Monday, 18 December 2006

Olimpiade Astronomi

Olimpiade Astronomi

I  A  O

The International Astronomy Olympiad

The International Astronomy Olympiad is a scientific-educating event for students of the junior high school classes - 14-18 years old, which includes an intellectual competition between these students. The style of the problems is aimed at developing the imagination, creativity and independent thinking.

The Olympiad is carried out in the spirit of friendship, tolerance, where the competition is a stimulus for showing the participants' capabilities but contacts, exchange of ideas and collaboration between students, teachers and scientists from various countries have a primary importance. It should be emphasized that the competition at the International Astronomy Olympiad is not the most important part of the programme, in other words the competition with its points, places and Diploma is only a tool but not the purpose of the Olympiad, and the Olympiad is not "selection of the best students" and not "examinations". 

IAO takes place each year in the first two months of the astronomical autumn (i.e. September 22 - November 22) in either an observatory, a scientific research centre (town) or at an institute of the participating countries. IAO is organized by the Authorized National Representative (Astronomical) Organization (ANRAO) of the organizing country.

ANRAO of every participating country may send five students, who are the winners of the respective national olympiads or easy competitions: three students for the junior group (14-15 years old) and two students to the senior group (16-18 years old). They are to be accompanied by two team leaders as representatives of each country.
Countries which did not registered their ANRAO may send up to three students: two students for the junior group (14-15 years old) and one student to the senior group (16-18 years old). They are also to be accompanied by two team leaders.

Observers may be at the Olympiad as well.

The Olympiad is intended for students of the adolescent age when the interest to astronomy is being formed most actively. Then preparation for participation is still a useful and purposeful education not yet having turned training into an end in itself. This is a principal idea of the Olympiad.

A considerable part of the knowledge which is necessary for participation in the Olympiad goes beyond the frames of the school curriculum in most of the countries in the world. Preparation for the Olympiad needs extra-curricular activities of various types, whose promoting is one of the Olympiad's aims.

The total amount of knowledge is comparatively not too large and acquiring it is completely within the reach of the students of the above-mentioned age. It is also the age for which the spirit of the Olympiad as an entertaining competition is most appropriate. Involving students of the senior classes would lead to enlarging and complicating the necessary material, which would be senseless. For such students it already turns into a kind of sport and is significantly less useful from an educational point of view. For students 17-20 years old more serious events are preferable like conferences and competitions of research projects.

The International Olympiad is also a meeting of teachers and scientists, where exchange of ideas takes place, methods for refining the astronomical education are discussed and international collaboration in this field is established. 

Aims of the IAO

In recognition that:
  • astronomy plays a fundamental role in the human progress and will acquire more global importance in the 21 century;
  • astronomical knowledge is an important part of the culture of our civilization and an essential factor in forming the view of life and the way of thinking of the young people in the modern world;
The Olympiad sets itself the aims of:
1) Attracting the most talented young people to professional careers in the field of astronomy:
  • giving an opportunity for showing their capabilities and encouraging the best students;
  • helping the young people in choosing a profession;
  • acquainting the participants with the real working conditions and the nature of the research work in the hosting observatory (institute), exchange of ideas and knowledge between astronomers, teachers and students.
2) Spreading astronomical knowledge among as many students as possible and improving the astronomical education:
  • provoking an interest in astronomy, physics and astronautics in a greater number of children and young people;
  • popularization of natural science and the scientific approach in astronomy and related sciences;
  • encouraging teachers to work for improving, enriching and enlarging school astronomy education and including more children in it;
  • activating the astronomy education in the junior classes of the high school;
  • promoting extracurricular activities in amateur clubs, scientific societies, circles etc.;
  • stimulating organization of national astronomical Olympiads in different countries;
  • enhancing international contacts in the field of astronomy and physics education in schools.
3) Stimulating the imagination and creativity of children:
  • the character of the astronomical tasks enables putting the students in non-standard situations, very close to those of a real scientific research; they can require creating hypotheses, assuming approximations, choosing between a multitude of factors that could be taken into account or neglected, and independent decision about the form of the needed answers;
  • the astronomical tasks can set unusual problems for which solving a free way of thinking, fantasy and inventiveness are necessary;
  • astronomical tasks allow a variety of completely different, although correct solutions applying different original approaches.
4) Cultivating a spirit of correctness and friendship:
  • the International Astronomy Olympiad is a meeting of young people from different countries - future colleagues in the scientific exploration who will have to work in cooperation;
  • in the time of the Olympiad favourable conditions are created for active contacts between students, teachers and specialists from the hosting observatory (institute) aimed at an intense exchange of knowledge, education and upbringing of the children during all stages of the event.



Competitions for young amateur astronomers, called "Olympiads", are carried out in former USSR for many years. The first attempts to organize a competition in an international (i.e. multilanguage) scale have been done in scientific town of Soviet Academy of Sciences "Chernogolovka" in 1990-1991 as a part of the so called Olympiad of Naukograds and Scientific Centres (NSC Olympiad). Teams of Russia, Moscow, Latvia and Estonia participated. These events provided ground for the future IAO proper.

Positive experience during international olympiads in other natural sciences (Physics, Chemistry, Biology), Mathematics and Informatics led to the idea of starting an International Astronomy Olympiad.

Juridical founding of the International Astronomy Olympiad has been done by Euro-Asian Astronomical Society on June, 7, 1996, as an annual astronomy competition for high-school students. 

Immediately after the founding the Olympic Coordinating Council was established.

The first official IAO was held in the Special Astrophysical Observatory of Russian Academy of Sciences in November 1996 as a part of programme of the Autumn Astronomy School for Young Astrophysists. Notwithstanding some initial difficulties, this olympiad was a great success and it was decided to continue with the IAO. In subsequent olympiads the number of participating states increased.

So the Olympiad of Naukograds and Scientific Centres and Autumn Astronomy School of Young Astrophysists in the Special Astrophysical Observatory may be considered as father and mother of the International Astronomy Olympiad.


Saturday, 18 November 2006

Olimpiade Astronomi

Olimpiade Astronomi

Dalam seleksi olimpiade Provinsi di Hotel Bumi Makmur Indah, saya mendapatkan berbagai macam pengalaman yang berharga, banyak teman dan suasana baru yang saya dapatkan.

Saya belum berhasil melewati seleksi tingkat provinsi tersebut, namun saya berharap suatu hari nanti ada adik-adik saya yang berhasil melewati seleksi tersebut.

Sayapun mempunyai impian untuk memperdalam ilmu astronomi ini.

Wednesday, 18 October 2006

Olimpiade Astronomi

Olimpiade Astronomi

Akhirnya sayapun berhasil melewati seleksi sekolah dan seleksi kota dalam menyelesaian pertandingan olimpaide astronomi tersebut, hal ini merupakan pengalaman yang menyenangkan.

Tuesday, 10 October 2006

The Discovery of the Blackbody Form and Anisotropy of the Cosmic Microwave Background Radiation

The Royal Swedish Academy of Sciences has decided to award the Nobel Prize in Physics for 2006 jointly to

John C. Mather
NASA Goddard Space Flight Center, Greenbelt, MD, USA,

George F. Smoot
University of California, Berkeley, CA, USA

"for their discovery of the blackbody form and anisotropy of the cosmic microwave background radiation".


Pictures of a newborn Universe

This year the Physics Prize is awarded for work that looks back into the infancy of the Universe and attempts to gain some understanding of the origin of galaxies and stars. It is based on measurements made with the help of the COBE satellite launched by NASA in 1989.

The COBE results provided increased support for the Big Bang scenario for the origin of the Universe, as this is the only scenario that predicts the kind of cosmic microwave background radiation measured by COBE. These measurements also marked the inception of cosmology as a precise science. It was not long before it was followed up, for instance by the WMAP satellite, which yielded even clearer images of the background radiation. Very soon the European Planck satellite will be launched in order to study the radiation in even greater detail.

According to the Big Bang scenario, the cosmic microwave background radiation is a relic of the earliest phase of the Universe. Immediately after the big bang itself, the Universe can be compared to a glowing "body emitting radiation in which the distribution across different wavelengths depends solely on its temperature. The shape of the spectrum of this kind of radiation has a special form known as blackbody radiation. When it was emitted the temperature of the Universe was almost 3,000 degrees Centigrade. Since then, according to the Big Bang scenario, the radiation has gradually cooled as the Universe has expanded. The background radiation we can measure today corresponds to a temperature that is barely 2.7 degrees above absolute zero. The Laureates were able to calculate this temperature thanks to the blackbody spectrum revealed by the COBE measurements.

COBE also had the task of seeking small variations of temperature in different directions (which is what the term 'anisotropy' refers to). Extremely small differences of this kind in the temperature of the cosmic background radiation – in the range of a hundred-thousandth of a degree – offer an important clue to how the galaxies came into being. The variations in temperature show us how the matter in the Universe began to "aggregate". This was necessary if the galaxies, stars and ultimately life like us were to be able to develop. Without this mechanism matter would have taken a completely different form, spread evenly throughout the Universe.

COBE was launched using its own rocket on 18 November 1989. The first results were received after nine minutes of observations: COBE had registered a perfect blackbody spectrum. When the curve was later shown at an astronomy conference the results received a standing ovation. 

The success of COBE was the outcome of prodigious team work involving more than 1,000 researchers, engineers and other participants. John Mather coordinated the entire process and also had primary responsibility for the experiment that revealed the blackbody form of the microwave background radiation measured by COBE. George Smoot had main responsibility for measuring the small variations in the temperature of the radiation.

Read more about this year's prize 
Information for the Public (pdf)
Scientific Background (pdf)
To read the text you need Acrobat Reader.
Links and Further Reading

John C. Mather, born 1946 (60), (US citizen). PhD in Physics in 1974 from the University of California at Berkeley, CA, USA. Senior Astrophysicist at NASA's Goddard Space Flight Center, Greenbelt, MD, USA. 

George F. Smoot, born 1945 (61) in Yukon, FL, USA, (US citizen). PhD in Physics in 1970 from MIT, Cambridge, MA, USA. Professor of Physics at the University of California, Berkeley, CA, USA.

Prize amount:
SEK 10 million to be shared equally between the Laureates

Contact persons:

Malin Lindgren, Information Officer, Phone +46 8 673 95 22, +46 709 88 60 04,
Ulrika Björkstén, Scientific Editor, Phone +46 8 673 95 00, +46 702 06 67 50,

Monday, 18 September 2006

Olimpiade Astronomi

Olimpiade Astronomi.

Alhamdulilah saya mengikuti seleksi peserta olimpiade astronomi tingkat sekolah, dan saya berusaha belajar banyak mengenai ilmu yang sangat menyenagkan ini.

Saya berusaha mempelajarinya dengan seksama bersama teman-teman sekelas saya.

Sejarah Singkat

Pada bagian awal sejarahnya, astronomi memerlukan hanya pengamatan dan ramalan gerakan benda di langit yang bisa dilihat dengan mata telanjang. Rigveda menunjuk kepada ke-27 rasi bintang yang dihubungkan dengan gerakan matahari dan juga ke-12 Zodiak pembagian langit. Yunani kuno membuatkan sumbangan penting sampai astronomi, di antara mereka definisi dari sistem magnitudo.

Alkitab berisi sejumlah pernyataan atas posisi tanah di alam semesta dan sifat bintang dan planet, kebanyakan di antaranya puitis daripada harfiah; melihat Kosmologi Biblikal. Pada tahun 500 M, Aryabhata memberikan sistem matematis yang mengambil tanah untuk berputar atas porosnya dan mempertimbangkan gerakan planet dengan rasa hormat ke matahari.

Penelitian astronomi hampir berhenti selama abad pertengahan, kecuali penelitian astronom Arab. Pada akhir abad ke-9 astronom Muslim al-Farghani (Abu'l-Abbas Ahmad ibn Muhammad ibn Kathir al-Farghani) menulis secara ekstensif tentang gerakan benda langit.

Karyanya diterjemahkan ke dalam bahasa Latin di abad ke-12. Pada akhir abad ke-10, observatorium yang sangat besar dibangun di dekat Teheran, Iran, oleh astronom al-Khujandi yang mengamati rentetan transit garis bujur Matahari, yang membolehkannya untuk menghitung sudut miring dari gerhana.

Di Parsi, Umar Khayyām (Ghiyath al-Din Abu'l-Fath Umar ibn Ibrahim al-Nisaburi al-Khayyami) menyusun banyak tabel astronomis dan melakukan reformasi kalender yang lebih tepat daripada Kalender Julian dan mirip dengan Kalender Gregorian.

Selama Renaisans Copernicus mengusulkan model heliosentris dari Tata Surya. Kerjanya dipertahankan, dikembangkan, dan diperbaiki oleh Galileo Galilei dan Johannes Kepler. Kepler adalah yang pertama untuk memikirkan sistem yang menggambarkan dengan benar detail gerakan planet dengan Matahari di pusat. Tetapi, Kepler tidak mengerti sebab di belakang hukum yang ia tulis. Hal itu kemudian diwariskan kepada Isaac Newton yang akhirnya dengan penemuan dinamika langit dan hukum gravitasinya dapat menerangkan gerakan planet.

Bintang adalah benda yang sangat jauh. Dengan munculnya spektroskop terbukti bahwa mereka mirip matahari kita sendiri, tetapi dengan berbagai temperatur, massa dan ukuran. Keberadaan galaksi kita, Bima Sakti, dan beberapa kelompok bintang terpisah hanya terbukti pada abad ke-20, serta keberadaan galaksi "eksternal", dan segera sesudahnya, perluasan Jagad Raya dilihat di resesi kebanyakan galaksi dari kita.

Kosmologi membuat kemajuan sangat besar selama abad ke-20, dengan model Ledakan Dahsyat yang didukung oleh pengamatan astronomi dan eksperimen fisika, seperti radiasi kosmik gelombang mikro latar belakang, Hukum Hubble dan Elemen Kosmologikal. Untuk sejarah astronomi yang lebih terperinci, lihat sejarah astronomi.

Semoga Bermanfaat

Monday, 21 August 2006

Fisika SMA

Pengalaman Belajar Fisika di SMAN BI 1 Banjar

Ada Apa Dengan Fisika?


In the paper airplane activity students select and build one of five different paper airplane designs and test them for distance and for time aloft. Part of this activity is designed to explore NASA developed software, FoilSim, with respect to the lift of an airfoil and the surface area of a wing.


Technology Needed

  • Internet Access
  • Graphing Calculator (optional)

Time Required

  • 2-3 class periods

Classroom Organization

  • Students should work in groups of 3 or 4.


  1. Give students a sheet of unlined paper and instructions for construction of a paper airplane (See download above).

  2. Students should give their plane a name using the aviation alphabet. (Example N 831 FE represents November 831 Foxtrot Echo. Identification numbers and letters must not exceed 7; and the identification must begin with N, which stands for the United States.)

  3. Students should determine the area of the wings of their planes. If students are able, have them unfold their planes and lay out basic geometric shapes to fill the wing area. Then have them calculate the total area from the sum of the areas of the shapes. (See example. Use "back arrow" to return here.)

    If students are not able to calculate geometric areas, they could make a duplicate plane, cut off the wings, and lay the wings onto measured grids or pieces of graph paper and count the total squares covered, estimating partial squares.

    A variation of this technique that eliminates a duplicate plane and cutting wings is to draw or trace a grid on a blank transparency with a sharpie marker and then hold the clear grid over the wings to count squares covered.

  4. Have students fly their planes in the gym or hallway or other large indoors area (to eliminate wind effects) five times, each time trying for maximum distance. Stress trying to duplicate the same launch angle and speed. Now do another five trials, this time trying for maximum time aloft. Students should record their distances and times and average the three longest distances and the three longest times.

  5. Have students put their data onto a graph for the class, one graph of time aloft vs. wing area and the other of distance vs. wing area.

  6. Discuss the results from the graphs as a class, and then ask for predictions as to what would happen if the wings were made smaller.

  7. Have the students draw a line two centimeters from and parallel to the trailing edges of their wings, and then cut that 2 cm portion off the wings (Shown in red).
    Picture of paper airplane with

    tail section removed.
    The cut off part should be tucked on the inside of the plane when it is refolded in order to keep mass constant. You might ask the class to provide an explanation for doing this.

  8. Repeat steps three through six.

  9. Have the students investigate their results using FoilSim. They should set the ANGLE OF ATTACK to 5 degrees and then vary only the area of the wing and note the effect on the value of LIFT. They can compare these results to their own experimental results.

  10. ADDITIONAL QUESTION: "Why don't all planes have the biggest wing area possible? Why do some fighter jets have small wings?" (ANSWER: There are other factors that contribute to lift, such as velocity and shape of the wing. The weight of a plane is also very important.) Students can investigate these other factors by going through the lessons that are part of FoilSim.

Extension Activity

Assessment Strategies/Evaluation

  1. Each group could make a presentation on their airplane and what made its design successful.
  2. Students could individually graph the experimental data and make a report.
  3. Challenge students to fold a better plane and explain the reasons for changes in design.
  4. Students could write a summary of experimental results and relate the variables tested.

Supplementary Resources



Friday, 18 August 2006

Astronomi Indonesia

Masyarakat tradisional

Seperti kebudayaan-kebudayaan lain di dunia, masyarakat asli Indonesia sudah sejak lama menaruh perhatian pada langit. Keterbatasan pengetahuan membuat kebanyakan pengamatan dilakukan untuk keperluan astrologi. Pada tingkatan praktis, pengamatan langit digunakan dalam pertanian dan pelayaran. Dalam masyarakat Jawa misalnya dikenal pranatamangsa, yaitu peramalan musim berdasarkan gejala-gejala alam, dan umumnya berhubungan dengan tata letak bintang di langit.

Nama-nama asli daerah untuk penyebutan obyek-obyek astronomi juga memperkuat fakta bahwa pengamatan langit telah dilakukan oleh masyarakat tradisional sejak lama. Lintang Waluku adalah sebutan masyarakat Jawa tradisional untuk menyebut tiga bintang dalam sabuk Orion dan digunakan sebagai pertanda dimulainya masa tanam. Gubuk Penceng adalah nama lain untuk rasi Salib Selatan dan digunakan oleh para nelayan Jawa tradisional dalam menentukan arah selatan. Joko Belek adalah sebutan untuk Planet Mars, sementara lintang kemukus adalah sebutan untuk komet. Sebuah bentangan nebula raksasa dengan fitur gelap di tengahnya disebut sebagai Bimasakti.

Masa modern

Pelaut-pelaut Belanda pertama yang mencapai Indonesia pada akhir abad-16 dan awal abad-17 adalah juga astronom-astronom ulung, seperti Pieter Dirkszoon Keyser dan Frederick de Houtman. Lebih 150 tahun kemudian setelah era penjelajahan tersebut, misionaris Belanda kelahiran Jerman yang menaruh perhatian pada bidang astronomi, Johan Maurits Mohr, mendirikan observatorium pertamanya di Batavia pada 1765. James Cook, seorang penjelajah Inggris, dan Louis Antoine de Bougainville, seorang penjelajah Perancis, bahkan pernah mengunjungi Mohr di observatoriumnya untuk mengamati transit Planet Venus pada 1769[1].
Ilmu astronomi modern makin berkembang setelah pata tahun 1928, atas kebaikan Karel Albert Rudolf Bosscha, seorang pengusaha perkebunan teh di daerah Malabar, dipasang beberapa teleskop besar di Lembang, Jawa Barat, yang menjadi cikal bakal Observatorium Bosscha, sebagaimana dikenal pada masa kini.

Penelitian astronomi yang dilakukan pada masa kolonial diarahkan pada pengamatan bintang ganda visual dan survei langit di belahan selatan ekuator bumi, karena pada masa tersebut belum banyak observatorium untuk pengamatan daerah selatan ekuator.

Setelah Indonesia memperoleh kemerdekaan, bukan berarti penelitian astronomi terhenti, karena penelitian astronomi masih dilakukan dan mulai adanya rintisan astronom pribumi. Untuk membuka jalan kemajuan astronomi di Indonesia, pada tahun 1959, secara resmi dibuka Pendidikan Astronomi di Institut Teknologi Bandung.

Pendidikan Astronomi di Indonesia secara formal dilakukan di Departemen Astronomi, Institut Teknologi Bandung. Departemen Astronomi berada dalam lingkungan Fakultas Matematika dan Ilmu Pengetahuan Alam (FMIPA) dan secara langsung terkait dengan penelitian dan pengamatan di Observatorium Bosscha.
Lembaga negara yang terlibat secara aktif dalam perkembangan astronomi di Indonesia adalah Lembaga Penerbangan dan Antariksa Nasional (LAPAN).

Selain pendidikan formal, terdapat wadah informal penggemar astronomi, seperti Himpunan Astronomi Amatir Jakarta, serta tersedianya planetarium di Taman Ismail Marzuki, Jakarta yang selalu ramai dipadati pengunjung.

Perkembangan astronomi di Indonesia mengalami pertumbuhan yang pesat, dan mendapat pengakuan di tingkat Internasional, seiring dengan semakin banyaknya pakar astronomi asal Indonesia yang terlibat dalam kegiatan astronomi di seluruh dunia, serta banyaknya siswa SMU yang memenangi Olimpiade Astronomi Internasional maupun Olimpiade Astronomi Asia Pasific.

Demikian juga dengan adanya salah seorang putra terbaik bangsa dalam bidang astronomi di tingkat Internasional, yaitu Profesor Bambang Hidayat yang pernah menjabat sebagai vice president IAU (International Astronomical Union).

Semoga Bermanfaat

Thursday, 17 August 2006

Pembuktian Tuanya Peradaban Khatulistiwa

Tulisan ini saya ambil dari sebuah forum bersama tulisan dari seseorang yang mengunakan nickname sharsono, yang kemudian banyak blog dan site yang memuat nya dikarenakan “artikel” tersebut tulisannya yang berbobot disampaikan dengan cerdas dan relative mudah dicerna…..

Bukan bermaksud untuk menghargai diri sendiri (budaya negeri sendiri) secara berlebihan, tetapi sudah selayaknya kita tidak kehilangan harga diri ketika kita merasa budaya kita berumur sudah sangat tua, tidak ada sama sekali untuk niatan bersombong – sombong ria tetapi untuk membangkitkan motivasi dari rasa rendah diri kita, ketika kita merasa tidak tidak mempunyai pijakan untuk berdiri sama tinggi dengan budaya yang lain.

Memang artikel tersebut cukup panjang tapi nila-nilai yang terkandung dalam artikel tersebut layak untuk dibaca dan mendapat kan nilai dari saya tooooooop markotop binti maknyuuuuus :D

Asimilasi Agama dan Budaya

Perpaduan unsur agama yang terjadi dinegara kita Indonesia terutama dalam lingkungan masyarakat Jawa, sudah lazim dan tidak asing lagi, yang nalurinya sampai sekarang ini masih sering kita jumpai.

Misalnya unsur agama Hindu berpadu dengan agama Budha sehingga diantaranya terdapat sebutan Sang Hyang Siwa Budha, Naluri Jawa berpadu dengan unsure agama Hindu dan Agama Islam, misalnya soal Slametan / Kenduri dengan doa Islam. Itu semua menjadi kenyataan sejarah.

Jadi kesimpulannya bahwa para leluhur orang Indonesia sejak jaman purba sebelum bermacam-macam agama datang menggenangi pulau Jawa khususnya dan bumi Nusantara umumnya, mereka sudah mempunyai system PENYEMBAH kepada Tuhan Yang Maha Esa, yang sekarang lazim disebut dengan istilah PAWUKON.

Dengan demikian sejak jaman Purba leluhur kita pada hakekatnya sudah mengenal Tuhan, tidak seperti perkiraan dan pendapat beberapa orang jaman sekarang, bahwa pada waktu sebelum datangnya Agama Hindu dan Budha di Jawa, nenek moyang kita dikira belum mengenal Tuhan dan menyembah Roh. Menurut saya itu salah terka.

Pawukon itu tidak hanya sarana untuk menyembah Tuhan tetapi digunakan juga untuk pedoman mengetahui watak seseorang yang mirip horoscope. Oleh karena itu mungkin kita perlu mengungkap cerita tentang pengetahuan ilmu perbintangan peninggalan leluhur kita yang ada hubungannya dengan soal Pawukon.

Gule Angsa dan Kambing

Wah keliatannya mesti agak serius nih mengungkap pengetahuan ilmu perbintangan peninggalan leluhur kita yang erat hubungannya dengan Pawukon..

Dalam buku “Pustoko Rojo Purwo” gubahan Mas Ronggowarsito ada cerita lakon “WATUGUNUNG” yang mengisahkan hidupnya sejak kecil sampai menjadi raja di Negara Gilingwesi yang diuraikan sbb:

Dewi Sinto yang keluar dari dalam tanah, untuk dapat bertemu dengan Begawan Wrahaspati harus memasak daging BANYAK (Angsa) dan daging Wedhus Pedro (Domba).

Cerita tersebut sesungguhnya adalah kiasan sandi ilmu perbintangan, yaitu Wrahaspati adalah Planet Jupiter, Dewi Sinto yang keluar dari tanah melambangkan Bumi (Dunia). Banyak/Angsa adalah lambang gugusan bintang “Banyak Angrem” yaitu bintang Scorpio. Sedangkan Wedhus Pedro atau Domba adalah bintang Aries.

Bagi orang awam yang membaca ceritanya tanpa mengetahui bahasa samarannya (sandi) pasti mengira bahwa Dewi Sinto itu bener-bener harus memasak campuran daging kambing dan daging angsa untuk menemui Wrahaspati. Bikin gule buat sajen … ha ..ha

Dengan sedikit bersabar dan menggunakan nalar kita sekarang bisa melihat bahwa cerita roman tersebut sesungguhnya adalah pelajaran ilmu astrologi.

Selanjutnya dalam cerita diterangkan bahwa pertemuan Wrahaspati dengan Dewi Sinto menghasilkan anak yang namanya RADITE yaitu Matahari yang dibesarkan oleh Begawan Radi titisannya Batara Surya. Pertemuan Wrahaspati dengan Dewi Landep menghasilkan anak R.Wukir. Pertemuan Wrahaspati dengan Dewi Somo (Bulan) menghasilkan anak Anggoro (Mars), Sukro (Venus) dan Budo (Mercury).

Pertemuan Dewi Somo dengan RADITE menghasilkan anak Dewi Tumpak (Saturnus). Raden Radite ini akhirnya menjadi raja Watugunung.

Sinto, Landep, Wukir dan Watugunung termasuk menjadi nama Wuku dan ke 26 nama wuku lainnya adalah nama anak Wrahaspati yang lahir dari beberapa istri lainnya. He..he..he… rupanya Wrahaspati ini jagoan ditempat tidur, … jadi kalau ada istri tidak puas jangan cari Jin … tapi carilah Wrahaspati ditanggung memuaskan.

Nah, sekarang jelaskan bahwa cerita Watugunung itu sesungguhnya adalah cerita sandi lambing ilmu perbintangan yang menyebutkan semua nama planit dengan lingkaran zodiac dan equator langit. Jelas bahwa leluhur orang Indonesia jaman purba sudah mempunyai dan mengerti ilmu perbintangan yang lengkap, baik dalam bidang astronomi maupun astrologi seperti yang terdapat dalam Pawukon yang juga dipakai untuk mengetahui watak seseorang.

Pertanyaan selanjutnya adalah berapa tahunkah umur ilmu perbintangan bangsa Indonesia itu ?

Sesuai dengan makna yang tersirat dibelakang cerita Sinto ketemu Wrahaspati, persilangan dua garis lingkaran yang disebut equinox itu adalah menggesernya Banyak Angrem ke Wedhus Pedro atau Scorpio ke Aries, yaitu Bintang No.1 ke Bintang No.8.

Bergesernya Equinox dari arah Scorpio menuju ke Aries adalah melalui 7 gugus bintang yaitu Libra, Virgo, Leo, Cancer, Gemini, Taurus dan Aries.

Lamanya perjalanan waktu bergeser untuk tiap-tiap gugus bintang adalah 2156 tahun. Jadi total waktu perjalanan Scorpio ke Aries adalah 7 X 2156 tahun = 15.092 tahun.

Menurut ketentuan dalam ilmu perbintangan, untuk mengetahui umur berkembangnya ilmu perbintangan diberbagai Negara, ialah dengan memperhitungkan letak Equinox bagi ilmu perbintangan milik sesuatu Negara pada kedudukan mulai saat berkembangnya, kemudian dihitung jalan menggesernya sampai ke letak bintang Aries, yang menjadi pangkal menghitung posisi bintang pada garis lingkar zodiac.

Sesuai uraian diatas, dalam ilmu perbintangan kuno Indonesia, equinoxnya pada permulaannya berada diarah bintang Banyak Angrem (Scorpio) yang dalam perjalanannya ke Wedhus Pedro (Aries) melalui 7 gugus yang membutuhkan waktu 15.092 tahun lamanya. Jadi sesuai dengan pedoman ketentuan dari ilmu perbintangan maka umur ilmu perbintangan Indonesia sudah lebih dari 15.000 tahun.

Nol derajat equinox Aries itu terjadi pada tahun 108 Sebelum Masehi. Sedangkan sekarang ini pada saat Pak Biduanmanstrip membuat catatan ini adalah tahun 2007.

Jadi sampai waktu sekarang ini equinox tersebut sudah berjalan menggeser dari letak Aries selama (108 + 2007) 2115 tahun mendekati letak bintang Pisces. Dengan demikian berkat sandinya Pak Ronggowarsito maka kita dapat menghitung dengan pasti bahwa ilmu perbintangan Indonesia yang disebut Pawukon ini sudah 

mencapai (15.092 + 2115) 17.207 tahun lamanya.
Meskipun banyak yang sudah tidak mengerti artinya, tetapi ilmu perbintangan Indonesia yang disebut Pawukon ini sampai sekarang masih dikenal sebagian besar masyarakat, karena itu masih dicantumkan dalam “Tanggalan” untuk berbagai kepentingan adat dan keagamaan.

Perbandingan Pawukon dengan ilmu perbintangan bangsa lain
  1. Bulatan ditengah D = Dunia
    2. Lingkaran 1-R-7-S = Equator Langit
    3. Lingkaran 1-0-7-P = Zodiac terdiri dari 12 gugus bintang.
    4. Letak Aries adalah pangkal untuk menghitung jalan menggesernya equinox
    5. Panah B = arah jalan menggesernya equinox dari scorpio ke Aries.
    6. Equinox atau Titik Lente berada di Aries tahun 108 SMDengan terungkapnya umur ilmu perbintangan Pawukon yang sudah mencapai 17.000 tahun, ini menambah bukti lagi bahwa bangsa Indonesia adalah penduduk asli Bumi Nusantara yang sudah memiliki peradaban sendiri. Bukan keturunan Turki, Baghwan atau Mocetung yang katanya datang 300 tahun SM.Bangsa Indian
    Melalui penghitungan bergesernya equinox tersebut, diperhitungkan bahwa penduduk tertua di Peru telah memiliki ilmu perbintangan yang umurnya 15.000 tahun karena titik lentenya berada pada Libra. Pergeseran dari Libra sampai Aries melalui 6 gugus bintang yaitu 6 X 2156 = 12.936 + Letak equinox kini (108+2007) = 15.051 Tahun
    Dengan demikian terbukti melalui Pawukon ini bahwa Peradaban di Indonesia sudah ada ketika berkembangnya peradaban Bangsa Kulit Merah (Indian) yang hidup sejak jaman Lemuria dan Atlantis di bagian bumi sebelah barat. Sedangkan Bangsa Indonesia adalah bangsa Kulit Sawo Matang dengan peradabannya berkembang dibagian bumi sebelah timur.Bangsa Cina
    Ilmu perbintangan tertua dari bangsa Cina adalah 13000 tahun, dilihat pada titik lentenya yang berada pada perumahan bintang Virgo. Perjalanan dari Virgo – Leo – Cancer – Gemini – Taurus – Aries melalui 5 gugus bintang, yang membutuhkan waktu 5 X 2156 = 10.780 + letak equinox kini (108+2007) = 12.895 Tahun.
Bangsa Babilonia
Ilmu perbintangannya berumur 6500 tahun, dilihat dari letak titik lente yang berada pada bintang Gemini. Pergeseran dari Gemini sampai Aries melalui 2 gugus bintang Taurus dan aries yang membutuhkan waktu 2 X 2156 = 4312 + letak equinox kini (108+2007) = 6427 tahun.
Bangsa India (Hindia Belakang)
Ilmu perbintangannya disebut Surya Sidhanta dan titik lentenya dimulai dari Taurus yang dalam pergerakannya menuju ke Aries membutuhkan waktu 2156 tahun lamanya. Sampai dengan tahun ini dimana Pak Musha melakukan pencerahan Nasional yaitu tahun 2007 maka ilmu perbintangan Raden Mas AjiSaka beru mencapai 2156 + (108+2007) = 4271 tahun umurnya.


Terima Kasih

Monday, 31 July 2006

Fisika SMA

Pengalaman Belajar Fisika di SMAN BI 1 Banjar

Ada Apa Dengan Fisika?
Lift Formula

SUBJECT: Aeronautics
DESCRIPTION: A set of problems dealing with the aerodynamic lift equation.
CONTRIBUTED BY: Carol Hodanbosi
EDITED BY: Jonathan G. Fairman - August 1996


To understand for lift formula that determines aircraft lift capabilities.


An aircraft's lift capabilities can be measured from the following formula:
L = (1/2) d v2 s CL
  • L = Lift, which must equal the airplane's weight in pounds
  • d = density of the air. This will change due to altitude. These values can be found in a I.C.A.O. Standard Atmosphere Table.
  • v = velocity of an aircraft expressed in feet per second
  • s = the wing area of an aircraft in square feet
  • CL = Coefficient of lift , which is determined by the type of airfoil and angle of attack.
The angle of attack and CL are related and can be found using a Velocity Relationship Curve Graph (see Chart B below).


  1. You are flying an F-117A fully equipped, which means that your aircraft weighs 52,500 pounds. You want to maintain equilibrium in straight and level flight at an altitude of 30,000 feet, cruising at 400 knots to conserve fuel. The aircraft's wing area is 1,140 square feet. At what angle of attack should the F-117A be set to maintain equilibrium? ** Additional information you need to know to solve this problem:
    How are knots related to mph to ft/s? See conversion below
    What is air density at 30,000 feet? See following chart (I.C.A.O.). chart A
    What is the relationship between CL and angle of attack?
    Use following Chart B to determine this value.

    1 knot = 1 nautical mile per hour = 6,076 ft per hour
    1 mph = 1 mile per hour = 5,280 feet per hour
  2. To maintain an altitude of 45,000 feet at a constant speed requires an angle of attack of 4° . Your F-22 weighs 56,450 pounds, with a wing area of 830 square feet. To stay in level flight , what should your air speed indicate? The new F-22 air speed is indicated in Mach speed, not in knots. ** Additional information you need to know to solve this problem:
    How does Mach speed convert from feet per second or from knots?
    An angle of attack of 4° converts to what CL?
    What is the density of air at 45,000 feet?
    Mach 1 is the speed of sound, which varies with the air density. See the following I.C.A.O. chart (chart A) below. Mach 2 is twice the speed of sound, Mach 3 is three times the speed of sound.
    Use the Velocity Relationship Curve Graph chart below, Chart B, to find the appropriate CL from the given angle of attack.
  3. At what velocity should you be traveling in the F-22 to maintain an 0° angle of attack at an altitude of 42,000 feet? Your weight is 57,600 pounds
  4. Using the same plane (F-22) the same weight (57,600 pounds) and angle of attack, what should be its cruising speed at 36,000 feet? While descending from 42,000 feet to 36,000 feet the plane covers a horizontal distance of 40 nautical miles. What is the decent angle in degrees?

Chart A
I.C.A.O. Standard Atmosphere Table
Altitude Density Speed of Sound (Feet) (d) (Knots) 0 .002377 661.7 1,000 .002308 659.5 2,000 .002241 657.2 3,000 .002175 654.9 4,000 .002111 652.6 5,000 .002048 650.3 6,000 .001987 647.9 7,000 .001927 645.6 8,000 .001868 643.3 9,000 .001811 640.9 10,000 .001755 638.6 15,000 .001496 626.7 20,000 .001266 614.6 25,000 .001065 602.2 30,000 .000889 589.5 35,000 .000737 576.6 36,089* .000706 573.8 40,000 .000585 573.8 45,000 .000460 573.8 50,000 .000362 573.8 55,000 .000285 573.8
* Geopotential of Tropopause

Chart B
Velocity Relationship Curve Graph

Friday, 21 July 2006

Fisika SMA

Pengalaman Belajar Fisika di SMAN BI 1 Banjar

Ada Apa Dengan Fisika?
Experiment on Fluids:
Finding the Velocity of a Fluid in a Confined Container

SUBJECT: Aeronautics
TOPIC: Fluid Velocity
DESCRIPTION: A set of mathematics problems dealing with fluid velocity.
CONTRIBUTED BY: Carol Hodanbosi
EDITED BY: Jonathan G. Fairman - August 1996


To calculate the velocity of a confined fluid, given the cross-section area and velocity of another region.


The drawing below is a cross-section of a circular cone attached to a circular cylinder.

When a fluid (liquid or gas) is in a confined space, with no change in pressure or temperature, one can use the equation of continuity to find the velocity of the fluid if one knows the cross-section area and velocity in one of the regions. The formula for this is A1*V1 = A2*V2, where A is the cross-section area of one location and V is the velocity for that location.


Given three different locations in a confined container, A, B, and C, all having different radii, can you find the other two velocities of the fluid, if the velocity at A is given?

  1. If the cross-section at A has a radius of 6 meters, can you find the area of the slice through the cone
    (Area = pi * r 2)?
  2. If the velocity of the fluid at location A is 10.0 m/s, and the radius at location B is 4.2 meters, can you find the velocity at location B?
  3. If the velocity at location C is 8.6 m/s, can you find the radius at location C?
    (answer) Extension:
  4. If the radius of a fourth location, D, is one-half the radius of A , how would the velocity at location D compare to the velocity of the fluid at A? If D had one-third the radius of A, compare the velocity of the fluid at D to A. Explain your reasoning showing calculations.



Tuesday, 18 July 2006

Fisika SMA

Pengalaman Belajar Fisika di SMAN BI 1 Banjar

Fluids Pressure and Depth

SUBJECT: Aeronautics
TOPIC: Hydrostatic Pressure
DESCRIPTION: A set of mathematics problems dealing with hydrostatics.
CONTRIBUTED BY: Carol Hodanbosi
EDITED BY: Jonathan G. Fairman - August 1996

A fluid is a substance that flows easily. Gases and liquids are fluids, although sometimes the dividing line between liquids and solids is not always clear. Because of their ability to flow, fluids can exert buoyant forces, multiply forces in a hydraulic systems, allow aircraft to fly and ships to float.
The topic that this page will explore will be pressure and depth. If a fluid is within a container then the depth of an object placed in that fluid can be measured. The deeper the object is placed in the fluid, the more pressure it experiences. This is because is the weight of the fluid above it. The more dense the fluid above it, the more pressure is exerted on the object that is submerged, due to the weight of the fluid.
The formula that gives the P pressure on an object submerged in a fluid is:

P = r * g * h
  • r (rho) is the density of the fluid,
  • g is the acceleration of gravity
  • h is the height of the fluid above the object
If the container is open to the atmosphere above, the added pressure must be included if one is to find the total pressure on an object. The total pressure is the same as absolute pressure on pressure gauges readings, while the gauge pressure is the same as the fluid pressure alone, not including atmospheric pressure.

Ptotal = Patmosphere + Pfluid

Ptotal = Patmosphere + ( r * g * h )

A Pascal is the unit of pressure in the metric system. It represents 1 newton/m2
Find the pressure on a scuba diver when she is 12 meters below the surface of the ocean. Assume standard atmospheric conditions.
The density of sea water is 1.03 X 10 3 kg/m3 and the atmospheric pressure is 1.01 x 105 N/m2.

Pfluid = r g h = (1.03 x10 3 kg/m3) (9.8 m/s2) (12 m) = 1.21 x 105 Newtons/m2
Ptotal = Patmosphere + Pfluid = (1.01 x 105) + (1.21 x 105 ) Pa = 2.22 x 10 2 kPa (kilo Pascals)

Exercises :

  • What is the pressure experienced at a point on the bottom of a swimming pool 9 meters in depth? The density of water is 1.00 x 103 kg/m3.

  • The interior of a submarine located at a depth of 45 meters is maintained at normal atmospheric conditions. Find the total force exerted on a 20 cm by 20 cm square window. Use the density of sea water given above.

  • How many atmospheres is a depth of 100 meters of ocean water?

  • If the weight density of pure water is 62 pounds/ft3, find the weight of water in a swimming pool whose dimensions are 20 ft by 10 ft by 6 feet.

  • An airplane in level flight whose mass is 20,000 kg has a wing area of 60 m2. What is the pressure difference between the upper and lower surfaces of its wing? Express your answer in atmospheres.

  • Sumber:

    Sunday, 18 June 2006

    Fisika SMA

    Pengalaman Belajar Fisika di SMAN BI 1 Banjar

    Ada Apa Dengan Fisika?

    Pascal's Principle and Hydraulics

    SUBJECT: Physics
    TOPIC: Hydraulics
    DESCRIPTION: A set of mathematics problems dealing with hydraulics.
    CONTRIBUTED BY: Carol Hodanbosi
    EDITED BY: Jonathan G. Fairman - August 1996

    Hydraulic systems use a incompressible fluid, such as oil or water, to transmit forces from one location to another within the fluid. Most aircraft use hydraulics in the braking systems and landing gear. Pneumatic systems use compressible fluid, such as air, in their operation. Some aircraft utilize pneumatic systems for their brakes, landing gear and movement of flaps.
    Pascal's law states that when there is an increase in pressure at any point in a confined fluid, there is an equal increase at every other point in the container.
    A container, as shown below, contains a fluid. There is an increase in pressure as the length of the column of liquid increases, due to the increased mass of the fluid above.
    For example, in the figure below, P3 would be the highest value of the three pressure readings, because it has the highest level of fluid above it.

    If the above container had an increase in overall pressure, that same added pressure would affect each of the gauges (and the liquid throughout) the same. For example P1, P2, P3 were originally 1, 3, 5 units of pressure, and 5 units of pressure were added to the system, the new readings would be 6, 8, and 10.
    Applied to a more complex system below, such as a hydraulic car lift, Pascal's law allows forces to be multiplied. The cylinder on the left shows a cross-section area of 1 square inch, while the cylinder on the right shows a cross-section area of 10 square inches. The cylinder on the left has a weight (force) on 1 pound acting downward on the piston, which lowers the fluid 10 inches. As a result of this force, the piston on the right lifts a 10 pound weight a distance of 1 inch.
    The 1 pound load on the 1 square inch area causes an increase in pressure on the fluid in the system. This pressure is distributed equally throughout and acts on every square inch of the 10 square inch area of the large piston. As a result, the larger piston lifts up a 10 pound weight. The larger the cross-section area of the second piston, the larger the mechanical advantage, and the more weight it lifts.

    The formulas that relate to this are shown below:
    P1 = P2 (since the pressures are equal throughout).
    Since pressure equals force per unit area, then it follows that
    F1/A1 = F2/A2
    It can be shown by substitution that the values shown above are correct,
    1 pound / 1 square inches = 10 pounds / 10 square inches
    Because the volume of fluid pushed down on the left side equals the volume of fluid that is lifted up on the right side, the following formula is also true.
    V1 = V2
    by substitution,
    A1 D1 = A2 D2
    • A = cross sectional area
    • D = the distance moved
    A1/A2= D2/D1
    This system can be thought of as a simple machine (lever), since force is multiplied.The mechanical advantage can be found by rearranging terms in the above equation to
    Mechanical Advantage(IMA) = D1/D2 = A2/A1
    For the sample problem above, the IMA would be 10:1 (10 inches/ 1 inch or 10 square inches / 1 square inch).
    Given these simple formulas, try to answer the questions below.

    1. A hydraulic press has an input cylinder 1 inch in diameter and an output cylinder 6 inches in diameter.
      1. Assuming 100% efficiency, find the force exerted by the output piston when a force of 10 pounds is applied to the input piston.

      2. If the input piston is moved through 4 inches, how far is the output piston moved?

    2. A hydraulic system is said to have a mechanical advantage of 40. Mechanical advantage (MA) is FR (output) / FE (input). If the input piston, with a 12 inch radius, has a force of 65 pounds pushing downward a distance of 20 inches, find

      1. the volume of fluid that has been displaced

      2. the upward force on the output piston

      3. the radius of the output piston

      4. the distance the output piston moves

    3. What pressure does a 130 pound woman exert on the floor when she balances on one of her heels? Her heels have an average radius of 0.5 inch.

    4. A car has a weight of 2500 pounds and rests on four tires, each having a surface area of contact with the ground of 14 square inches. What is the pressure the ground experiences beneath the tires that is due to the car?

      Extension :
    5. The input and output pistons of a hydraulic jack are respectively 1 cm and 4 cm in diameter. A lever with a mechanical advantage of 6 is used to apply force to the input piston. How much mass can the jack lift if a force of 180 N is applied to the lever and efficiency is 80%?



    Thursday, 18 May 2006

    Fisika SMA

    Pengalaman Belajar Fisika di SMAN BI 1 Banjar

    Buoyancy: Archimedes Principle

    SUBJECT: Physics
    TOPIC: Buoyancy
    DESCRIPTION: A set of mathematics problems dealing with buoyancy.
    CONTRIBUTED BY: Carol Hodanbosi
    EDITED BY: Jonathan G. Fairman - August 1996

    There are two types of flying machines that allow for lift to overcome gravity. The first type, called the aerodynamic machines such as helicopters and airplanes, rely on thrust and forward speed to produce lift. The second type, aerostatic machines, such as hot air balloons and lighter than air-type craft, rely on the differences in air density for lift.
    This lesson is concerned with the second type, the type that are dependent on buoyancy.
    If a cubic centimeter of aluminum was suspended in a fluid such as water with a very thin and negligible thread, the metal cube would have the fluid exerting pressure on the cube. Try to imagine that if the cube were to disappear, and the fluid would magically replace the cube, then the surrounding water would support this cube that is now containing water, so that the cube of water would be motionless. That is, the forces would be balanced. The cube of water would push out on the surrounding water and the surrounding water would push back on the cube. The fluid would be static, or stationary. Now replace this same cube of water with the original cube of aluminum. The surrounding water would not 'know' that the cube has been replaced with another substance. It would still push inward and upward and downward with the same force that it pushed on the cube of water. The sideways forces would be balanced and oppose each other equally, but the upward and downward forces would not be the same. The pressure at the bottom of the cube is greater than the pressure at the top of the cube, because pressure increases with increased depth. The difference between the upward and downward forces acting on the bottom and the top of the cube, respectively,is called buoyancy.
    Using the aluminum as our example, it has a specific gravity of 2.8. Water has a specific gravity of 1.0. This means that a cubic centimeter of water would have a mass of 1.0 grams, while aluminum of the same size would have a mass of 2.8 grams. Since the aluminum cube displaces 1 cubic centimeter of water, it has a buoyancy of 1.0 grams. Since buoyancy is a force and not a mass, it must be converted to the proper units, which when multiplied by the acceleration of gravity (980 cm/s2) gives the units of dynes. That is,
    (1.0 grams) (980 cm/s2) = 980 grams cm /s2 or dynes
    So our aluminum cube immersed in water would not 'weigh' (2.8 x 980) dynes or 2744 dynes. It would weigh less due to the fact it has a buoyant force of (1 x 980) dynes from the water. So it would weigh (2744-980) dynes or 1764 dynes while immersed in the water.
    Archimedes Principle states that the buoyant force on a submerged object is equal to the weight of the fluid that is displaced by the object.
    Hot air balloons rise into the air because the density of the air (warmer air) inside the balloon is less dense than the air outside the balloon (cooler air). The balloon and the basket displaces a fluid that is heavier than the balloon and the basket, so it has a buoyant force acting on the system. Balloons tend to fly better in the morning, when the surrounding air is cool.


    1. Find the weight of the air in a room with dimensions of 20 ft x 12 ft x 15 ft. The weight density of air at sea level is 0.08 pounds /ft3.
    2. An iron anchor weighs 250 pounds in air and has a weight density of 480 lbs/ft3. If it is immersed in sea water that has a weight density of 64 lbs/ft3, how much force would be required to lift it while it is immersed?
    3. An aluminum bar weighs 17 pounds in air. How much force is required to lift the bar while it is immersed in gasoline? The weight density of aluminum is 170 pounds /ft3 and that of gasoline is 42 pounds /ft3.
    4. How much does a 20 ft x 10 ft x 8 ft swimming pool filled with water weigh? Assume the water has a weight density of 62 lbs/ft3.
    5. A balloon weighing 80 kg has a capacity of 1200 m3. If it is filled with helium, how great a payload can it support? The density of helium is 0.18 kg/ m3 and the density of air is 1.30 kg/ m3. Express your answer in newtons.