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• Challenger Men - Doubles
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• Gerch Lucas|Sperle John x Matias Penaranda/Leonardo Suarez Statistik

Gerch Lucas|Sperle John vs Matias Penaranda/Leonardo Suarez statistik 19.06.2025

• 19/06/25 15:10
• 

  Lucas Gerch/ John Sperle
  • 2 : 0
  • Selesai
  

  Matias Penaranda/ Leonardo Suarez

T

1

2

Gerch Lucas|Sperle John

2

6

6

Matias Penaranda/Leonardo Suarez

0

3

1

Gerch Lucas|Sperle John

Stats

Matias Penaranda/Leonardo Suarez

• Pertandingan
• Tetapkan 1
• Tetapkan 2

Melayani

1

Aces

0

1

Kesalahan ganda

2

73%

Persentase servis pertama

57%

24/3275%

Poin servis pertama %

15/2756%

8/1267%

Poin servis kedua %

8/2040%

4/4100%

Poin break %

2/633%

Kembali

12/2744%

Poin pengembalian pertama %

9/3228%

12/2060%

Poin pengembalian kedua %

4/1233%

4/667%

Poin break %

0/40%

Poin

0/00%

Poin di net %

0/00%

32/4473%

Poin servis %

23/4749%

24/4751%

Poin pengembalian %

12/4427%

56/9162%

Poin total %

35/9138%

Permainan

8/8100%

Permainan servis %

4/850%

4/850%

Permainan pengembalian %

0/80%

12/1675%

Permainan total %

4/1625%

Gerch Lucas|Sperle John

Н2Н

Matias Penaranda/Leonardo Suarez

Semua turnamen All matches

Tidak ada yang cocok

Ini adalah pertandingan pertama dari para pemain ini

Hasil terakhir: Gerch Lucas|Sperle John

Challenger Men - Doubles Lima 2

WP 26/06/25 16:10

Gerch Lucas|Sperle John

• 4
• 4

0 Pruchya Isaro/Kelsey Stevenson

• 6
• 6

2

WP 24/06/25 19:00

Gerch Lucas|Sperle John 0 Nicolas Mejia/Johan Alexander Rodriguez 0

Challenger Men - Doubles Santa Cruz 2

Dib. 19/06/25 19:10

Gerch Lucas|Sperle John Kestelboim Mariano|Villanueva Gonzalo

WP 19/06/25 15:10

Gerch Lucas|Sperle John

• 6
• 6

2 Matias Penaranda/Leonardo Suarez

• 3
• 1

0

Challenger Men - Doubles Santa Fe 1

WP 12/06/25 13:50

Britto Luis|Carou Ignacio

• 7 ⁷
• 6

2 Gerch Lucas|Sperle John

• 6 ³
• 4

0

Hasil terakhir: Matias Penaranda/Leonardo Suarez

Challenger Men - Doubles Santa Cruz 1

WP 19/06/25 15:10

Gerch Lucas|Sperle John

• 6
• 6

2 Matias Penaranda/Leonardo Suarez

• 3
• 1

0