• Tenis
• ITF Men - Singles
• M15 Ahmedabad
• Manish Sureshkumar x Siddharth Vishwakarma Statistik

Manish Sureshkumar vs Siddharth Vishwakarma statistik 18.03.2025

• 18/03/25 02:00
• 

  Manish Sureshkumar
  • 1 : 2
  • Selesai
  

  Siddharth Vishwakarma

T

1

2

3

Manish Sureshkumar

1

4

6

1

Siddharth Vishwakarma

2

6

4

6

Manish Sureshkumar

Н2Н

Siddharth Vishwakarma

Semua turnamen M15 Lucknow M15 Ahmedabad M25 Mandya 5 pertandingan 5 pertandingan 5 pertandingan 5 pertandingan

• 0menang
• 100% 3menang

• 2
• Skor
• 6

• 0
• menang
• 3

Kepala ke kepala

ITF Men - Singles M15 Ahmedabad 1

WP 18/03/25 02:00

Manish Sureshkumar

• 1
• 4
• 6

1 Siddharth Vishwakarma

• 6
• 6
• 4

2

ITF Men - Singles M25 Mandya 1

WP 11/01/24 03:00

Siddharth Vishwakarma

• 7
• 6

2 Manish Sureshkumar

• 5
• 4

0

ITF Men - Singles M15 Lucknow 1

WP 03/03/21 23:00

Siddharth Vishwakarma

• 7 ⁷
• 3
• 6

2 Manish Sureshkumar

• 6 ⁴
• 6
• 0

1

Hasil terakhir: Manish Sureshkumar

Challenger Men - Singles Bengaluru 3 2

WP 20/05/26 02:50

Manish Sureshkumar

• 1
• 1

0 Ilya Ivashka

• 6
• 6

2

WP 19/05/26 07:55

Keegan Smith

• 4
• 6
• 2

1 Manish Sureshkumar

• 6
• 2
• 6

2

Challenger Men - Singles Bengaluru 2 3

WP 14/05/26 04:30

Manish Sureshkumar

• 3
• 3

0 Ilya Ivashka

• 6
• 6

2

Dib. 13/05/26 04:50

Alastair Gray Manish Sureshkumar

WP 12/05/26 03:35

Ronit Karki

• 6 ⁷
• 4

0 Manish Sureshkumar

• 7 ⁹
• 6

2

Hasil terakhir: Siddharth Vishwakarma

ITF Men - Singles M25 Sharm ElSheikh 4 1

WP 01/10/25 06:25

Kris van Wyk

• 6
• 6

2 Siddharth Vishwakarma

• 3
• 4

0

ITF Men - Singles M25 Nakhon Pathom 1

WP 01/09/25 22:00

Seokhyeon Chu

• 6
• 4
• 7 ⁷

2 Siddharth Vishwakarma

• 2
• 6
• 6 ⁵

1

ITF Men - Singles M15 Nakhon Pathom 5 1

WP 26/08/25 23:30

Siddharth Vishwakarma

• 3
• 2

0 Max Basing

• 6
• 6

2

ITF Men - Singles M15 Nakhon Pathom 4 2

WP 20/08/25 23:30

Siddharth Vishwakarma

• 6 ⁴
• 6
• 3

1 Thando Longwe-Smit

• 7 ⁷
• 4
• 6

2

WP 19/08/25 22:00

Siddharth Vishwakarma

• 6
• 6

2 Ryo Minakata

• 4
• 1

0