis should be confirmed by a polymerase chain reaction test. However, at poultry farms, a simple antigen-antibody reaction is used. It takes time for the influenza virus antibody titer to reach a positive level. In settings with a lot of chickens, if the cardiac output and respiratory pattern of poultry could be monitored without restraint and contact, infected birds could be screened early and isolated from the population, preventing the culling of many birds.

The hosts of avian influenza viruses are mainly waterfowl such as ducks and swans, not chickens. A low-virulence influenza virus replicates slowly in the gut rather than the lungs, with a potential prevalence of up to 30% in nature. In poultry, on the other hand, the influenza virus has been found to spread rapidly from the intestine to the lungs, and it has been reported that even low pathogenicity causes mild respiratory symptoms. Screening for early mild respiratory distress is therefore important in the poultry industry. We recommend a system that can continuously monitor cardiac output and breathing patterns without contact.

## V. CONCLUSION

We developed a device that can monitor the cardiac output and anterior thoracic air sac pressure in birds using transmission-type microwaves (2.4−2.5 GHz). Compared to reflector-type radar Doppler devices, the proposed device can visualize blood flow and continuously measure breathing patterns that are correlated with the anterior thoracic air sac pressure. Data collection allows the early screening of unwell birds, which can be isolated to prevent mass infections in poultry houses and prevent pandemics.

Table 4. Approval number and target birds related to this exper-iment (FY-2013-2022 Tokai University, Japan).
Year Approval number Target animal
FY-2011 111114 Chicken
111115 Quail
111117 Crow
FY2012 111114 continued Chicken
122020 Chicken
122021 Quail
122022 Crow
121083 Pheasant
121111 *
FY-2013 133009 Chicken
133007 Quail
133008 Crow
132036 *
FY-2014 144009 Chicken
141007 Pheasant
141006 Crow
FY-2015 151021 Chicken
151022 Pheasant
151023 Crow
151077 Pigeon
FY-2016 162049 Chicken
162050 Pigeon
162027 Pheasant
162048 Crow
16081 Pigeon+
FY-2017 172032 Pigeon+
173034 Crow
173027 Chicken
173028 Pigeon
173032 Pheasant
FY-2018 184019 Pigeon
184020 Pheasant
183013 Pigeon+
184039 Chicken+
184040 Crow
FY2019 195020 Chicken
195021 Pigeon
194017 Pigeon+
1950027 Crow
FY-2020 201053 Crow
201054 Chicken
FY-2021 211055 Chicken
FY-2022 221073 Chicken

*: Verification of prototype portable inhalation anesthesia device.

+: Verification of airflow between air sacs with special electronic device.

## ACKNOWLEDGMENTS

The intellectual property rights of the equipment used in this research are owned by Tasada Works Inc. in Takaoka City, Toyama Prefecture (President: Mr. Masuhisa Ta). We would like to thank Dr. Kaoru Nakata, Ms. Noriko Numata, and Ms. Miyoshi Tanaka of the Tokai University School of Medicine, and Mr. Kokuryo Mitsuhashi and Ms. Megumi Amano of Seisa University for their assistance in the experiment. Pigeons (racing pigeons) were obtained courtesy of Yamazaki Pigeon (Mr. Kiyoshi Yamazaki) in Ueda City. Regarding chicken acquisition, we would like to thank Mr. Masatake Kamata (Takaoka Hikarien Poultry Farm), Veterinary Doctor Hisayoshi Nakamura (Takaoka City), Mr. Shigeru Yamamoto (Takaoka City), and Ms. Hideko Nagahara (Dog Friends) for their excellent support. Dr. Kiyoaki Ozaki and Mr. Noboru Nakamura from the Yamashina Institute for Ornithology provided guidance and suggestions.

Since 2012, this study has been submitted the Tokai University ‘Animal Experiment Plan and has undergone appropriate review. This research was funded by grants-in-aid for scientific research (grants 21241042, 23651169, 24241058, 26630165, and 15H01791) from the Ministry of Education, Culture, Sports, Science and Technology. The authors have no conflicts of interest to declare.

Every year, we have submitted experiment plans to Tokai University based on the regulations of the Japanese Ministry of Health, Labor and Welfare and the Ministry of the Environment on experimental animals, which were reviewed and approved by an expert committee. The themes covered in these applications are “Acquisition of avian physiological digital data, including anesthesia, surgical procedures, and insert electronic devices, also including free space flight data collections of crow, pheasant, and pigeon with electronic devices.”(Table 4).

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[12].

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[14].

W. J. K. Raymond, C. K. Chakrabarty, G. C. Hock, and A. B. Ghani,” Complex permittivity measurement using capacitance method from 300 kHz to 50 MHz,” Measurement, vol. 46, no. 10, pp. 3796-3801, 2013.

[15].

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[16].

M. Aweda, R. Meindinyo, S. Gbenebitse, and A. Z. Ibitoye, “Microwave radiation exposures affect cardiovascular system and antioxidants modify the effects,” Pelagia Research Library Advances in Applied Science Research, vol. 2, no. 2, pp. 246-251, 2011.

[17].

H. Tan, D. Qiao, and Y. Li, “Non-contact heart rate tracking using Doppler radar,” in Proceedings of the 2012 International Conference on Systems and Informatics (ICSAI2012), 2012, pp. 1711-1714.

[18].

M. Alizadeh, G. Shaker, and S. Safavi-Naeini, “Remote heart rate sensing with mm-wave radar,” in Proceedings of the2018 18th International Symposium on Antenna Technology and Applied Electromagnetics (ANTEM), 2018, pp. 1-2.

[19].

D. Obeid, S. Sadek, G. Zaharia, and G. E. Zein, “Doppler radar for heartbeat rate and heart rate variability extraction,” in Proceedings of the2011 E-Health and Bioengineering Conference (EHB), 2011.

[20].

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[21].

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[22].

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[23].

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[24].

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[25].

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[26].

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[27].

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[28].

P. F. Battley, Contrasting Extreme Long-Distance Migration Patterns in Bar-Tailed Godwits Limosa lapponica, 2012. https://www.researchgate.net/figure/Tracks-of-satellite-tagged-bar-tailed-godwits-on-southward-migration-with-place-names_fig1_263574008.

## AUTHORS

Isao Nakajima

He is a specially appointed professor of Seisa University and a visiting professor of Nakajima Labo. at the Dept. of Emergency Medicine and Critical Care, Tokai University School of Medicine. He got the Doctor of Applied Informatics (Ph.D.), Graduate School of Applied Informatics University of Hyogo 2009, and the Doctor of Medicine (Ph.D.), Post Graduate School of Medical Science Tokai University 1988, and the Medical Doctor (M.D.) from Tokai University School of Medicine 1980. He has been aiming to send huge multimedia data from moving ambulance via communications satellite to assist patient’s critical condition. He is a board member of the Pacific Science Congress, a Rapporteur for eHealth of ITU-D SG2 (2014-2021), and a specially appointed researcher at the Yamashina Institute for Ornithology, Japan.

Ichiro Kuwahira

He has received his MD degree from Tokai University School of Medicine, Kanagawa, in 1980. In 1984, he received a PhD degree in Tokai University Graduate School of Medicine. Since then, he has been mainly involved in respiratory medicine and physiology. He was appointed to a professor of medicine, Tokai University School of Medicine in 2005, and to a specially-appointed professor in 2020. He acted as a hospital director of Tokai University Tokyo Hospital from 2004 to 2011.

Jun-ichi Hata

He has received his MD degree from Keio University School of Medicine, Tokyo, in 1966. In 1971, he received a PhD degree in Keio University Post Graduate School of Medicine. Since then, he has been mainly involved in molecular pathology for pediatric tumors and embryonal pathology. He was appointed to a professor of pathology, Keio University School of Medicine in 1990. In 2013, he has been appointed as research director of Central Institute for Experimental Animals.


text:
[texhax] (no subject)

doug waud douglas.waud at umassmed.edu
Fri Jun 20 08:08:49 CEST 2003

Hi

Tzipmyster at aol.com wrote:
> I'm new to TeX and would like to pose the following question: How would
> I put a frame around some text {in this case, mathematical text} with a
> rule of .5pt? Is it possible to do this with plain TeX without using any

Here, I believe, is an example which does what you want:

% test file to put a box around text
\vbox{%
\hrule height 0.5 pt
\hbox{\vrule width 0.5 pt\hfill $A=B+C^2$\hfill\vrule width 0.5pt}
\hrule height 0.5 pt}

And a nicer version with some extra space

\noindent (notice that you have to
nest h and v boxes to get it right!):

\vbox{%
\hrule height 0.5 pt
\hbox{\vrule width 0.5 pt
\vbox{\vskip 1pt\hbox{\hfill\hskip 1pt
$A=B+C^2$\hskip 1pt\hfill}
\vskip 1pt}\vrule width 0.5pt}
\hrule height 0.5 pt}
\end

doug


text:
# A Formula for the Phase Difference Between the Electric and Magnetic Wave Oscillations?

A) Is there a formula for the phase difference between the electric and magnetic field oscillations, in vacuum, in an electromagnetic wave emitted from an antenna, as a function of the frequency the distance from the antenna?

B) Does the formula depend on the antenna type and on the direction of the radiation?

The electric and magnetic fields are always in-phase if the wave can be treated as a plane wave (which simply means it cannot be too close to the source), and in vacuum or any medium with linear response, such as air.

Boundary conditions of wave guides change this relationship, and must be solved for each specific case. If the wave guide is large enough, you will only see effects near the surfaces.

Transmission through a conductor results in phase-lag of the magnetic field, and a rapid extinction of the propagating wave; see skin depth.

The methods used are nicely described here: Chapter 9: Electromagnetic Waves - MIT OpenCourseWare

Electromagnetic plane wave: electric and magnetic fields are always in phase.

You can study the mathematical proof, based on Maxwell's equations for an arbitrary plane wave, here.

So the direct answers are: (A) No, except that they are always in phase in the "far field"; (B) No, see (A).

In the near field, for a dipole antenna, see Chapter 10: Antennas and Radiation - MIT OpenCourseWare, especially section 10.2.

This is why you ordinarily only need to consider the electric field with radio or light transmission, except in a wave guide, or non-linear media.

Occasionally somebody claims that the electric and magnetic fields are out of phase for circular polarization. This is not quite correct: the quarter wave plate accepts linearly polarized light at some angle wrt the fast axis of the QWP; the orthogonal components of the electric field vector under go different amounts of optical delay, resulting in two distinct electromagnetic waves, with the same phase delay in their electric and magnetic fields. See this animation of circular polarization; the input light is linearly polarized, with the red and blue representing the two components wrt the fast axis of the quarter wave plate, QWP.

• I am not following your argument about electric and magnetic fields being in phase for an electromagnetic plane wave. They must be out of phase according to Maxwell's equations (e.g., $\nabla \times \mathbf{E} = -\partial_{t} \mathbf{B}$), unless you mean something different when referring to phase. If both are linear, they oscillate along orthogonal directions to each other and to the wavevector. But they could reach their maximum at the same phase... Is that to what you refer? – honeste_vivere May 28 '16 at 16:42
• I'm interested exactly in the transition between close-field and far-field. Would there be such a formula if I concentrated specifically on a half-wave dipole antenna? – Lior May 28 '16 at 17:28
• @honeste_vivere: Yes, that is exactly what is meant; see illustration. The proof can be seen by following the link; the equation given, the Maxwell-Faraday form of Faraday's Law of Induction directly relates spatial and temporal aspects of the fields; when one goes through the details, for plane waves, the fields are in phase. – Peter Diehr May 28 '16 at 17:35
• @Lior: See additional link to Chapter 10: Antennas and Radiation - MIT OpenCourseWare, section 10.2. Equations are provided for a specific case, 10.2.2 Near fields of a Hertzian dipole, on p. 310. – Peter Diehr May 28 '16 at 17:41


text:
# [NTG-context] Overlapping terms in description lists

Tom Harrop thomas.harrop at ird.fr
Sat Apr 23 16:15:48 CEST 2016

> On 21 Apr 2016, at 1:18 pm, luigi scarso <luigi.scarso at gmail.com> wrote:
>
> the title is in  a \vtop and not a \vbox, and its height is fixed to \strutht (and the depth to \strutdp)
> so basically the title is one line.

OK, I’ll do my best to stick to one line.

I guess it would be better to use a paragraphs or column environment for long labels but the ConTeXt is generated by pandoc, which writes 'definition lists' as \startdescription \stopdescription.
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• DILIP KUMAR DAS

Articles written in Journal of Earth System Science

• Understanding the interactions between meteorological and soil moisture drought over Indian region

The present study aims at analysing the relationship between meteorological and soil moisture droughts over different meteorological regions of India based on change-point, trends and probability analysis. The long term gridded standardized-precipitation-evapotranspiration-index (SPEI) and standardized-soil-moisture-index (SSMI) data during 1980–2015 were used to address the meteorological and soil-moisture drought, respectively. The significant changes in the time-series SPEI and SSMI data at different time-scales were estimated using Pettitt’s test. The variations in change-points among different time-scales of SSMI were much lower than SPEI, specifically in parts of NW and central India. The inter-annual variability of drought parameters, i.e., duration and magnitude, showed good agreement between meteorological and soil moisture droughts for mild drought conditions. However, under moderate drought conditions, the behaviour of meteorological and soil moisture drought was different, specifically over peninsular India. The trend analysis revealed that SSMI based drought occurrences were significantly decreasing over all the four meteorological regions, whereas in case of SPEI, significant trends were observed only over peninsular India for mild drought condition. The probability of occurrences of both meteorological and soil moisture droughts were higher over major parts of India in case of mild droughts, whereas the values were lower for moderate drought conditions.

$\bf{Highlights}$

$\bullet$The Pettitt’s change point analysis revealed that variations of the period of significant changes among different time scale SSMI was much lower comparison to SPEI.

$\bullet$In major parts of the country, good agreement between SPEI and SSMI derived drought parameters, i.e., duration and magnitude, were observed for mild drought conditions. However, it was quite low in moderate drought conditions, more specifically over peninsular and central India.

$\bullet$The trend analysis revealed that the soil moisture drought was decreasing during the study period. Similar observations were recorded in meteorological drought also, but with some increasing drought duration and magnitude in parts of NW and NE India.

$\bullet$The probability of occurrences of both meteorological and soil moisture droughts were higher over major parts of India in case of mild droughts, whereas the values were lower for moderate drought conditions.

• # Journal of Earth System Science

Volume 129, 2020
All articles
Continuous Article Publishing mode

• # Editorial Note on Continuous Article Publication

Posted on July 25, 2019